{"id":1669,"date":"2018-01-11T20:35:30","date_gmt":"2018-01-11T20:35:30","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/continuity\/"},"modified":"2018-02-01T17:48:51","modified_gmt":"2018-02-01T17:48:51","slug":"continuity","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/chapter\/continuity\/","title":{"raw":"2.4 Continuity","rendered":"2.4 Continuity"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Explain the three conditions for continuity at a point.<\/li>\r\n \t<li>Describe three kinds of discontinuities.<\/li>\r\n \t<li>Define continuity on an interval.<\/li>\r\n \t<li>State the theorem for limits of composite functions.<\/li>\r\n \t<li>Provide an example of the intermediate value theorem.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170570996716\">Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Such functions are called <em>continuous<\/em>. Other functions have points at which a break in the graph occurs, but satisfy this property over intervals contained in their domains. They are continuous on these intervals and are said to have a <strong>discontinuity at a point<\/strong> where a break occurs.<\/p>\r\n<p id=\"fs-id1170573595201\">We begin our investigation of continuity by exploring what it means for a function to have <strong>continuity at a point<\/strong>. Intuitively, a function is continuous at a particular point if there is no break in its graph at that point.<\/p>\r\n\r\n<div id=\"fs-id1170573262889\" class=\"bc-section section\">\r\n<h1>Continuity at a Point<\/h1>\r\n<p id=\"fs-id1170573324888\">Before we look at a formal definition of what it means for a function to be continuous at a point, let\u2019s consider various functions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditions that prevent such failures.<\/p>\r\n<p id=\"fs-id1170573390124\">Our first function of interest is shown in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_001\">(Figure)<\/a>. We see that the graph of [latex]f(x)[\/latex] has a hole at [latex]a[\/latex]. In fact, [latex]f(a)[\/latex] is undefined. At the very least, for [latex]f(x)[\/latex] to be continuous at [latex]a[\/latex], we need the following condition:<\/p>\r\n\r\n<div id=\"fs-id1170570997076\" class=\"equation unnumbered\">[latex]\\text{i.}f(a)\\text{is defined.}[\/latex]<\/div>\r\n<div id=\"CNX_Calc_Figure_02_04_001\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203502\/CNX_Calc_Figure_02_04_001.jpg\" alt=\"A graph of an increasing linear function f(x) which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. A point a greater than zero is marked on the x axis. The point on the function f(x) above a is an open circle; the function is not defined at a.\" width=\"325\" height=\"277\" \/> Figure 1. The function [latex]f(x)[\/latex] is not continuous at a because [latex]f(a)[\/latex] is undefined.[\/caption]<\/div>\r\n<p id=\"fs-id1170573750564\">However, as we see in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_002\">(Figure)<\/a>, this condition alone is insufficient to guarantee continuity at the point [latex]a[\/latex]. Although [latex]f(a)[\/latex] is defined, the function has a gap at [latex]a[\/latex]. In this example, the gap exists because [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] does not exist. We must add another condition for continuity at [latex]a[\/latex]\u2014namely,<\/p>\r\n\r\n<div id=\"fs-id1170570997548\" class=\"equation unnumbered\">[latex]\\text{ii.}\\underset{x\\to a}{\\text{lim}}f(x)\\text{exists.}[\/latex]<\/div>\r\n<div id=\"CNX_Calc_Figure_02_04_002\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203504\/CNX_Calc_Figure_02_04_002.jpg\" alt=\"The graph of a piecewise function f(x) with two parts. The first part is an increasing linear function that crosses from quadrant three to quadrant one at the origin. A point a greater than zero is marked on the x axis. At fa. on this segment, there is a solid circle. The other segment is also an increasing linear function. It exists in quadrant one for values of x greater than a. At x=a, this segment has an open circle.\" width=\"325\" height=\"277\" \/> Figure 2. The function [latex]f(x)[\/latex] is not continuous at a because [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] does not exist.[\/caption]<\/div>\r\n<p id=\"fs-id1170573493837\">However, as we see in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_003\">(Figure)<\/a>, these two conditions by themselves do not guarantee continuity at a point. The function in this figure satisfies both of our first two conditions, but is still not continuous at [latex]a[\/latex]. We must add a third condition to our list:<\/p>\r\n\r\n<div id=\"fs-id1170573419105\" class=\"equation unnumbered\">[latex]\\text{iii.}\\underset{x\\to a}{\\text{lim}}f(x)=f(a).[\/latex]<\/div>\r\n<div id=\"CNX_Calc_Figure_02_04_003\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203506\/CNX_Calc_Figure_02_04_003.jpg\" alt=\"The graph of a piecewise function with two parts. The first part is an increasing linear function that crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. A point a greater than zero is marked on the x axis. At this point, there is an open circle on the linear function. The second part is a point at x=a above the line.\" width=\"325\" height=\"277\" \/> Figure 3. The function [latex]f(x)[\/latex] is not continuous at a because [latex]\\underset{x\\to a}{\\text{lim}}f(x)\\ne f(a).[\/latex][\/caption]<\/div>\r\n<p id=\"fs-id1170573502301\">Now we put our list of conditions together and form a definition of continuity at a point.<\/p>\r\n\r\n<div id=\"fs-id1170571048312\" class=\"textbox key-takeaways\">\r\n<div class=\"title\">\r\n<h3>Definition<\/h3>\r\n<\/div>\r\n<p id=\"fs-id1170573381317\">A function [latex]f(x)[\/latex] is continuous at a point [latex]a[\/latex] if and only if the following three conditions are satisfied:<\/p>\r\n\r\n<ol id=\"fs-id1170571094915\">\r\n \t<li>[latex]f(a)[\/latex] is defined<\/li>\r\n \t<li>[latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists<\/li>\r\n \t<li>[latex]\\underset{x\\to a}{\\text{lim}}f(x)=f(a)[\/latex]<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1170573370662\">A function is discontinuous at a point [latex]a[\/latex] if it fails to be continuous at [latex]a[\/latex].<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170573281579\">The following procedure can be used to analyze the continuity of a function at a point using this definition.<\/p>\r\n\r\n<div id=\"fs-id1170573398041\" class=\"textbox key-takeaways problem-solving\">\r\n<h3>Problem-Solving Strategy: Determining Continuity at a Point<\/h3>\r\n<ol id=\"fs-id1170571103215\">\r\n \t<li>Check to see if [latex]f(a)[\/latex] is defined. If [latex]f(a)[\/latex] is undefined, we need go no further. The function is not continuous at [latex]a[\/latex]. If [latex]f(a)[\/latex] is defined, continue to step 2.<\/li>\r\n \t<li>Compute [latex]\\underset{x\\to a}{\\text{lim}}f(x).[\/latex] In some cases, we may need to do this by first computing [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)[\/latex] and [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x).[\/latex] If [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] does not exist (that is, it is not a real number), then the function is not continuous at [latex]a[\/latex] and the problem is solved. If [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists, then continue to step 3.<\/li>\r\n \t<li>Compare [latex]f(a)[\/latex] and [latex]\\underset{x\\to a}{\\text{lim}}f(x).[\/latex] If [latex]\\underset{x\\to a}{\\text{lim}}f(x)\\ne f(a),[\/latex] then the function is not continuous at [latex]a[\/latex]. If [latex]\\underset{x\\to a}{\\text{lim}}f(x)=f(a),[\/latex] then the function is continuous at [latex]a[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<p id=\"fs-id1170573570770\">The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given point. These examples illustrate situations in which each of the conditions for continuity in the definition succeed or fail.<\/p>\r\n\r\n<div id=\"fs-id1170573442080\" class=\"textbox examples\">\r\n<h3>Determining Continuity at a Point, Condition 1<\/h3>\r\n<div id=\"fs-id1170573368010\" class=\"exercise\">\r\n<div id=\"fs-id1170573411938\" class=\"textbox\">\r\n<p id=\"fs-id1170573258704\">Using the definition, determine whether the function [latex]f(x)=({x}^{2}-4)\\text{\/}(x-2)[\/latex] is continuous at [latex]x=2.[\/latex] Justify the conclusion.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573331408\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573331408\"]\r\n<p id=\"fs-id1170573331408\">Let\u2019s begin by trying to calculate [latex]f(2).[\/latex] We can see that [latex]f(2)=0\\text{\/}0,[\/latex] which is undefined. Therefore, [latex]f(x)=\\frac{{x}^{2}-4}{x-2}[\/latex] is discontinuous at 2 because [latex]f(2)[\/latex] is undefined. The graph of [latex]f(x)[\/latex] is shown in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_004\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"CNX_Calc_Figure_02_04_004\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203508\/CNX_Calc_Figure_02_04_004.jpg\" alt=\"A graph of the given function. There is a line which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. At a point in quadrant one, there is an open circle where the function is not defined.\" width=\"487\" height=\"425\" \/> Figure 4. The function [latex]f(x)[\/latex] is discontinuous at 2 because [latex]f(2)[\/latex] is undefined. [\/hidden-answer][\/caption]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573389760\" class=\"textbox examples\">\r\n<h3>Determining Continuity at a Point, Condition 2<\/h3>\r\n<div id=\"fs-id1170571098221\" class=\"exercise\">\r\n<div id=\"fs-id1170573426610\" class=\"textbox\">\r\n<p id=\"fs-id1170573370721\">Using the definition, determine whether the function [latex]f(x)=\\bigg\\{\\begin{array}{ll}-{x}^{2}+4&amp; \\text{ if }x\\le 3\\hfill \\\\ 4x-8&amp; \\text{ if }x&gt;3\\hfill \\end{array}[\/latex] is continuous at [latex]x=3.[\/latex] Justify the conclusion.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573349668\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573349668\"]\r\n<p id=\"fs-id1170573349668\">Let\u2019s begin by trying to calculate [latex]f(3).[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170573405058\" class=\"equation unnumbered\">[latex]f(3)=-({3}^{2})+4=-5.[\/latex]<\/div>\r\n<p id=\"fs-id1170573307347\">Thus, [latex]f(3)[\/latex] is defined. Next, we calculate [latex]\\underset{x\\to 3}{\\text{lim}}f(x).[\/latex] To do this, we must compute [latex]\\underset{x\\to {3}^{-}}{\\text{lim}}f(x)[\/latex] and [latex]\\underset{x\\to {3}^{+}}{\\text{lim}}f(x)\\text{:}[\/latex]<\/p>\r\n\r\n<div class=\"equation unnumbered\">[latex]\\underset{x\\to {3}^{-}}{\\text{lim}}f(x)=-({3}^{2})+4=-5[\/latex]<\/div>\r\n<p id=\"fs-id1170573400834\">and<\/p>\r\n\r\n<div class=\"equation unnumbered\">[latex]\\underset{x\\to {3}^{+}}{\\text{lim}}f(x)=4(3)-8=4.[\/latex]<\/div>\r\n<p id=\"fs-id1170573414763\">Therefore, [latex]\\underset{x\\to 3}{\\text{lim}}f(x)[\/latex] does not exist. Thus, [latex]f(x)[\/latex] is not continuous at 3. The graph of [latex]f(x)[\/latex] is shown in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_005\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"CNX_Calc_Figure_02_04_005\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203511\/CNX_Calc_Figure_02_04_005.jpg\" alt=\"A graph of the given piecewise function, which has two parts. The first is a downward opening parabola which is symmetric about the y axis. Its vertex is on the y axis, greater than zero. There is a closed circle on the parabola for x=3. The second part is an increasing linear function in the first quadrant, which exists for values of x &gt; 3. There is an open circle at the end of the line where x would be 3.\" width=\"487\" height=\"575\" \/> Figure 5. The function [latex]f(x)[\/latex] is not continuous at 3 because [latex]\\underset{x\\to 3}{\\text{lim}}f(x)[\/latex] does not exist. [\/hidden-answer][\/caption]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573429945\" class=\"textbox examples\">\r\n<h3>Determining Continuity at a Point, Condition 3<\/h3>\r\n<div id=\"fs-id1170571048204\" class=\"exercise\">\r\n<div id=\"fs-id1170573268754\" class=\"textbox\">\r\n<p id=\"fs-id1170573502415\">Using the definition, determine whether the function [latex]f(x)=\\bigg\\{\\begin{array}{ll}\\frac{ \\sin x}{x}\\hfill &amp; \\text{ if }x\\ne 0\\hfill \\\\ \\hfill 1&amp; \\text{ if }x=0\\hfill \\end{array}[\/latex] is continuous at [latex]x=0.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573382461\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573382461\"]\r\n<p id=\"fs-id1170573382461\">First, observe that<\/p>\r\n\r\n<div id=\"fs-id1170573569453\" class=\"equation unnumbered\">[latex]f(0)=1.[\/latex]<\/div>\r\n<p id=\"fs-id1170573367184\">Next,<\/p>\r\n\r\n<div id=\"fs-id1170573442773\" class=\"equation unnumbered\">[latex]\\underset{x\\to 0}{\\text{lim}}f(x)=\\underset{x\\to 0}{\\text{lim}}\\frac{ \\sin x}{x}=1.[\/latex]<\/div>\r\n<p id=\"fs-id1170573394612\">Last, compare [latex]f(0)[\/latex] and [latex]\\underset{x\\to 1}{\\text{lim}}f(x).[\/latex] We see that<\/p>\r\n\r\n<div id=\"fs-id1170573332534\" class=\"equation unnumbered\">[latex]f(0)=1=\\underset{x\\to 0}{\\text{lim}}f(x).[\/latex]<\/div>\r\n<p id=\"fs-id1170573394599\">Since all three of the conditions in the definition of continuity are satisfied, [latex]f(x)[\/latex] is continuous at [latex]x=0.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170570976463\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1170573288457\" class=\"exercise\">\r\n<div id=\"fs-id1170573435480\" class=\"textbox\">\r\n<p id=\"fs-id1170573406319\">Using the definition, determine whether the function [latex]f(x)=\\Bigg\\{\\begin{array}{cc}2x+1\\hfill &amp; \\text{ if }x&lt;1\\hfill \\\\ \\hfill 2&amp; \\text{ if }x=1\\hfill \\\\ -x+4\\hfill &amp; \\text{ if }x&gt;1\\hfill \\end{array}[\/latex] is continuous at [latex]x=1.[\/latex] If the function is not continuous at 1, indicate the condition for continuity at a point that fails to hold.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573402451\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573402451\"]\r\n<p id=\"fs-id1170573402451\">[latex]f[\/latex] is not continuous at 1 because [latex]f(1)=2\\ne 3=\\underset{x\\to 1}{\\text{lim}}f(x).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572589102\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170573418814\">Check each condition of the definition.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170573513079\">By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can state the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1170573430304\" class=\"textbox key-takeaways theorem\">\r\n<h3>Continuity of Polynomials and Rational Functions<\/h3>\r\n<p id=\"fs-id1170573255225\">Polynomials and rational functions are continuous at every point in their domains.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170573429973\" class=\"bc-section section\">\r\n<h2>Proof<\/h2>\r\n<p id=\"fs-id1170573326736\">Previously, we showed that if [latex]p(x)[\/latex] and [latex]q(x)[\/latex] are polynomials, [latex]\\underset{x\\to a}{\\text{lim}}p(x)=p(a)[\/latex] for every polynomial [latex]p(x)[\/latex] and [latex]\\underset{x\\to a}{\\text{lim}}\\frac{p(x)}{q(x)}=\\frac{p(a)}{q(a)}[\/latex] as long as [latex]q(a)\\ne 0.[\/latex] Therefore, polynomials and rational functions are continuous on their domains.<\/p>\r\n<p id=\"fs-id1170573276680\">\u25a1<\/p>\r\n<p id=\"fs-id1170573335424\">We now apply <a class=\"autogenerated-content\" href=\"#fs-id1170573430304\">(Figure)<\/a> to determine the points at which a given rational function is continuous.<\/p>\r\n\r\n<div id=\"fs-id1170573381194\" class=\"textbox examples\">\r\n<h3>Continuity of a Rational Function<\/h3>\r\n<div id=\"fs-id1170573326117\" class=\"exercise\">\r\n<div id=\"fs-id1170573307687\" class=\"textbox\">\r\n<p id=\"fs-id1170573365757\">For what values of [latex]x[\/latex] is [latex]f(x)=\\frac{x+1}{x-5}[\/latex] continuous?<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573397031\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573397031\"]\r\n<p id=\"fs-id1170573397031\">The rational function [latex]f(x)=\\frac{x+1}{x-5}[\/latex] is continuous for every value of [latex]x[\/latex] except [latex]x=5.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573434500\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1170573427664\" class=\"exercise\">\r\n<div id=\"fs-id1170573400271\" class=\"textbox\">\r\n<p id=\"fs-id1170570999873\">For what values of [latex]x[\/latex] is [latex]f(x)=3{x}^{4}-4{x}^{2}[\/latex] continuous?<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573246209\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573246209\"]\r\n<p id=\"fs-id1170573246209\">[latex]f(x)[\/latex] is continuous at every real number.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572209021\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170573431750\">Use <a class=\"autogenerated-content\" href=\"#fs-id1170573430304\">(Figure)<\/a><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573352138\" class=\"bc-section section\">\r\n<h1>Types of Discontinuities<\/h1>\r\n<p id=\"fs-id1170573435662\">As we have seen in <a class=\"autogenerated-content\" href=\"#fs-id1170573442080\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#fs-id1170573389760\">(Figure)<\/a>, discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a<strong> removable discontinuity<\/strong> is a discontinuity for which there is a hole in the graph, a<strong> jump discontinuity<\/strong> is a noninfinite discontinuity for which the sections of the function do not meet up, and an <strong>infinite discontinuity<\/strong> is a discontinuity located at a vertical asymptote. <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_006\">(Figure)<\/a> illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories.<\/p>\r\n\r\n<div id=\"CNX_Calc_Figure_02_04_006\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203514\/CNX_Calc_Figure_02_04_006.jpg\" alt=\"Three graphs, each showing a different discontinuity. The first is removable discontinuity. Here, the given function is a line with positive slope. At a point x=a, where a&gt;0, there is an open circle on the line and a closed circle a few units above the line. The second is a jump discontinuity. Here, there are two lines with positive slope. The first line exists for x&lt;=a, and the second exists for x&gt;a, where a&gt;0. The first line ends at a solid circle where x=a, and the second begins a few units up with an open circle at x=a. The third discontinuity type is infinite discontinuity. Here, the function has two parts separated by an asymptote x=a. The first segment is a curve stretching along the x axis to 0 as x goes to negative infinity and along the y axis to infinity as x goes to zero. The second segment is a curve stretching along the y axis to negative infinity as x goes to zero and along the x axis to 0 as x goes to infinity.\" width=\"975\" height=\"315\" \/> Figure 6. Discontinuities are classified as (a) removable, (b) jump, or (c) infinite.[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1170573400557\">These three discontinuities are formally defined as follows:<\/p>\r\n\r\n<div id=\"fs-id1170573419172\" class=\"textbox key-takeaways\">\r\n<div class=\"title\">\r\n<h3>Definition<\/h3>\r\n<\/div>\r\nIf [latex]f(x)[\/latex] is discontinuous at [latex]a[\/latex], then\r\n<ol id=\"fs-id1170573216342\">\r\n \t<li>[latex]f[\/latex] has a <strong>removable discontinuity<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists. (Note: When we state that [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists, we mean that [latex]\\underset{x\\to a}{\\text{lim}}f(x)=L,[\/latex] where <em>L<\/em> is a real number.)<\/li>\r\n \t<li>[latex]f[\/latex] has a <strong>jump discontinuity<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)[\/latex] and [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)[\/latex] both exist, but [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)\\ne \\underset{x\\to {a}^{+}}{\\text{lim}}f(x).[\/latex] (Note: When we state that [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)[\/latex] and [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)[\/latex] both exist, we mean that both are real-valued and that neither take on the values \u00b1\u221e.)<\/li>\r\n \t<li>[latex]f[\/latex] has an <strong>infinite discontinuity<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)=\\text{\u00b1}\\infty [\/latex] or [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)=\\text{\u00b1}\\infty .[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1170570976348\" class=\"textbox examples\">\r\n<h3>Classifying a Discontinuity<\/h3>\r\n<div id=\"fs-id1170573248707\" class=\"exercise\">\r\n<div id=\"fs-id1170573399643\" class=\"textbox\">\r\n<p id=\"fs-id1170573424344\">In <a class=\"autogenerated-content\" href=\"#fs-id1170573442080\">(Figure)<\/a>, we showed that [latex]f(x)=\\frac{{x}^{2}-4}{x-2}[\/latex] is discontinuous at [latex]x=2.[\/latex] Classify this discontinuity as removable, jump, or infinite.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573402024\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573402024\"]\r\n<p id=\"fs-id1170573402024\">To classify the discontinuity at 2 we must evaluate [latex]\\underset{x\\to 2}{\\text{lim}}f(x)\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170571103554\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}\\hfill \\underset{x\\to 2}{\\text{lim}}f(x)&amp; =\\underset{x\\to 2}{\\text{lim}}\\frac{{x}^{2}-4}{x-2}\\hfill \\\\ &amp; =\\underset{x\\to 2}{\\text{lim}}\\frac{(x-2)(x+2)}{x-2}\\hfill \\\\ &amp; =\\underset{x\\to 2}{\\text{lim}}(x+2)\\hfill \\\\ &amp; =4.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571136107\">Since [latex]f[\/latex] is discontinuous at 2 and [latex]\\underset{x\\to 2}{\\text{lim}}f(x)[\/latex] exists, [latex]f[\/latex] has a removable discontinuity at [latex]x=2.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573426544\" class=\"textbox examples\">\r\n<h3>Classifying a Discontinuity<\/h3>\r\n<div id=\"fs-id1170573507547\" class=\"exercise\">\r\n<div id=\"fs-id1170573493638\" class=\"textbox\">\r\n<p id=\"fs-id1170573586256\">In <a class=\"autogenerated-content\" href=\"#fs-id1170573389760\">(Figure)<\/a>, we showed that [latex]f(x)=\\bigg\\{\\begin{array}{cc}-{x}^{2}+4&amp; \\text{ if }x\\le 3\\hfill \\\\ 4x-8&amp; \\text{ if }x&gt;3\\hfill \\end{array}[\/latex] is discontinuous at [latex]x=3.[\/latex] Classify this discontinuity as removable, jump, or infinite.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571095481\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571095481\"]\r\n<p id=\"fs-id1170571095481\">Earlier, we showed that [latex]f[\/latex] is discontinuous at 3 because [latex]\\underset{x\\to 3}{\\text{lim}}f(x)[\/latex] does not exist. However, since [latex]\\underset{x\\to {3}^{-}}{\\text{lim}}f(x)=-5[\/latex] and [latex]\\underset{x\\to {3}^{-}}{\\text{lim}}f(x)=4[\/latex] both exist, we conclude that the function has a jump discontinuity at 3.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573355402\" class=\"textbox examples\">\r\n<div id=\"fs-id1170573750481\" class=\"exercise\">\r\n<div class=\"textbox\">\r\n<h3>Classifying a Discontinuity<\/h3>\r\n<p id=\"fs-id1170573732461\">Determine whether [latex]f(x)=\\frac{x+2}{x+1}[\/latex] is continuous at \u22121. If the function is discontinuous at \u22121, classify the discontinuity as removable, jump, or infinite.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571000111\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571000111\"]\r\n<p id=\"fs-id1170571000111\">The function value [latex]f(-1)[\/latex] is undefined. Therefore, the function is not continuous at \u22121. To determine the type of discontinuity, we must determine the limit at \u22121. We see that [latex]\\underset{x\\to {-1}^{-}}{\\text{lim}}\\frac{x+2}{x+1}=\\text{\u2212}\\infty [\/latex] and [latex]\\underset{x\\to {-1}^{+}}{\\text{lim}}\\frac{x+2}{x+1}=\\text{+}\\infty .[\/latex] Therefore, the function has an infinite discontinuity at \u22121.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571047778\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1170570998828\" class=\"exercise\">\r\n<div id=\"fs-id1170573351595\" class=\"textbox\">\r\n<p id=\"fs-id1170573361594\">For [latex]f(x)=\\bigg\\{\\begin{array}{cc}{x}^{2}\\hfill &amp; \\text{ if }x\\ne 1\\hfill \\\\ \\hfill 3&amp; \\text{ if }x=1\\hfill \\end{array},[\/latex] decide whether [latex]f[\/latex] is continuous at 1. If [latex]f[\/latex] is not continuous at 1, classify the discontinuity as removable, jump, or infinite.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573355512\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573355512\"]\r\n<p id=\"fs-id1170573355512\">Discontinuous at 1; removable<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571648706\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170573436674\">Follow the steps in <a class=\"autogenerated-content\" href=\"#fs-id1170573398041\">(Figure)<\/a>. If the function is discontinuous at 1, look at [latex]\\underset{x\\to 1}{\\text{lim}}f(x)[\/latex] and use the definition to determine the type of discontinuity.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573494299\" class=\"bc-section section\">\r\n<h1>Continuity over an Interval<\/h1>\r\n<p id=\"fs-id1170573612125\">Now that we have explored the concept of continuity at a point, we extend that idea to <strong>continuity over an interval<\/strong>. As we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function is continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a point.<\/p>\r\n\r\n<div id=\"fs-id1170573581958\" class=\"textbox key-takeaways\">\r\n<h3>Continuity from the Right and from the Left<\/h3>\r\n<p id=\"fs-id1170573352232\">A function [latex]f(x)[\/latex] is said to be <strong>continuous from the right<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)=f(a).[\/latex]<\/p>\r\n<p id=\"fs-id1170571120467\">A function [latex]f(x)[\/latex] is said to be<strong> continuous from the left<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)=f(a).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170570967081\">A function is continuous over an open interval if it is continuous at every point in the interval. A function [latex]f(x)[\/latex] is continuous over a closed interval of the form [latex]\\left[a,b\\right][\/latex] if it is continuous at every point in [latex](a,b)[\/latex] and is continuous from the right at [latex]a[\/latex] and is continuous from the left at [latex]b[\/latex]. Analogously, a function [latex]f(x)[\/latex] is continuous over an interval of the form [latex](a,b\\right][\/latex] if it is continuous over [latex](a,b)[\/latex] and is continuous from the left at [latex]b[\/latex]. Continuity over other types of intervals are defined in a similar fashion.<\/p>\r\n<p id=\"fs-id1170573533852\">Requiring that [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)=f(a)[\/latex] and [latex]\\underset{x\\to {b}^{-}}{\\text{lim}}f(x)=f(b)[\/latex] ensures that we can trace the graph of the function from the point [latex](a,f(a))[\/latex] to the point [latex](b,f(b))[\/latex] without lifting the pencil. If, for example, [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)\\ne f(a),[\/latex] we would need to lift our pencil to jump from [latex]f(a)[\/latex] to the graph of the rest of the function over [latex](a,b\\right].[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170573395559\" class=\"textbox examples\">\r\n<h3>Continuity on an Interval<\/h3>\r\n<div id=\"fs-id1170573395562\" class=\"exercise\">\r\n<div id=\"fs-id1170571096791\" class=\"textbox\">\r\n<p id=\"fs-id1170570990867\">State the interval(s) over which the function [latex]f(x)=\\frac{x-1}{{x}^{2}+2x}[\/latex] is continuous.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573426588\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573426588\"]\r\n<p id=\"fs-id1170573426588\">Since [latex]f(x)=\\frac{x-1}{{x}^{2}+2x}[\/latex] is a rational function, it is continuous at every point in its domain. The domain of [latex]f(x)[\/latex] is the set [latex](\\text{\u2212}\\infty ,-2)\\cup (-2,0)\\cup (0,\\text{+}\\infty ).[\/latex] Thus, [latex]f(x)[\/latex] is continuous over each of the intervals [latex](\\text{\u2212}\\infty ,-2),(-2,0),[\/latex] and [latex](0,\\text{+}\\infty ).[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573387892\" class=\"textbox examples\">\r\n<h3>Continuity over an Interval<\/h3>\r\n<div id=\"fs-id1170573387894\" class=\"exercise\">\r\n<div id=\"fs-id1170573404298\" class=\"textbox\">\r\n<p id=\"fs-id1170573406820\">State the interval(s) over which the function [latex]f(x)=\\sqrt{4-{x}^{2}}[\/latex] is continuous.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573439420\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573439420\"]\r\n<p id=\"fs-id1170573439420\">From the limit laws, we know that [latex]\\underset{x\\to a}{\\text{lim}}\\sqrt[]{4-{x}^{2}}=\\sqrt{4-{a}^{2}}[\/latex] for all values of [latex]a[\/latex] in [latex](-2,2).[\/latex] We also know that [latex]\\underset{x\\to {-2}^{+}}{\\text{lim}}\\sqrt{4-{x}^{2}}=0[\/latex] exists and [latex]\\underset{x\\to {2}^{-}}{\\text{lim}}\\sqrt{4-{x}^{2}}=0[\/latex] exists. Therefore, [latex]f(x)[\/latex] is continuous over the interval [latex]\\left[-2,2\\right].[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573403253\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1170573361460\" class=\"exercise\">\r\n<div id=\"fs-id1170573361462\" class=\"textbox\">\r\n<p id=\"fs-id1170571120474\">State the interval(s) over which the function [latex]f(x)=\\sqrt{x+3}[\/latex] is continuous.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571287079\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571287079\"]\r\n<p id=\"fs-id1170571287079\">[latex]\\left[-3,\\text{+}\\infty )[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572542876\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170570999934\">Use <a class=\"autogenerated-content\" href=\"#fs-id1170573387892\">(Figure)<\/a> as a guide for solving.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170573361765\">The <a class=\"autogenerated-content\" href=\"#fs-id1170573352212\">(Figure)<\/a> allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.<\/p>\r\n\r\n<div id=\"fs-id1170573352212\" class=\"textbox key-takeaways theorem\">\r\n<h3>Composite Function Theorem<\/h3>\r\n<p id=\"fs-id1170573762934\">If [latex]f(x)[\/latex] is continuous at <em>L<\/em> and [latex]\\underset{x\\to a}{\\text{lim}}g(x)=L,[\/latex] then<\/p>\r\n\r\n<div id=\"fs-id1170573400410\" class=\"equation unnumbered\">[latex]\\underset{x\\to a}{\\text{lim}}f(g(x))=f(\\underset{x\\to a}{\\text{lim}}g(x))=f(L).[\/latex]<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170573573603\">Before we move on to <a class=\"autogenerated-content\" href=\"#fs-id1170573718134\">(Figure)<\/a>, recall that earlier, in the section on limit laws, we showed [latex]\\underset{x\\to 0}{\\text{lim}} \\cos x=1= \\cos (0).[\/latex] Consequently, we know that [latex]f(x)= \\cos x[\/latex] is continuous at 0. In <a class=\"autogenerated-content\" href=\"#fs-id1170573718134\">(Figure)<\/a> we see how to combine this result with the composite function theorem.<\/p>\r\n\r\n<div id=\"fs-id1170573718134\" class=\"textbox examples\">\r\n<h3>Limit of a Composite Cosine Function<\/h3>\r\n<div id=\"fs-id1170570998131\" class=\"exercise\">\r\n<div id=\"fs-id1170570998133\" class=\"textbox\">\r\n\r\nEvaluate [latex]\\underset{x\\to \\pi \\text{\/}2}{\\text{lim}} \\cos (x-\\frac{\\pi }{2}).[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573408578\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573408578\"]\r\n<p id=\"fs-id1170573408578\">The given function is a composite of [latex] \\cos x[\/latex] and [latex]x-\\frac{\\pi }{2}.[\/latex] Since [latex]\\underset{x\\to \\pi \\text{\/}2}{\\text{lim}}(x-\\frac{\\pi }{2})=0[\/latex] and [latex] \\cos x[\/latex] is continuous at 0, we may apply the composite function theorem. Thus,<\/p>\r\n\r\n<div id=\"fs-id1170573570975\" class=\"equation unnumbered\">[latex]\\underset{x\\to \\pi \\text{\/}2}{\\text{lim}} \\cos (x-\\frac{\\pi }{2})= \\cos (\\underset{x\\to \\pi \\text{\/}2}{\\text{lim}}(x-\\frac{\\pi }{2}))= \\cos (0)=1.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573732417\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1170573732420\" class=\"exercise\">\r\n<div id=\"fs-id1170571131879\" class=\"textbox\">\r\n<p id=\"fs-id1170571131881\">Evaluate [latex]\\underset{x\\to \\pi }{\\text{lim}} \\sin (x-\\pi ).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573362583\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573362583\"]\r\n<p id=\"fs-id1170573362583\">0<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571623332\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170573359398\">[latex]f(x)= \\sin x[\/latex] is continuous at 0. Use <a class=\"autogenerated-content\" href=\"#fs-id1170573718134\">(Figure)<\/a> as a guide.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571101566\">The proof of the next theorem uses the composite function theorem as well as the continuity of [latex]f(x)= \\sin x[\/latex] and [latex]g(x)= \\cos x[\/latex] at the point 0 to show that trigonometric functions are continuous over their entire domains.<\/p>\r\n\r\n<div id=\"fs-id1170573418920\" class=\"textbox key-takeaways theorem\">\r\n<h3>Continuity of Trigonometric Functions<\/h3>\r\n<p id=\"fs-id1170573512311\">Trigonometric functions are continuous over their entire domains.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573512316\" class=\"bc-section section\">\r\n<h1>Proof<\/h1>\r\n<p id=\"fs-id1170573429466\">We begin by demonstrating that [latex] \\cos x[\/latex] is continuous at every real number. To do this, we must show that [latex]\\underset{x\\to a}{\\text{lim}} \\cos x= \\cos a[\/latex] for all values of [latex]a[\/latex].<\/p>\r\n<p id=\"fs-id1170571099777\">[latex]\\begin{array}{ccccc}\\underset{x\\to a}{\\text{lim}} \\cos x\\hfill &amp; =\\underset{x\\to a}{\\text{lim}} \\cos ((x-a)+a)\\hfill &amp; &amp; &amp; \\text{rewrite}x=x-a+a\\hfill \\\\ &amp; =\\underset{x\\to a}{\\text{lim}}( \\cos (x-a) \\cos a- \\sin (x-a) \\sin a)\\hfill &amp; &amp; &amp; \\text{apply the identity for the cosine of the sum of two angles}\\hfill \\\\ &amp; = \\cos (\\underset{x\\to a}{\\text{lim}}(x-a)) \\cos a- \\sin (\\underset{x\\to a}{\\text{lim}}(x-a)) \\sin a\\hfill &amp; &amp; &amp; \\underset{x\\to a}{\\text{lim}}(x-a)=0,\\text{ and } \\sin x\\text{ and } \\cos x\\text{are continuous at 0}\\hfill \\\\ &amp; = \\cos (0) \\cos a- \\sin (0) \\sin a\\hfill &amp; &amp; &amp; \\text{evaluate cos(0) and sin(0) and simplify}\\hfill \\\\ &amp; =1\u00b7 \\cos a-0\u00b7 \\sin a= \\cos a.\\hfill \\end{array}[\/latex]<\/p>\r\n<p id=\"fs-id1170571216459\">The proof that [latex] \\sin x[\/latex] is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of [latex] \\sin x[\/latex] and [latex] \\cos x,[\/latex] their continuity follows from the quotient limit law.<\/p>\r\n<p id=\"fs-id1170573370281\">\u25a1<\/p>\r\n<p id=\"fs-id1170573370285\">As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions. As we continue our study of calculus, we revisit this theorem many times.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571100154\" class=\"bc-section section\">\r\n<h1>The Intermediate Value Theorem<\/h1>\r\n<p id=\"fs-id1170573717714\">Functions that are continuous over intervals of the form [latex]\\left[a,b\\right],[\/latex] where [latex]a[\/latex] and [latex]b[\/latex] are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the <strong>Intermediate Value Theorem<\/strong>.<\/p>\r\n\r\n<div id=\"fs-id1170571120551\" class=\"textbox key-takeaways theorem\">\r\n<h3>The Intermediate Value Theorem<\/h3>\r\n<p id=\"fs-id1170573534418\">Let [latex]f[\/latex] be continuous over a closed, bounded interval [latex]\\left[a,b\\right].[\/latex] If [latex]z[\/latex] is any real number between [latex]f(a)[\/latex] and [latex]f(b),[\/latex] then there is a number [latex]c[\/latex] in [latex]\\left[a,b\\right][\/latex] satisfying [latex]f(c)=z[\/latex] in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_007\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"CNX_Calc_Figure_02_04_007\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"417\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203518\/CNX_Calc_Figure_02_04_007.jpg\" alt=\"A diagram illustrating the intermediate value theorem. There is a generic continuous curved function shown over the interval [a,b]. The points fa. and fb. are marked, and dotted lines are drawn from a, b, fa., and fb. to the points (a, fa.) and (b, fb.). A third point, c, is plotted between a and b. Since the function is continuous, there is a value for fc. along the curve, and a line is drawn from c to (c, fc.) and from (c, fc.) to fc., which is labeled as z on the y axis.\" width=\"417\" height=\"385\" \/> Figure 7. There is a number [latex]c\\in \\left[a,b\\right][\/latex] that satisfies [latex]f(c)=z.[\/latex][\/caption]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571120881\" class=\"textbox examples\">\r\n<h3>Application of the Intermediate Value Theorem<\/h3>\r\n<div id=\"fs-id1170571120883\" class=\"exercise\">\r\n<div id=\"fs-id1170570997452\" class=\"textbox\">\r\n<p id=\"fs-id1170570997458\">Show that [latex]f(x)=x- \\cos x[\/latex] has at least one zero.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571136342\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571136342\"]\r\n<p id=\"fs-id1170571136342\">Since [latex]f(x)=x- \\cos x[\/latex] is continuous over [latex](\\text{\u2212}\\infty ,\\text{+}\\infty ),[\/latex] it is continuous over any closed interval of the form [latex]\\left[a,b\\right].[\/latex] If you can find an interval [latex]\\left[a,b\\right][\/latex] such that [latex]f(a)[\/latex] and [latex]f(b)[\/latex] have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number [latex]c[\/latex] in [latex](a,b)[\/latex] that satisfies [latex]f(c)=0.[\/latex] Note that<\/p>\r\n\r\n<div id=\"fs-id1170571239008\" class=\"equation unnumbered\">[latex]f(0)=0- \\cos (0)=-1&lt;0[\/latex]<\/div>\r\n<p id=\"fs-id1170573424328\">and<\/p>\r\n\r\n<div id=\"fs-id1170573424331\" class=\"equation unnumbered\">[latex]f(\\frac{\\pi }{2})=\\frac{\\pi }{2}- \\cos \\frac{\\pi }{2}=\\frac{\\pi }{2}&gt;0.[\/latex]<\/div>\r\n<p id=\"fs-id1170573753266\">Using the Intermediate Value Theorem, we can see that there must be a real number [latex]c[\/latex] in [latex]\\left[0,\\pi \\text{\/}2\\right][\/latex] that satisfies [latex]f(c)=0.[\/latex] Therefore, [latex]f(x)=x- \\cos x[\/latex] has at least one zero.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573439386\" class=\"textbox examples\">\r\n<h3>When Can You Apply the Intermediate Value Theorem?<\/h3>\r\n<div id=\"fs-id1170573439388\" class=\"exercise\">\r\n<div id=\"fs-id1170573413593\" class=\"textbox\">\r\n<p id=\"fs-id1170573413598\">If [latex]f(x)[\/latex] is continuous over [latex]\\left[0,2\\right],f(0)&gt;0[\/latex] and [latex]f(2)&gt;0,[\/latex] can we use the Intermediate Value Theorem to conclude that [latex]f(x)[\/latex] has no zeros in the interval [latex]\\left[0,2\\right]\\text{?}[\/latex] Explain.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571262087\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571262087\"]\r\n<p id=\"fs-id1170571262087\">No. The Intermediate Value Theorem only allows us to conclude that we can find a value between [latex]f(0)[\/latex] and [latex]f(2);[\/latex] it doesn\u2019t allow us to conclude that we can\u2019t find other values. To see this more clearly, consider the function [latex]f(x)={(x-1)}^{2}.[\/latex] It satisfies [latex]f(0)=1&gt;0,f(2)=1&gt;0,[\/latex] and [latex]f(1)=0.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170570996536\" class=\"textbox examples\">\r\n<h3>When Can You Apply the Intermediate Value Theorem?<\/h3>\r\n<div id=\"fs-id1170573586870\" class=\"exercise\">\r\n<div id=\"fs-id1170573586872\" class=\"textbox\">\r\n<p id=\"fs-id1170573586819\">For [latex]f(x)=1\\text{\/}x,f(-1)=-1&lt;0[\/latex] and [latex]f(1)=1&gt;0.[\/latex] Can we conclude that [latex]f(x)[\/latex] has a zero in the interval [latex]\\left[-1,1\\right]?[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571138880\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571138880\"]\r\n<p id=\"fs-id1170571138880\">No. The function is not continuous over [latex]\\left[-1,1\\right].[\/latex] The Intermediate Value Theorem does not apply here.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571100300\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1170571100303\" class=\"exercise\">\r\n<div id=\"fs-id1170571048764\" class=\"textbox\">\r\n<p id=\"fs-id1170571048766\">Show that [latex]f(x)={x}^{3}-{x}^{2}-3x+1[\/latex] has a zero over the interval [latex]\\left[0,1\\right].[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573382710\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573382710\"]\r\n<p id=\"fs-id1170573382710\">[latex]f(0)=1&gt;0,f(1)=-2&lt;0;f(x)[\/latex] is continuous over [latex]\\left[0,1\\right].[\/latex] It must have a zero on this interval.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572203444\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170573569887\">Find [latex]f(0)[\/latex] and [latex]f(1).[\/latex] Apply the Intermediate Value Theorem.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573391172\" class=\"textbox key-takeaways\">\r\n<h3>Key Concepts<\/h3>\r\n<ul id=\"fs-id1170573574290\">\r\n \t<li>For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point.<\/li>\r\n \t<li>Discontinuities may be classified as removable, jump, or infinite.<\/li>\r\n \t<li>A function is continuous over an open interval if it is continuous at every point in the interval. It is continuous over a closed interval if it is continuous at every point in its interior and is continuous at its endpoints.<\/li>\r\n \t<li>The composite function theorem states: If [latex]f(x)[\/latex] is continuous at <em>L<\/em> and [latex]\\underset{x\\to a}{\\text{lim}}g(x)=L,[\/latex] then [latex]\\underset{x\\to a}{\\text{lim}}f(g(x))=f(\\underset{x\\to a}{\\text{lim}}g(x))=f(L).[\/latex]<\/li>\r\n \t<li>The Intermediate Value Theorem guarantees that if a function is continuous over a closed interval, then the function takes on every value between the values at its endpoints.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1170573397457\" class=\"textbox exercises\">\r\n<p id=\"fs-id1170573397460\">For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.<\/p>\r\n\r\n<div id=\"fs-id1170571246283\" class=\"exercise\">\r\n<div id=\"fs-id1170571246285\" class=\"textbox\">\r\n<p id=\"fs-id1170571246287\">[latex]f(x)=\\frac{1}{\\sqrt{x}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"solution\">\r\n<p id=\"fs-id1170571301576\">The function is defined for all [latex]x[\/latex] in the interval [latex](0,\\infty ).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571103211\" class=\"exercise\">\r\n<div id=\"fs-id1170570973770\" class=\"textbox\">\r\n<p id=\"fs-id1170570973773\">[latex]f(x)=\\frac{2}{{x}^{2}+1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573367933\" class=\"exercise\">\r\n<div id=\"fs-id1170573367935\" class=\"textbox\">\r\n<p id=\"fs-id1170573367937\">[latex]f(x)=\\frac{x}{{x}^{2}-x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573590405\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573590405\"]\r\n<p id=\"fs-id1170573590405\">Removable discontinuity at [latex]x=0;[\/latex] infinite discontinuity at [latex]x=1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573734927\" class=\"exercise\">\r\n<div id=\"fs-id1170573734930\" class=\"textbox\">\r\n<p id=\"fs-id1170573404382\">[latex]g(t)={t}^{-1}+1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573581616\" class=\"exercise\">\r\n<div id=\"fs-id1170573581618\" class=\"textbox\">\r\n<p id=\"fs-id1170573581620\">[latex]f(x)=\\frac{5}{{e}^{x}-2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573586367\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573586367\"]\r\n<p id=\"fs-id1170573586367\">Infinite discontinuity at [latex]x=\\text{ln}2[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573403108\" class=\"exercise\">\r\n<div id=\"fs-id1170573403110\" class=\"textbox\">\r\n<p id=\"fs-id1170573403112\">[latex]f(x)=\\frac{|x-2|}{x-2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573361738\" class=\"exercise\">\r\n<div id=\"fs-id1170573361740\" class=\"textbox\">\r\n<p id=\"fs-id1170573361742\">[latex]H(x)= \\tan 2x[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571047553\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571047553\"]\r\n<p id=\"fs-id1170571047553\">Infinite discontinuities at [latex]x=\\frac{(2k+1)\\pi }{4},[\/latex] for [latex]k=0,\u00b11,\u00b12,\u00b13\\text{,\u2026}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573581931\" class=\"exercise\">\r\n<div id=\"fs-id1170573581933\" class=\"textbox\">\r\n<p id=\"fs-id1170573408756\">[latex]f(t)=\\frac{t+3}{{t}^{2}+5t+6}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170573750406\">For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?<\/p>\r\n\r\n<div id=\"fs-id1170573750411\" class=\"exercise\">\r\n<div id=\"fs-id1170573750413\" class=\"textbox\">\r\n<p id=\"fs-id1170573593159\">[latex]\\frac{2{x}^{2}-5x+3}{x-1}[\/latex] at [latex]x=1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573331459\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573331459\"]\r\n<p id=\"fs-id1170573331459\">No. It is a removable discontinuity.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170570999796\" class=\"exercise\">\r\n<div id=\"fs-id1170570999798\" class=\"textbox\">\r\n<p id=\"fs-id1170570999800\">[latex]h(\\theta )=\\frac{ \\sin \\theta - \\cos \\theta }{ \\tan \\theta }[\/latex] at [latex]\\theta =\\pi [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573580625\" class=\"exercise\">\r\n<div id=\"fs-id1170573580627\" class=\"textbox\">\r\n<p id=\"fs-id1170573580629\">[latex]g(u)=\\bigg\\{\\begin{array}{ll}\\frac{6{u}^{2}+u-2}{2u-1}&amp; \\text{ if }u\\ne \\frac{1}{2}\\hfill \\\\ \\frac{7}{2}\\hfill &amp; \\text{ if }u=\\frac{1}{2}\\hfill \\end{array},[\/latex] at [latex]u=\\frac{1}{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571120270\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571120270\"]\r\n<p id=\"fs-id1170571120270\">Yes. It is continuous.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571120275\" class=\"exercise\">\r\n<div id=\"fs-id1170573590166\" class=\"textbox\">\r\n<p id=\"fs-id1170573590169\">[latex]f(y)=\\frac{ \\sin (\\pi y)}{ \\tan (\\pi y)},[\/latex] at [latex]y=1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571120812\" class=\"exercise\">\r\n<div id=\"fs-id1170571130844\" class=\"textbox\">\r\n<p id=\"fs-id1170571130846\">[latex]f(x)=\\bigg\\{\\begin{array}{ll}{x}^{2}-{e}^{x}&amp; \\text{ if }x&lt;0\\hfill \\\\ x-1&amp; \\text{ if }x\\ge 0\\hfill \\end{array},[\/latex] at [latex]x=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573381211\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573381211\"]\r\n<p id=\"fs-id1170573381211\">Yes. It is continuous.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573381217\" class=\"exercise\">\r\n<div id=\"fs-id1170573381219\" class=\"textbox\">\r\n<p id=\"fs-id1170571285463\">[latex]f(x)=\\bigg\\{\\begin{array}{l}x \\sin (x)\\text{ if }x\\le \\pi \\\\ x \\tan (x)\\text{ if }x&gt;\\pi \\end{array},[\/latex] at [latex]x=\\pi [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571123088\">In the following exercises, find the value(s) of [latex]k[\/latex] that makes each function continuous over the given interval.<\/p>\r\n\r\n<div id=\"fs-id1170573413991\" class=\"exercise\">\r\n<div id=\"fs-id1170573413993\" class=\"textbox\">\r\n<p id=\"fs-id1170571137944\">[latex]f(x)=\\bigg\\{\\begin{array}{cc}3x+2,\\hfill &amp; x&lt;k\\hfill \\\\ 2x-3,\\hfill &amp; k\\le x\\le 8\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573440208\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573440208\"]\r\n<p id=\"fs-id1170573440208\">[latex]k=-5[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571068186\" class=\"exercise\">\r\n<div id=\"fs-id1170571068188\" class=\"textbox\">\r\n<p id=\"fs-id1170573760712\">[latex]f(\\theta )=\\bigg\\{\\begin{array}{ll}\\hfill \\sin \\theta ,&amp; 0\\le \\theta &lt;\\frac{\\pi }{2}\\hfill \\\\ \\cos (\\theta +k),\\hfill &amp; \\frac{\\pi }{2}\\le \\theta \\le \\pi \\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573575226\" class=\"exercise\">\r\n<div id=\"fs-id1170573575228\" class=\"textbox\">\r\n<p id=\"fs-id1170573575230\">[latex]f(x)=\\bigg\\{\\begin{array}{cc}\\frac{{x}^{2}+3x+2}{x+2},\\hfill &amp; x\\ne -2\\hfill \\\\ \\hfill k,&amp; x=-2\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573502717\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573502717\"]\r\n<p id=\"fs-id1170573502717\">[latex]k=-1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"exercise\">\r\n<div class=\"textbox\">\r\n<p id=\"fs-id1170573589735\">[latex]f(x)=\\bigg\\{\\begin{array}{cc}\\hfill {e}^{kx},&amp; 0\\le x&lt;4\\hfill \\\\ x+3,\\hfill &amp; 4\\le x\\le 8\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573449551\" class=\"exercise\">\r\n<div id=\"fs-id1170573449554\" class=\"textbox\">\r\n<p id=\"fs-id1170573413580\">[latex]f(x)=\\bigg\\{\\begin{array}{cc}\\hfill \\sqrt{kx},&amp; 0\\le x\\le 3\\hfill \\\\ x+1,\\hfill &amp; 3&lt;x\\le 10\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571050054\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571050054\"]\r\n<p id=\"fs-id1170571050054\">[latex]k=\\frac{16}{3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571121702\">In the following exercises, use the Intermediate Value Theorem (IVT).<\/p>\r\n\r\n<div id=\"fs-id1170571121705\" class=\"exercise\">\r\n<div id=\"fs-id1170573541393\" class=\"textbox\">\r\n<p id=\"fs-id1170573541395\">Let [latex]h(x)=\\bigg\\{\\begin{array}{ll}3{x}^{2}-4,&amp; x\\le 2\\hfill \\\\ 5+4x,&amp; x&gt;2\\hfill \\end{array}[\/latex] Over the interval [latex]\\left[0,4\\right],[\/latex] there is no value of [latex]x[\/latex] such that [latex]h(x)=10,[\/latex] although [latex]h(0)&lt;10[\/latex] and [latex]h(4)&gt;10.[\/latex] Explain why this does not contradict the IVT.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571087121\" class=\"exercise\">\r\n<div id=\"fs-id1170571087123\" class=\"textbox\">\r\n<p id=\"fs-id1170571087125\">A particle moving along a line has at each time [latex]t[\/latex] a position function [latex]s(t),[\/latex] which is continuous. Assume [latex]s(2)=5[\/latex] and [latex]s(5)=2.[\/latex] Another particle moves such that its position is given by [latex]h(t)=s(t)-t.[\/latex] Explain why there must be a value [latex]c[\/latex] for [latex]2&lt;c&lt;5[\/latex] such that [latex]h(c)=0.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170570998991\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170570998991\"]\r\n<p id=\"fs-id1170570998991\">Since both [latex]s[\/latex] and [latex]y=t[\/latex] are continuous everywhere, then [latex]h(t)=s(t)-t[\/latex] is continuous everywhere and, in particular, it is continuous over the closed interval [latex]\\left[2,5\\right].[\/latex] Also, [latex]h(2)=3&gt;0[\/latex] and [latex]h(5)=-3&lt;0.[\/latex] Therefore, by the IVT, there is a value [latex]x=c[\/latex] such that [latex]h(c)=0.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573590374\" class=\"exercise\">\r\n<div id=\"fs-id1170573590376\" class=\"textbox\">\r\n<p id=\"fs-id1170573590378\"><strong>[T]<\/strong> Use the statement \u201cThe cosine of [latex]t[\/latex] is equal to [latex]t[\/latex] cubed.\u201d<\/p>\r\n\r\n<ol id=\"fs-id1170571132593\" style=\"list-style-type: lower-alpha\">\r\n \t<li>Write a mathematical equation of the statement.<\/li>\r\n \t<li>Prove that the equation in part a. has at least one real solution.<\/li>\r\n \t<li>Use a calculator to find an interval of length 0.01 that contains a solution.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170570982565\" class=\"exercise\">\r\n<div id=\"fs-id1170570982567\" class=\"textbox\">\r\n<p id=\"fs-id1170570982569\">Apply the IVT to determine whether [latex]{2}^{x}={x}^{3}[\/latex] has a solution in one of the intervals [latex]\\left[1.25,1.375\\right][\/latex] or [latex]\\left[1.375,1.5\\right].[\/latex] Briefly explain your response for each interval.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170570998210\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170570998210\"]\r\n<p id=\"fs-id1170570998210\">The function [latex]f(x)={2}^{x}-{x}^{3}[\/latex] is continuous over the interval [latex]\\left[1.25,1.375\\right][\/latex] and has opposite signs at the endpoints.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573423919\" class=\"exercise\">\r\n<div id=\"fs-id1170573423921\" class=\"textbox\">\r\n<p id=\"fs-id1170573423923\">Consider the graph of the function [latex]y=f(x)[\/latex] shown in the following graph.<\/p>\r\n<span id=\"fs-id1170573581162\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203520\/CNX_Calc_Figure_02_04_201.jpg\" alt=\"A diagram illustrating the intermediate value theorem. There is a generic continuous curved function shown over the interval [a,b]. The points fa. and fb. are marked, and dotted lines are drawn from a, b, fa., and fb. to the points (a, fa.) and (b, fb.). A third point, c, is plotted between a and b. Since the function is continuous, there is a value for fc. along the curve, and a line is drawn from c to (c, fc.) and from (c, fc.) to fc., which is labeled as z on the y axis.\" \/><\/span>\r\n<ol id=\"fs-id1170571053591\" style=\"list-style-type: lower-alpha\">\r\n \t<li>Find all values for which the function is discontinuous.<\/li>\r\n \t<li>For each value in part a., state why the formal definition of continuity does not apply.<\/li>\r\n \t<li>Classify each discontinuity as either jump, removable, or infinite.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571290233\" class=\"exercise\">\r\n<div id=\"fs-id1170571290235\" class=\"textbox\">\r\n<p id=\"fs-id1170571290237\">Let [latex]f(x)=\\bigg\\{\\begin{array}{c}3x,x&gt;1\\\\ {x}^{3},x&lt;1\\end{array}.[\/latex]<\/p>\r\n\r\n<ol id=\"fs-id1170571101024\" style=\"list-style-type: lower-alpha\">\r\n \t<li>Sketch the graph of [latex]f[\/latex].<\/li>\r\n \t<li>Is it possible to find a value [latex]k[\/latex] such that [latex]f(1)=k,[\/latex] which makes [latex]f(x)[\/latex] continuous for all real numbers? Briefly explain.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571100273\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571100273\"]\r\n<p id=\"fs-id1170571100273\">a.<\/p>\r\n<span id=\"fs-id1170571100281\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203523\/CNX_Calc_Figure_02_04_202.jpg\" alt=\"A graph of the given piecewise function containing two segments. The first, x^3, exists for x &lt; 1 and ends with an open circle at (1,1). The second, 3x, exists for x &gt; 1. It beings with an open circle at (1,3).\" \/><\/span>\r\nb. It is not possible to redefine [latex]f(1)[\/latex] since the discontinuity is a jump discontinuity.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573359564\" class=\"exercise\">\r\n<div id=\"fs-id1170573359566\" class=\"textbox\">\r\n<p id=\"fs-id1170573359568\">Let [latex]f(x)=\\frac{{x}^{4}-1}{{x}^{2}-1}[\/latex] for [latex]x\\ne -1,1.[\/latex]<\/p>\r\n\r\n<ol id=\"fs-id1170573633982\" style=\"list-style-type: lower-alpha\">\r\n \t<li>Sketch the graph of [latex]f[\/latex].<\/li>\r\n \t<li>Is it possible to find values [latex]{k}_{1}[\/latex] and [latex]{k}_{2}[\/latex] such that [latex]f(-1)=k[\/latex] and [latex]f(1)={k}_{2},[\/latex] and that makes [latex]f(x)[\/latex] continuous for all real numbers? Briefly explain.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573512268\" class=\"exercise\">\r\n<div id=\"fs-id1170573439551\" class=\"textbox\">\r\n<p id=\"fs-id1170573439554\">Sketch the graph of the function [latex]y=f(x)[\/latex] with properties i. through vii.<\/p>\r\n\r\n<ol id=\"fs-id1170573397517\">\r\n \t<li>The domain of [latex]f[\/latex] is [latex](\\text{\u2212}\\infty ,\\text{+}\\infty ).[\/latex]<\/li>\r\n \t<li>[latex]f[\/latex] has an infinite discontinuity at [latex]x=-6.[\/latex]<\/li>\r\n \t<li>[latex]f(-6)=3[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to {-3}^{-}}{\\text{lim}}f(x)=\\underset{x\\to {-3}^{+}}{\\text{lim}}f(x)=2[\/latex]<\/li>\r\n \t<li>[latex]f(-3)=3[\/latex]<\/li>\r\n \t<li>[latex]f[\/latex] is left continuous but not right continuous at [latex]x=3.[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to -\\infty }{\\text{lim}}f(x)=\\text{\u2212}\\infty [\/latex] and [latex]\\underset{x\\to +\\infty }{\\text{lim}}f(x)=\\text{+}\\infty [\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170570982616\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170570982616\"]\r\n<p id=\"fs-id1170570982616\">Answers may vary; see the following example:<\/p>\r\n<span id=\"fs-id1170571130672\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203525\/CNX_Calc_Figure_02_04_207.jpg\" alt=\"A graph of a piecewise function with several segments. The first is an increasing line that exists for x &lt; -8. It ends at an open circle at (-8,-8). The second is an increasing curve that exists from -8 &lt;= x &lt; -6. It begins with a closed circle at (-8, 0 ) and goes to infinity as x goes to -6 from the left. The third is a closed circle at the point (-6, 3). The fourth is a line that exists from -6 &lt; x &lt;= 3. It begins with an open circle at (-6, 2) and ends with a closed circle at (3,2). The fifth is an increasing line starting with an open circle at (3,3). It exists for x &gt; 3.\" \/><\/span>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571130669\" class=\"exercise\">\r\n<div id=\"fs-id1170571073840\" class=\"textbox\">\r\n<p id=\"fs-id1170571073842\">Sketch the graph of the function [latex]y=f(x)[\/latex] with properties i. through iv.<\/p>\r\n\r\n<ol id=\"fs-id1170573388516\">\r\n \t<li>The domain of [latex]f[\/latex] is [latex]\\left[0,5\\right].[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to {1}^{+}}{\\text{lim}}f(x)[\/latex] and [latex]\\underset{x\\to {1}^{-}}{\\text{lim}}f(x)[\/latex] exist and are equal.<\/li>\r\n \t<li>[latex]f(x)[\/latex] is left continuous but not continuous at [latex]x=2,[\/latex] and right continuous but not continuous at [latex]x=3.[\/latex]<\/li>\r\n \t<li>[latex]f(x)[\/latex] has a removable discontinuity at [latex]x=1,[\/latex] a jump discontinuity at [latex]x=2,[\/latex] and the following limits hold: [latex]\\underset{x\\to {3}^{-}}{\\text{lim}}f(x)=\\text{\u2212}\\infty [\/latex] and [latex]\\underset{x\\to {3}^{+}}{\\text{lim}}f(x)=2.[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170573425311\">In the following exercises, suppose [latex]y=f(x)[\/latex] is defined for all [latex]x[\/latex]. For each description, sketch a graph with the indicated property.<\/p>\r\n\r\n<div id=\"fs-id1170571096173\" class=\"exercise\">\r\n<div id=\"fs-id1170571096175\" class=\"textbox\">\r\n<p id=\"fs-id1170571096177\">Discontinuous at [latex]x=1[\/latex] with [latex]\\underset{x\\to -1}{\\text{lim}}f(x)=-1[\/latex] and [latex]\\underset{x\\to 2}{\\text{lim}}f(x)=4[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571123483\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571123483\"]\r\n<p id=\"fs-id1170571123483\">Answers may vary; see the following example:<\/p>\r\n<span id=\"fs-id1170571123491\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203529\/CNX_Calc_Figure_02_04_205.jpg\" alt=\"The graph of a piecewise function with two parts. The first part is an increasing curve that exists for x &lt; 1. It ends at (1,1). The second part is an increasing line that exists for x &gt; 1. It begins at (1,3).\" \/><\/span>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571123500\" class=\"exercise\">\r\n<div id=\"fs-id1170571123502\" class=\"textbox\">\r\n<p id=\"fs-id1170573401181\">Discontinuous at [latex]x=2[\/latex] but continuous elsewhere with [latex]\\underset{x\\to 0}{\\text{lim}}f(x)=\\frac{1}{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170573422225\">Determine whether each of the given statements is true. Justify your response with an explanation or counterexample.<\/p>\r\n\r\n<div id=\"fs-id1170573422229\" class=\"exercise\">\r\n<div id=\"fs-id1170573422231\" class=\"textbox\">\r\n<p id=\"fs-id1170573422234\">[latex]f(t)=\\frac{2}{{e}^{t}-{e}^{-t}}[\/latex] is continuous everywhere.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573404350\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573404350\"]\r\n<p id=\"fs-id1170573404350\">False. It is continuous over [latex](\\text{\u2212}\\infty ,0)\\cup (0,\\infty ).[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573624587\" class=\"exercise\">\r\n<div id=\"fs-id1170573624589\" class=\"textbox\">\r\n<p id=\"fs-id1170573624591\">If the left- and right-hand limits of [latex]f(x)[\/latex] as [latex]x\\to a[\/latex] exist and are equal, then [latex]f[\/latex] cannot be discontinuous at [latex]x=a.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571287009\" class=\"exercise\">\r\n<div id=\"fs-id1170571287012\" class=\"textbox\">\r\n<p id=\"fs-id1170571287014\">If a function is not continuous at a point, then it is not defined at that point.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571287020\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571287020\"]\r\n<p id=\"fs-id1170571287020\">False. Consider [latex]f(x)=\\bigg\\{\\begin{array}{l}x\\text{ if }x\\ne 0\\\\ 4\\text{ if }x=0\\end{array}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571131867\" class=\"exercise\">\r\n<div id=\"fs-id1170571131869\" class=\"textbox\">\r\n<p id=\"fs-id1170571131872\">According to the IVT, [latex] \\cos x- \\sin x-x=2[\/latex] has a solution over the interval [latex]\\left[-1,1\\right].[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571131236\" class=\"exercise\">\r\n<div id=\"fs-id1170571131238\" class=\"textbox\">\r\n<p id=\"fs-id1170571131241\">If [latex]f(x)[\/latex] is continuous such that [latex]f(a)[\/latex] and [latex]f(b)[\/latex] have opposite signs, then [latex]f(x)=0[\/latex] has exactly one solution in [latex]\\left[a,b\\right].[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573519289\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573519289\"]\r\n<p id=\"fs-id1170573519289\">False. Consider [latex]f(x)= \\cos (x)[\/latex] on [latex]\\left[-\\pi ,2\\pi \\right].[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571099130\" class=\"exercise\">\r\n<div id=\"fs-id1170571099132\" class=\"textbox\">\r\n<p id=\"fs-id1170571099134\">The function [latex]f(x)=\\frac{{x}^{2}-4x+3}{{x}^{2}-1}[\/latex] is continuous over the interval [latex]\\left[0,3\\right].[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170570998957\" class=\"exercise\">\r\n<div id=\"fs-id1170570998960\" class=\"textbox\">\r\n<p id=\"fs-id1170570998962\">If [latex]f(x)[\/latex] is continuous everywhere and [latex]f(a),f(b)&gt;0,[\/latex] then there is no root of [latex]f(x)[\/latex] in the interval [latex]\\left[a,b\\right].[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571197414\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571197414\"]\r\n<p id=\"fs-id1170571197414\">False. The IVT does <em>not<\/em> work in reverse! Consider [latex]{(x-1)}^{2}[\/latex] over the interval [latex]\\left[-2,2\\right].[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170573506388\"><strong>[T]<\/strong> The following problems consider the scalar form of Coulomb\u2019s law, which describes the electrostatic force between two point charges, such as electrons. It is given by the equation [latex]F(r)={k}_{e}\\frac{|{q}_{1}{q}_{2}|}{{r}^{2}},[\/latex] where [latex]{k}_{e}[\/latex] is Coulomb\u2019s constant, [latex]{q}_{i}[\/latex] are the magnitudes of the charges of the two particles, and [latex]r[\/latex] is the distance between the two particles.<\/p>\r\n\r\n<div id=\"fs-id1170573393357\" class=\"exercise\">\r\n<div id=\"fs-id1170573393359\" class=\"textbox\">\r\n<p id=\"fs-id1170573393361\">To simplify the calculation of a model with many interacting particles, after some threshold value [latex]r=R,[\/latex] we approximate <em>F<\/em> as zero.<\/p>\r\n\r\n<ol id=\"fs-id1170573750444\" style=\"list-style-type: lower-alpha\">\r\n \t<li>Explain the physical reasoning behind this assumption.<\/li>\r\n \t<li>What is the force equation?<\/li>\r\n \t<li>Evaluate the force <em>F<\/em> using both Coulomb\u2019s law and our approximation, assuming two protons with a charge magnitude of [latex]1.6022\u00d7{10}^{-19}\\text{coulombs (C)},[\/latex] and the Coulomb constant [latex]{k}_{e}=8.988\u00d7{10}^{9}{\\text{Nm}}^{2}\\text{\/}{\\text{C}}^{2}[\/latex] are 1 m apart. Also, assume [latex]R&lt;1\\text{m}.[\/latex] How much inaccuracy does our approximation generate? Is our approximation reasonable?<\/li>\r\n \t<li>Is there any finite value of <em>R<\/em> for which this system remains continuous at <em>R<\/em>?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573438919\" class=\"exercise\">\r\n<div id=\"fs-id1170573438922\" class=\"textbox\">\r\n<p id=\"fs-id1170573438924\">Instead of making the force 0 at <em>R<\/em>, instead we let the force be 10<sup>\u221220<\/sup> for [latex]r\\ge R.[\/latex] Assume two protons, which have a magnitude of charge [latex]1.6022\u00d7{10}^{-19}\\text{C},[\/latex] and the Coulomb constant [latex]{k}_{e}=8.988\u00d7{10}^{9}{\\text{Nm}}^{2}\\text{\/}{\\text{C}}^{2}.[\/latex] Is there a value <em>R<\/em> that can make this system continuous? If so, find it.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170573420081\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573420081\"]\r\n<p id=\"fs-id1170573420081\">[latex]R=0.0001519\\text{m}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571038053\">Recall the discussion on spacecraft from the chapter opener. The following problems consider a rocket launch from Earth\u2019s surface. The force of gravity on the rocket is given by [latex]F(d)=-mk\\text{\/}{d}^{2},[\/latex] where [latex]m[\/latex] is the mass of the rocket, [latex]d[\/latex] is the distance of the rocket from the center of Earth, and [latex]k[\/latex] is a constant.<\/p>\r\n\r\n<div id=\"fs-id1170573587225\" class=\"exercise\">\r\n<div id=\"fs-id1170570991016\" class=\"textbox\">\r\n<p id=\"fs-id1170570991018\"><strong>[T]<\/strong> Determine the value and units of [latex]k[\/latex] given that the mass of the rocket on Earth is 3 million kg. (<em>Hint<\/em>: The distance from the center of Earth to its surface is 6378 km.)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573571185\" class=\"exercise\">\r\n<div id=\"fs-id1170573571187\" class=\"textbox\">\r\n<p id=\"fs-id1170573571189\"><strong>[T]<\/strong> After a certain distance <em>D<\/em> has passed, the gravitational effect of Earth becomes quite negligible, so we can approximate the force function by [latex]F(d)=\\bigg\\{\\begin{array}{ll}-\\frac{mk}{{d}^{2}}\\hfill &amp; \\text{ if }d&lt;D\\hfill \\\\ 10,000\\hfill &amp; \\text{ if }d\\ge D\\hfill \\end{array}.[\/latex] Find the necessary condition <em>D<\/em> such that the force function remains continuous.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170570974552\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170570974552\"]\r\n<p id=\"fs-id1170570974552\">[latex]D=63.78\\text{km}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571124644\" class=\"exercise\">\r\n<div id=\"fs-id1170571124646\" class=\"textbox\">\r\n<p id=\"fs-id1170571124648\">As the rocket travels away from Earth\u2019s surface, there is a distance <em>D<\/em> where the rocket sheds some of its mass, since it no longer needs the excess fuel storage. We can write this function as [latex]F(d)=\\bigg\\{\\begin{array}{l}-\\frac{{m}_{1}k}{{d}^{2}}\\text{ if }d&lt;D\\hfill \\\\ -\\frac{{m}_{2}k}{{d}^{2}}\\text{ if }d\\ge D\\hfill \\end{array}.[\/latex] Is there a <em>D<\/em> value such that this function is continuous, assuming [latex]{m}_{1}\\ne {m}_{2}?[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571136676\">Prove the following functions are continuous everywhere<\/p>\r\n\r\n<div id=\"fs-id1170571136680\" class=\"exercise\">\r\n<div id=\"fs-id1170571276945\" class=\"textbox\">\r\n<p id=\"fs-id1170571276947\">[latex]f(\\theta )= \\sin \\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571276973\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571276973\"]\r\n<p id=\"fs-id1170571276973\">For all values of [latex]a,f(a)[\/latex] is defined, [latex]\\underset{\\theta \\to a}{\\text{lim}}f(\\theta )[\/latex] exists, and [latex]\\underset{\\theta \\to a}{\\text{lim}}f(\\theta )=f(a).[\/latex] Therefore, [latex]f(\\theta )[\/latex] is continuous everywhere.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573413789\" class=\"exercise\">\r\n<div id=\"fs-id1170573413791\" class=\"textbox\">\r\n<p id=\"fs-id1170573413793\">[latex]g(x)=|x|[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573414086\" class=\"exercise\">\r\n<div id=\"fs-id1170573414089\" class=\"textbox\">\r\n<p id=\"fs-id1170573414091\">Where is [latex]f(x)=\\bigg\\{\\begin{array}{l}0\\text{ if }x\\text{is irrational}\\\\ 1\\text{ if }x\\text{is rational}\\end{array}[\/latex] continuous?<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1170571258410\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571258410\"]\r\n<p id=\"fs-id1170571258410\">Nowhere<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1170571119822\" class=\"definition\">\r\n \t<dt>continuity at a point<\/dt>\r\n \t<dd id=\"fs-id1170571119828\">A function [latex]f(x)[\/latex] is continuous at a point [latex]a[\/latex] if and only if the following three conditions are satisfied: (1) [latex]f(a)[\/latex] is defined, (2) [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists, and (3) [latex]\\underset{x\\to a}{\\text{lim}}f(x)=f(a)[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170573574137\" class=\"definition\">\r\n \t<dt>continuity from the left<\/dt>\r\n \t<dd id=\"fs-id1170573574142\">A function is continuous from the left at [latex]b[\/latex] if [latex]\\underset{x\\to {b}^{-}}{\\text{lim}}f(x)=f(b)[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170571275835\" class=\"definition\">\r\n \t<dt>continuity from the right<\/dt>\r\n \t<dd id=\"fs-id1170571275840\">A function is continuous from the right at [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)=f(a)[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170573363352\" class=\"definition\">\r\n \t<dt>continuity over an interval<\/dt>\r\n \t<dd id=\"fs-id1170573363357\">a function that can be traced with a pencil without lifting the pencil; a function is continuous over an open interval if it is continuous at every point in the interval; a function [latex]f(x)[\/latex] is continuous over a closed interval of the form [latex]\\left[a,b\\right][\/latex] if it is continuous at every point in [latex](a,b),[\/latex] and it is continuous from the right at [latex]a[\/latex] and from the left at [latex]b[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170573573745\" class=\"definition\">\r\n \t<dt>discontinuity at a point<\/dt>\r\n \t<dd id=\"fs-id1170573573750\">A function is discontinuous at a point or has a discontinuity at a point if it is not continuous at the point<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170573573755\" class=\"definition\">\r\n \t<dt>infinite discontinuity<\/dt>\r\n \t<dd id=\"fs-id1170573573761\">An infinite discontinuity occurs at a point [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)=\\text{\u00b1}\\infty [\/latex] or [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)=\\text{\u00b1}\\infty [\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170571049355\" class=\"definition\">\r\n \t<dt>Intermediate Value Theorem<\/dt>\r\n \t<dd id=\"fs-id1170571049360\">Let [latex]f[\/latex] be continuous over a closed bounded interval [latex]\\left[\\text{a},\\text{b}\\right];[\/latex] if [latex]z[\/latex] is any real number between [latex]f(a)[\/latex] and [latex]f(b),[\/latex] then there is a number [latex]c[\/latex] in [latex]\\left[a,b\\right][\/latex] satisfying [latex]f(c)=z[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170570997709\" class=\"definition\">\r\n \t<dt>jump discontinuity<\/dt>\r\n \t<dd id=\"fs-id1170570997714\">A jump discontinuity occurs at a point [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)[\/latex] and [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)[\/latex] both exist, but [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)\\ne \\underset{x\\to {a}^{+}}{\\text{lim}}f(x)[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170571146508\" class=\"definition\">\r\n \t<dt>removable discontinuity<\/dt>\r\n \t<dd id=\"fs-id1170571146514\">A removable discontinuity occurs at a point [latex]a[\/latex] if [latex]f(x)[\/latex] is discontinuous at [latex]a[\/latex], but [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Explain the three conditions for continuity at a point.<\/li>\n<li>Describe three kinds of discontinuities.<\/li>\n<li>Define continuity on an interval.<\/li>\n<li>State the theorem for limits of composite functions.<\/li>\n<li>Provide an example of the intermediate value theorem.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170570996716\">Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Such functions are called <em>continuous<\/em>. Other functions have points at which a break in the graph occurs, but satisfy this property over intervals contained in their domains. They are continuous on these intervals and are said to have a <strong>discontinuity at a point<\/strong> where a break occurs.<\/p>\n<p id=\"fs-id1170573595201\">We begin our investigation of continuity by exploring what it means for a function to have <strong>continuity at a point<\/strong>. Intuitively, a function is continuous at a particular point if there is no break in its graph at that point.<\/p>\n<div id=\"fs-id1170573262889\" class=\"bc-section section\">\n<h1>Continuity at a Point<\/h1>\n<p id=\"fs-id1170573324888\">Before we look at a formal definition of what it means for a function to be continuous at a point, let\u2019s consider various functions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditions that prevent such failures.<\/p>\n<p id=\"fs-id1170573390124\">Our first function of interest is shown in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_001\">(Figure)<\/a>. We see that the graph of [latex]f(x)[\/latex] has a hole at [latex]a[\/latex]. In fact, [latex]f(a)[\/latex] is undefined. At the very least, for [latex]f(x)[\/latex] to be continuous at [latex]a[\/latex], we need the following condition:<\/p>\n<div id=\"fs-id1170570997076\" class=\"equation unnumbered\">[latex]\\text{i.}f(a)\\text{is defined.}[\/latex]<\/div>\n<div id=\"CNX_Calc_Figure_02_04_001\" class=\"wp-caption aligncenter\">\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203502\/CNX_Calc_Figure_02_04_001.jpg\" alt=\"A graph of an increasing linear function f(x) which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. A point a greater than zero is marked on the x axis. The point on the function f(x) above a is an open circle; the function is not defined at a.\" width=\"325\" height=\"277\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The function [latex]f(x)[\/latex] is not continuous at a because [latex]f(a)[\/latex] is undefined.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573750564\">However, as we see in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_002\">(Figure)<\/a>, this condition alone is insufficient to guarantee continuity at the point [latex]a[\/latex]. Although [latex]f(a)[\/latex] is defined, the function has a gap at [latex]a[\/latex]. In this example, the gap exists because [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] does not exist. We must add another condition for continuity at [latex]a[\/latex]\u2014namely,<\/p>\n<div id=\"fs-id1170570997548\" class=\"equation unnumbered\">[latex]\\text{ii.}\\underset{x\\to a}{\\text{lim}}f(x)\\text{exists.}[\/latex]<\/div>\n<div id=\"CNX_Calc_Figure_02_04_002\" class=\"wp-caption aligncenter\">\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203504\/CNX_Calc_Figure_02_04_002.jpg\" alt=\"The graph of a piecewise function f(x) with two parts. The first part is an increasing linear function that crosses from quadrant three to quadrant one at the origin. A point a greater than zero is marked on the x axis. At fa. on this segment, there is a solid circle. The other segment is also an increasing linear function. It exists in quadrant one for values of x greater than a. At x=a, this segment has an open circle.\" width=\"325\" height=\"277\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The function [latex]f(x)[\/latex] is not continuous at a because [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] does not exist.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573493837\">However, as we see in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_003\">(Figure)<\/a>, these two conditions by themselves do not guarantee continuity at a point. The function in this figure satisfies both of our first two conditions, but is still not continuous at [latex]a[\/latex]. We must add a third condition to our list:<\/p>\n<div id=\"fs-id1170573419105\" class=\"equation unnumbered\">[latex]\\text{iii.}\\underset{x\\to a}{\\text{lim}}f(x)=f(a).[\/latex]<\/div>\n<div id=\"CNX_Calc_Figure_02_04_003\" class=\"wp-caption aligncenter\">\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203506\/CNX_Calc_Figure_02_04_003.jpg\" alt=\"The graph of a piecewise function with two parts. The first part is an increasing linear function that crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. A point a greater than zero is marked on the x axis. At this point, there is an open circle on the linear function. The second part is a point at x=a above the line.\" width=\"325\" height=\"277\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The function [latex]f(x)[\/latex] is not continuous at a because [latex]\\underset{x\\to a}{\\text{lim}}f(x)\\ne f(a).[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573502301\">Now we put our list of conditions together and form a definition of continuity at a point.<\/p>\n<div id=\"fs-id1170571048312\" class=\"textbox key-takeaways\">\n<div class=\"title\">\n<h3>Definition<\/h3>\n<\/div>\n<p id=\"fs-id1170573381317\">A function [latex]f(x)[\/latex] is continuous at a point [latex]a[\/latex] if and only if the following three conditions are satisfied:<\/p>\n<ol id=\"fs-id1170571094915\">\n<li>[latex]f(a)[\/latex] is defined<\/li>\n<li>[latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists<\/li>\n<li>[latex]\\underset{x\\to a}{\\text{lim}}f(x)=f(a)[\/latex]<\/li>\n<\/ol>\n<p id=\"fs-id1170573370662\">A function is discontinuous at a point [latex]a[\/latex] if it fails to be continuous at [latex]a[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1170573281579\">The following procedure can be used to analyze the continuity of a function at a point using this definition.<\/p>\n<div id=\"fs-id1170573398041\" class=\"textbox key-takeaways problem-solving\">\n<h3>Problem-Solving Strategy: Determining Continuity at a Point<\/h3>\n<ol id=\"fs-id1170571103215\">\n<li>Check to see if [latex]f(a)[\/latex] is defined. If [latex]f(a)[\/latex] is undefined, we need go no further. The function is not continuous at [latex]a[\/latex]. If [latex]f(a)[\/latex] is defined, continue to step 2.<\/li>\n<li>Compute [latex]\\underset{x\\to a}{\\text{lim}}f(x).[\/latex] In some cases, we may need to do this by first computing [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)[\/latex] and [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x).[\/latex] If [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] does not exist (that is, it is not a real number), then the function is not continuous at [latex]a[\/latex] and the problem is solved. If [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists, then continue to step 3.<\/li>\n<li>Compare [latex]f(a)[\/latex] and [latex]\\underset{x\\to a}{\\text{lim}}f(x).[\/latex] If [latex]\\underset{x\\to a}{\\text{lim}}f(x)\\ne f(a),[\/latex] then the function is not continuous at [latex]a[\/latex]. If [latex]\\underset{x\\to a}{\\text{lim}}f(x)=f(a),[\/latex] then the function is continuous at [latex]a[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p id=\"fs-id1170573570770\">The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given point. These examples illustrate situations in which each of the conditions for continuity in the definition succeed or fail.<\/p>\n<div id=\"fs-id1170573442080\" class=\"textbox examples\">\n<h3>Determining Continuity at a Point, Condition 1<\/h3>\n<div id=\"fs-id1170573368010\" class=\"exercise\">\n<div id=\"fs-id1170573411938\" class=\"textbox\">\n<p id=\"fs-id1170573258704\">Using the definition, determine whether the function [latex]f(x)=({x}^{2}-4)\\text{\/}(x-2)[\/latex] is continuous at [latex]x=2.[\/latex] Justify the conclusion.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573331408\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573331408\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573331408\">Let\u2019s begin by trying to calculate [latex]f(2).[\/latex] We can see that [latex]f(2)=0\\text{\/}0,[\/latex] which is undefined. Therefore, [latex]f(x)=\\frac{{x}^{2}-4}{x-2}[\/latex] is discontinuous at 2 because [latex]f(2)[\/latex] is undefined. The graph of [latex]f(x)[\/latex] is shown in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_004\">(Figure)<\/a>.<\/p>\n<div id=\"CNX_Calc_Figure_02_04_004\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203508\/CNX_Calc_Figure_02_04_004.jpg\" alt=\"A graph of the given function. There is a line which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. At a point in quadrant one, there is an open circle where the function is not defined.\" width=\"487\" height=\"425\" \/> Figure 4. The function [latex]f(x)[\/latex] is discontinuous at 2 because [latex]f(2)[\/latex] is undefined. <\/div>\n<\/div>\n<p>[\/caption]<\/p><\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573389760\" class=\"textbox examples\">\n<h3>Determining Continuity at a Point, Condition 2<\/h3>\n<div id=\"fs-id1170571098221\" class=\"exercise\">\n<div id=\"fs-id1170573426610\" class=\"textbox\">\n<p id=\"fs-id1170573370721\">Using the definition, determine whether the function [latex]f(x)=\\bigg\\{\\begin{array}{ll}-{x}^{2}+4& \\text{ if }x\\le 3\\hfill \\\\ 4x-8& \\text{ if }x>3\\hfill \\end{array}[\/latex] is continuous at [latex]x=3.[\/latex] Justify the conclusion.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573349668\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573349668\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573349668\">Let\u2019s begin by trying to calculate [latex]f(3).[\/latex]<\/p>\n<div id=\"fs-id1170573405058\" class=\"equation unnumbered\">[latex]f(3)=-({3}^{2})+4=-5.[\/latex]<\/div>\n<p id=\"fs-id1170573307347\">Thus, [latex]f(3)[\/latex] is defined. Next, we calculate [latex]\\underset{x\\to 3}{\\text{lim}}f(x).[\/latex] To do this, we must compute [latex]\\underset{x\\to {3}^{-}}{\\text{lim}}f(x)[\/latex] and [latex]\\underset{x\\to {3}^{+}}{\\text{lim}}f(x)\\text{:}[\/latex]<\/p>\n<div class=\"equation unnumbered\">[latex]\\underset{x\\to {3}^{-}}{\\text{lim}}f(x)=-({3}^{2})+4=-5[\/latex]<\/div>\n<p id=\"fs-id1170573400834\">and<\/p>\n<div class=\"equation unnumbered\">[latex]\\underset{x\\to {3}^{+}}{\\text{lim}}f(x)=4(3)-8=4.[\/latex]<\/div>\n<p id=\"fs-id1170573414763\">Therefore, [latex]\\underset{x\\to 3}{\\text{lim}}f(x)[\/latex] does not exist. Thus, [latex]f(x)[\/latex] is not continuous at 3. The graph of [latex]f(x)[\/latex] is shown in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_005\">(Figure)<\/a>.<\/p>\n<div id=\"CNX_Calc_Figure_02_04_005\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203511\/CNX_Calc_Figure_02_04_005.jpg\" alt=\"A graph of the given piecewise function, which has two parts. The first is a downward opening parabola which is symmetric about the y axis. Its vertex is on the y axis, greater than zero. There is a closed circle on the parabola for x=3. The second part is an increasing linear function in the first quadrant, which exists for values of x &gt; 3. There is an open circle at the end of the line where x would be 3.\" width=\"487\" height=\"575\" \/> Figure 5. The function [latex]f(x)[\/latex] is not continuous at 3 because [latex]\\underset{x\\to 3}{\\text{lim}}f(x)[\/latex] does not exist. <\/div>\n<\/div>\n<p>[\/caption]<\/p><\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573429945\" class=\"textbox examples\">\n<h3>Determining Continuity at a Point, Condition 3<\/h3>\n<div id=\"fs-id1170571048204\" class=\"exercise\">\n<div id=\"fs-id1170573268754\" class=\"textbox\">\n<p id=\"fs-id1170573502415\">Using the definition, determine whether the function [latex]f(x)=\\bigg\\{\\begin{array}{ll}\\frac{ \\sin x}{x}\\hfill & \\text{ if }x\\ne 0\\hfill \\\\ \\hfill 1& \\text{ if }x=0\\hfill \\end{array}[\/latex] is continuous at [latex]x=0.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573382461\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573382461\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573382461\">First, observe that<\/p>\n<div id=\"fs-id1170573569453\" class=\"equation unnumbered\">[latex]f(0)=1.[\/latex]<\/div>\n<p id=\"fs-id1170573367184\">Next,<\/p>\n<div id=\"fs-id1170573442773\" class=\"equation unnumbered\">[latex]\\underset{x\\to 0}{\\text{lim}}f(x)=\\underset{x\\to 0}{\\text{lim}}\\frac{ \\sin x}{x}=1.[\/latex]<\/div>\n<p id=\"fs-id1170573394612\">Last, compare [latex]f(0)[\/latex] and [latex]\\underset{x\\to 1}{\\text{lim}}f(x).[\/latex] We see that<\/p>\n<div id=\"fs-id1170573332534\" class=\"equation unnumbered\">[latex]f(0)=1=\\underset{x\\to 0}{\\text{lim}}f(x).[\/latex]<\/div>\n<p id=\"fs-id1170573394599\">Since all three of the conditions in the definition of continuity are satisfied, [latex]f(x)[\/latex] is continuous at [latex]x=0.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170570976463\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1170573288457\" class=\"exercise\">\n<div id=\"fs-id1170573435480\" class=\"textbox\">\n<p id=\"fs-id1170573406319\">Using the definition, determine whether the function [latex]f(x)=\\Bigg\\{\\begin{array}{cc}2x+1\\hfill & \\text{ if }x<1\\hfill \\\\ \\hfill 2& \\text{ if }x=1\\hfill \\\\ -x+4\\hfill & \\text{ if }x>1\\hfill \\end{array}[\/latex] is continuous at [latex]x=1.[\/latex] If the function is not continuous at 1, indicate the condition for continuity at a point that fails to hold.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573402451\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573402451\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573402451\">[latex]f[\/latex] is not continuous at 1 because [latex]f(1)=2\\ne 3=\\underset{x\\to 1}{\\text{lim}}f(x).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1170572589102\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1170573418814\">Check each condition of the definition.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573513079\">By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can state the following theorem.<\/p>\n<div id=\"fs-id1170573430304\" class=\"textbox key-takeaways theorem\">\n<h3>Continuity of Polynomials and Rational Functions<\/h3>\n<p id=\"fs-id1170573255225\">Polynomials and rational functions are continuous at every point in their domains.<\/p>\n<\/div>\n<div id=\"fs-id1170573429973\" class=\"bc-section section\">\n<h2>Proof<\/h2>\n<p id=\"fs-id1170573326736\">Previously, we showed that if [latex]p(x)[\/latex] and [latex]q(x)[\/latex] are polynomials, [latex]\\underset{x\\to a}{\\text{lim}}p(x)=p(a)[\/latex] for every polynomial [latex]p(x)[\/latex] and [latex]\\underset{x\\to a}{\\text{lim}}\\frac{p(x)}{q(x)}=\\frac{p(a)}{q(a)}[\/latex] as long as [latex]q(a)\\ne 0.[\/latex] Therefore, polynomials and rational functions are continuous on their domains.<\/p>\n<p id=\"fs-id1170573276680\">\u25a1<\/p>\n<p id=\"fs-id1170573335424\">We now apply <a class=\"autogenerated-content\" href=\"#fs-id1170573430304\">(Figure)<\/a> to determine the points at which a given rational function is continuous.<\/p>\n<div id=\"fs-id1170573381194\" class=\"textbox examples\">\n<h3>Continuity of a Rational Function<\/h3>\n<div id=\"fs-id1170573326117\" class=\"exercise\">\n<div id=\"fs-id1170573307687\" class=\"textbox\">\n<p id=\"fs-id1170573365757\">For what values of [latex]x[\/latex] is [latex]f(x)=\\frac{x+1}{x-5}[\/latex] continuous?<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573397031\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573397031\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573397031\">The rational function [latex]f(x)=\\frac{x+1}{x-5}[\/latex] is continuous for every value of [latex]x[\/latex] except [latex]x=5.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573434500\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1170573427664\" class=\"exercise\">\n<div id=\"fs-id1170573400271\" class=\"textbox\">\n<p id=\"fs-id1170570999873\">For what values of [latex]x[\/latex] is [latex]f(x)=3{x}^{4}-4{x}^{2}[\/latex] continuous?<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573246209\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573246209\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573246209\">[latex]f(x)[\/latex] is continuous at every real number.<\/p>\n<\/div>\n<div id=\"fs-id1170572209021\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1170573431750\">Use <a class=\"autogenerated-content\" href=\"#fs-id1170573430304\">(Figure)<\/a><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573352138\" class=\"bc-section section\">\n<h1>Types of Discontinuities<\/h1>\n<p id=\"fs-id1170573435662\">As we have seen in <a class=\"autogenerated-content\" href=\"#fs-id1170573442080\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#fs-id1170573389760\">(Figure)<\/a>, discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a<strong> removable discontinuity<\/strong> is a discontinuity for which there is a hole in the graph, a<strong> jump discontinuity<\/strong> is a noninfinite discontinuity for which the sections of the function do not meet up, and an <strong>infinite discontinuity<\/strong> is a discontinuity located at a vertical asymptote. <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_006\">(Figure)<\/a> illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories.<\/p>\n<div id=\"CNX_Calc_Figure_02_04_006\" class=\"wp-caption aligncenter\">\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203514\/CNX_Calc_Figure_02_04_006.jpg\" alt=\"Three graphs, each showing a different discontinuity. The first is removable discontinuity. Here, the given function is a line with positive slope. At a point x=a, where a&gt;0, there is an open circle on the line and a closed circle a few units above the line. The second is a jump discontinuity. Here, there are two lines with positive slope. The first line exists for x&lt;=a, and the second exists for x&gt;a, where a&gt;0. The first line ends at a solid circle where x=a, and the second begins a few units up with an open circle at x=a. The third discontinuity type is infinite discontinuity. Here, the function has two parts separated by an asymptote x=a. The first segment is a curve stretching along the x axis to 0 as x goes to negative infinity and along the y axis to infinity as x goes to zero. The second segment is a curve stretching along the y axis to negative infinity as x goes to zero and along the x axis to 0 as x goes to infinity.\" width=\"975\" height=\"315\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. Discontinuities are classified as (a) removable, (b) jump, or (c) infinite.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573400557\">These three discontinuities are formally defined as follows:<\/p>\n<div id=\"fs-id1170573419172\" class=\"textbox key-takeaways\">\n<div class=\"title\">\n<h3>Definition<\/h3>\n<\/div>\n<p>If [latex]f(x)[\/latex] is discontinuous at [latex]a[\/latex], then<\/p>\n<ol id=\"fs-id1170573216342\">\n<li>[latex]f[\/latex] has a <strong>removable discontinuity<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists. (Note: When we state that [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists, we mean that [latex]\\underset{x\\to a}{\\text{lim}}f(x)=L,[\/latex] where <em>L<\/em> is a real number.)<\/li>\n<li>[latex]f[\/latex] has a <strong>jump discontinuity<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)[\/latex] and [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)[\/latex] both exist, but [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)\\ne \\underset{x\\to {a}^{+}}{\\text{lim}}f(x).[\/latex] (Note: When we state that [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)[\/latex] and [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)[\/latex] both exist, we mean that both are real-valued and that neither take on the values \u00b1\u221e.)<\/li>\n<li>[latex]f[\/latex] has an <strong>infinite discontinuity<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)=\\text{\u00b1}\\infty[\/latex] or [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)=\\text{\u00b1}\\infty .[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1170570976348\" class=\"textbox examples\">\n<h3>Classifying a Discontinuity<\/h3>\n<div id=\"fs-id1170573248707\" class=\"exercise\">\n<div id=\"fs-id1170573399643\" class=\"textbox\">\n<p id=\"fs-id1170573424344\">In <a class=\"autogenerated-content\" href=\"#fs-id1170573442080\">(Figure)<\/a>, we showed that [latex]f(x)=\\frac{{x}^{2}-4}{x-2}[\/latex] is discontinuous at [latex]x=2.[\/latex] Classify this discontinuity as removable, jump, or infinite.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573402024\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573402024\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573402024\">To classify the discontinuity at 2 we must evaluate [latex]\\underset{x\\to 2}{\\text{lim}}f(x)\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1170571103554\" class=\"equation unnumbered\">[latex]\\begin{array}{cc}\\hfill \\underset{x\\to 2}{\\text{lim}}f(x)& =\\underset{x\\to 2}{\\text{lim}}\\frac{{x}^{2}-4}{x-2}\\hfill \\\\ & =\\underset{x\\to 2}{\\text{lim}}\\frac{(x-2)(x+2)}{x-2}\\hfill \\\\ & =\\underset{x\\to 2}{\\text{lim}}(x+2)\\hfill \\\\ & =4.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571136107\">Since [latex]f[\/latex] is discontinuous at 2 and [latex]\\underset{x\\to 2}{\\text{lim}}f(x)[\/latex] exists, [latex]f[\/latex] has a removable discontinuity at [latex]x=2.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573426544\" class=\"textbox examples\">\n<h3>Classifying a Discontinuity<\/h3>\n<div id=\"fs-id1170573507547\" class=\"exercise\">\n<div id=\"fs-id1170573493638\" class=\"textbox\">\n<p id=\"fs-id1170573586256\">In <a class=\"autogenerated-content\" href=\"#fs-id1170573389760\">(Figure)<\/a>, we showed that [latex]f(x)=\\bigg\\{\\begin{array}{cc}-{x}^{2}+4& \\text{ if }x\\le 3\\hfill \\\\ 4x-8& \\text{ if }x>3\\hfill \\end{array}[\/latex] is discontinuous at [latex]x=3.[\/latex] Classify this discontinuity as removable, jump, or infinite.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571095481\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571095481\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571095481\">Earlier, we showed that [latex]f[\/latex] is discontinuous at 3 because [latex]\\underset{x\\to 3}{\\text{lim}}f(x)[\/latex] does not exist. However, since [latex]\\underset{x\\to {3}^{-}}{\\text{lim}}f(x)=-5[\/latex] and [latex]\\underset{x\\to {3}^{-}}{\\text{lim}}f(x)=4[\/latex] both exist, we conclude that the function has a jump discontinuity at 3.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573355402\" class=\"textbox examples\">\n<div id=\"fs-id1170573750481\" class=\"exercise\">\n<div class=\"textbox\">\n<h3>Classifying a Discontinuity<\/h3>\n<p id=\"fs-id1170573732461\">Determine whether [latex]f(x)=\\frac{x+2}{x+1}[\/latex] is continuous at \u22121. If the function is discontinuous at \u22121, classify the discontinuity as removable, jump, or infinite.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571000111\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571000111\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571000111\">The function value [latex]f(-1)[\/latex] is undefined. Therefore, the function is not continuous at \u22121. To determine the type of discontinuity, we must determine the limit at \u22121. We see that [latex]\\underset{x\\to {-1}^{-}}{\\text{lim}}\\frac{x+2}{x+1}=\\text{\u2212}\\infty[\/latex] and [latex]\\underset{x\\to {-1}^{+}}{\\text{lim}}\\frac{x+2}{x+1}=\\text{+}\\infty .[\/latex] Therefore, the function has an infinite discontinuity at \u22121.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571047778\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1170570998828\" class=\"exercise\">\n<div id=\"fs-id1170573351595\" class=\"textbox\">\n<p id=\"fs-id1170573361594\">For [latex]f(x)=\\bigg\\{\\begin{array}{cc}{x}^{2}\\hfill & \\text{ if }x\\ne 1\\hfill \\\\ \\hfill 3& \\text{ if }x=1\\hfill \\end{array},[\/latex] decide whether [latex]f[\/latex] is continuous at 1. If [latex]f[\/latex] is not continuous at 1, classify the discontinuity as removable, jump, or infinite.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573355512\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573355512\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573355512\">Discontinuous at 1; removable<\/p>\n<\/div>\n<div id=\"fs-id1170571648706\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1170573436674\">Follow the steps in <a class=\"autogenerated-content\" href=\"#fs-id1170573398041\">(Figure)<\/a>. If the function is discontinuous at 1, look at [latex]\\underset{x\\to 1}{\\text{lim}}f(x)[\/latex] and use the definition to determine the type of discontinuity.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573494299\" class=\"bc-section section\">\n<h1>Continuity over an Interval<\/h1>\n<p id=\"fs-id1170573612125\">Now that we have explored the concept of continuity at a point, we extend that idea to <strong>continuity over an interval<\/strong>. As we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function is continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a point.<\/p>\n<div id=\"fs-id1170573581958\" class=\"textbox key-takeaways\">\n<h3>Continuity from the Right and from the Left<\/h3>\n<p id=\"fs-id1170573352232\">A function [latex]f(x)[\/latex] is said to be <strong>continuous from the right<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)=f(a).[\/latex]<\/p>\n<p id=\"fs-id1170571120467\">A function [latex]f(x)[\/latex] is said to be<strong> continuous from the left<\/strong> at [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)=f(a).[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1170570967081\">A function is continuous over an open interval if it is continuous at every point in the interval. A function [latex]f(x)[\/latex] is continuous over a closed interval of the form [latex]\\left[a,b\\right][\/latex] if it is continuous at every point in [latex](a,b)[\/latex] and is continuous from the right at [latex]a[\/latex] and is continuous from the left at [latex]b[\/latex]. Analogously, a function [latex]f(x)[\/latex] is continuous over an interval of the form [latex](a,b\\right][\/latex] if it is continuous over [latex](a,b)[\/latex] and is continuous from the left at [latex]b[\/latex]. Continuity over other types of intervals are defined in a similar fashion.<\/p>\n<p id=\"fs-id1170573533852\">Requiring that [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)=f(a)[\/latex] and [latex]\\underset{x\\to {b}^{-}}{\\text{lim}}f(x)=f(b)[\/latex] ensures that we can trace the graph of the function from the point [latex](a,f(a))[\/latex] to the point [latex](b,f(b))[\/latex] without lifting the pencil. If, for example, [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)\\ne f(a),[\/latex] we would need to lift our pencil to jump from [latex]f(a)[\/latex] to the graph of the rest of the function over [latex](a,b\\right].[\/latex]<\/p>\n<div id=\"fs-id1170573395559\" class=\"textbox examples\">\n<h3>Continuity on an Interval<\/h3>\n<div id=\"fs-id1170573395562\" class=\"exercise\">\n<div id=\"fs-id1170571096791\" class=\"textbox\">\n<p id=\"fs-id1170570990867\">State the interval(s) over which the function [latex]f(x)=\\frac{x-1}{{x}^{2}+2x}[\/latex] is continuous.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573426588\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573426588\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573426588\">Since [latex]f(x)=\\frac{x-1}{{x}^{2}+2x}[\/latex] is a rational function, it is continuous at every point in its domain. The domain of [latex]f(x)[\/latex] is the set [latex](\\text{\u2212}\\infty ,-2)\\cup (-2,0)\\cup (0,\\text{+}\\infty ).[\/latex] Thus, [latex]f(x)[\/latex] is continuous over each of the intervals [latex](\\text{\u2212}\\infty ,-2),(-2,0),[\/latex] and [latex](0,\\text{+}\\infty ).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573387892\" class=\"textbox examples\">\n<h3>Continuity over an Interval<\/h3>\n<div id=\"fs-id1170573387894\" class=\"exercise\">\n<div id=\"fs-id1170573404298\" class=\"textbox\">\n<p id=\"fs-id1170573406820\">State the interval(s) over which the function [latex]f(x)=\\sqrt{4-{x}^{2}}[\/latex] is continuous.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573439420\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573439420\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573439420\">From the limit laws, we know that [latex]\\underset{x\\to a}{\\text{lim}}\\sqrt[]{4-{x}^{2}}=\\sqrt{4-{a}^{2}}[\/latex] for all values of [latex]a[\/latex] in [latex](-2,2).[\/latex] We also know that [latex]\\underset{x\\to {-2}^{+}}{\\text{lim}}\\sqrt{4-{x}^{2}}=0[\/latex] exists and [latex]\\underset{x\\to {2}^{-}}{\\text{lim}}\\sqrt{4-{x}^{2}}=0[\/latex] exists. Therefore, [latex]f(x)[\/latex] is continuous over the interval [latex]\\left[-2,2\\right].[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573403253\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1170573361460\" class=\"exercise\">\n<div id=\"fs-id1170573361462\" class=\"textbox\">\n<p id=\"fs-id1170571120474\">State the interval(s) over which the function [latex]f(x)=\\sqrt{x+3}[\/latex] is continuous.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571287079\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571287079\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571287079\">[latex]\\left[-3,\\text{+}\\infty )[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1170572542876\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1170570999934\">Use <a class=\"autogenerated-content\" href=\"#fs-id1170573387892\">(Figure)<\/a> as a guide for solving.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573361765\">The <a class=\"autogenerated-content\" href=\"#fs-id1170573352212\">(Figure)<\/a> allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.<\/p>\n<div id=\"fs-id1170573352212\" class=\"textbox key-takeaways theorem\">\n<h3>Composite Function Theorem<\/h3>\n<p id=\"fs-id1170573762934\">If [latex]f(x)[\/latex] is continuous at <em>L<\/em> and [latex]\\underset{x\\to a}{\\text{lim}}g(x)=L,[\/latex] then<\/p>\n<div id=\"fs-id1170573400410\" class=\"equation unnumbered\">[latex]\\underset{x\\to a}{\\text{lim}}f(g(x))=f(\\underset{x\\to a}{\\text{lim}}g(x))=f(L).[\/latex]<\/div>\n<\/div>\n<p id=\"fs-id1170573573603\">Before we move on to <a class=\"autogenerated-content\" href=\"#fs-id1170573718134\">(Figure)<\/a>, recall that earlier, in the section on limit laws, we showed [latex]\\underset{x\\to 0}{\\text{lim}} \\cos x=1= \\cos (0).[\/latex] Consequently, we know that [latex]f(x)= \\cos x[\/latex] is continuous at 0. In <a class=\"autogenerated-content\" href=\"#fs-id1170573718134\">(Figure)<\/a> we see how to combine this result with the composite function theorem.<\/p>\n<div id=\"fs-id1170573718134\" class=\"textbox examples\">\n<h3>Limit of a Composite Cosine Function<\/h3>\n<div id=\"fs-id1170570998131\" class=\"exercise\">\n<div id=\"fs-id1170570998133\" class=\"textbox\">\n<p>Evaluate [latex]\\underset{x\\to \\pi \\text{\/}2}{\\text{lim}} \\cos (x-\\frac{\\pi }{2}).[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573408578\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573408578\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573408578\">The given function is a composite of [latex]\\cos x[\/latex] and [latex]x-\\frac{\\pi }{2}.[\/latex] Since [latex]\\underset{x\\to \\pi \\text{\/}2}{\\text{lim}}(x-\\frac{\\pi }{2})=0[\/latex] and [latex]\\cos x[\/latex] is continuous at 0, we may apply the composite function theorem. Thus,<\/p>\n<div id=\"fs-id1170573570975\" class=\"equation unnumbered\">[latex]\\underset{x\\to \\pi \\text{\/}2}{\\text{lim}} \\cos (x-\\frac{\\pi }{2})= \\cos (\\underset{x\\to \\pi \\text{\/}2}{\\text{lim}}(x-\\frac{\\pi }{2}))= \\cos (0)=1.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573732417\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1170573732420\" class=\"exercise\">\n<div id=\"fs-id1170571131879\" class=\"textbox\">\n<p id=\"fs-id1170571131881\">Evaluate [latex]\\underset{x\\to \\pi }{\\text{lim}} \\sin (x-\\pi ).[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573362583\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573362583\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573362583\">0<\/p>\n<\/div>\n<div id=\"fs-id1170571623332\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1170573359398\">[latex]f(x)= \\sin x[\/latex] is continuous at 0. Use <a class=\"autogenerated-content\" href=\"#fs-id1170573718134\">(Figure)<\/a> as a guide.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571101566\">The proof of the next theorem uses the composite function theorem as well as the continuity of [latex]f(x)= \\sin x[\/latex] and [latex]g(x)= \\cos x[\/latex] at the point 0 to show that trigonometric functions are continuous over their entire domains.<\/p>\n<div id=\"fs-id1170573418920\" class=\"textbox key-takeaways theorem\">\n<h3>Continuity of Trigonometric Functions<\/h3>\n<p id=\"fs-id1170573512311\">Trigonometric functions are continuous over their entire domains.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573512316\" class=\"bc-section section\">\n<h1>Proof<\/h1>\n<p id=\"fs-id1170573429466\">We begin by demonstrating that [latex]\\cos x[\/latex] is continuous at every real number. To do this, we must show that [latex]\\underset{x\\to a}{\\text{lim}} \\cos x= \\cos a[\/latex] for all values of [latex]a[\/latex].<\/p>\n<p id=\"fs-id1170571099777\">[latex]\\begin{array}{ccccc}\\underset{x\\to a}{\\text{lim}} \\cos x\\hfill & =\\underset{x\\to a}{\\text{lim}} \\cos ((x-a)+a)\\hfill & & & \\text{rewrite}x=x-a+a\\hfill \\\\ & =\\underset{x\\to a}{\\text{lim}}( \\cos (x-a) \\cos a- \\sin (x-a) \\sin a)\\hfill & & & \\text{apply the identity for the cosine of the sum of two angles}\\hfill \\\\ & = \\cos (\\underset{x\\to a}{\\text{lim}}(x-a)) \\cos a- \\sin (\\underset{x\\to a}{\\text{lim}}(x-a)) \\sin a\\hfill & & & \\underset{x\\to a}{\\text{lim}}(x-a)=0,\\text{ and } \\sin x\\text{ and } \\cos x\\text{are continuous at 0}\\hfill \\\\ & = \\cos (0) \\cos a- \\sin (0) \\sin a\\hfill & & & \\text{evaluate cos(0) and sin(0) and simplify}\\hfill \\\\ & =1\u00b7 \\cos a-0\u00b7 \\sin a= \\cos a.\\hfill \\end{array}[\/latex]<\/p>\n<p id=\"fs-id1170571216459\">The proof that [latex]\\sin x[\/latex] is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of [latex]\\sin x[\/latex] and [latex]\\cos x,[\/latex] their continuity follows from the quotient limit law.<\/p>\n<p id=\"fs-id1170573370281\">\u25a1<\/p>\n<p id=\"fs-id1170573370285\">As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions. As we continue our study of calculus, we revisit this theorem many times.<\/p>\n<\/div>\n<div id=\"fs-id1170571100154\" class=\"bc-section section\">\n<h1>The Intermediate Value Theorem<\/h1>\n<p id=\"fs-id1170573717714\">Functions that are continuous over intervals of the form [latex]\\left[a,b\\right],[\/latex] where [latex]a[\/latex] and [latex]b[\/latex] are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the <strong>Intermediate Value Theorem<\/strong>.<\/p>\n<div id=\"fs-id1170571120551\" class=\"textbox key-takeaways theorem\">\n<h3>The Intermediate Value Theorem<\/h3>\n<p id=\"fs-id1170573534418\">Let [latex]f[\/latex] be continuous over a closed, bounded interval [latex]\\left[a,b\\right].[\/latex] If [latex]z[\/latex] is any real number between [latex]f(a)[\/latex] and [latex]f(b),[\/latex] then there is a number [latex]c[\/latex] in [latex]\\left[a,b\\right][\/latex] satisfying [latex]f(c)=z[\/latex] in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_02_04_007\">(Figure)<\/a>.<\/p>\n<div id=\"CNX_Calc_Figure_02_04_007\" class=\"wp-caption aligncenter\">\n<div style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203518\/CNX_Calc_Figure_02_04_007.jpg\" alt=\"A diagram illustrating the intermediate value theorem. There is a generic continuous curved function shown over the interval [a,b]. The points fa. and fb. are marked, and dotted lines are drawn from a, b, fa., and fb. to the points (a, fa.) and (b, fb.). A third point, c, is plotted between a and b. Since the function is continuous, there is a value for fc. along the curve, and a line is drawn from c to (c, fc.) and from (c, fc.) to fc., which is labeled as z on the y axis.\" width=\"417\" height=\"385\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. There is a number [latex]c\\in \\left[a,b\\right][\/latex] that satisfies [latex]f(c)=z.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571120881\" class=\"textbox examples\">\n<h3>Application of the Intermediate Value Theorem<\/h3>\n<div id=\"fs-id1170571120883\" class=\"exercise\">\n<div id=\"fs-id1170570997452\" class=\"textbox\">\n<p id=\"fs-id1170570997458\">Show that [latex]f(x)=x- \\cos x[\/latex] has at least one zero.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571136342\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571136342\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571136342\">Since [latex]f(x)=x- \\cos x[\/latex] is continuous over [latex](\\text{\u2212}\\infty ,\\text{+}\\infty ),[\/latex] it is continuous over any closed interval of the form [latex]\\left[a,b\\right].[\/latex] If you can find an interval [latex]\\left[a,b\\right][\/latex] such that [latex]f(a)[\/latex] and [latex]f(b)[\/latex] have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number [latex]c[\/latex] in [latex](a,b)[\/latex] that satisfies [latex]f(c)=0.[\/latex] Note that<\/p>\n<div id=\"fs-id1170571239008\" class=\"equation unnumbered\">[latex]f(0)=0- \\cos (0)=-1<0[\/latex]<\/div>\n<p id=\"fs-id1170573424328\">and<\/p>\n<div id=\"fs-id1170573424331\" class=\"equation unnumbered\">[latex]f(\\frac{\\pi }{2})=\\frac{\\pi }{2}- \\cos \\frac{\\pi }{2}=\\frac{\\pi }{2}>0.[\/latex]<\/div>\n<p id=\"fs-id1170573753266\">Using the Intermediate Value Theorem, we can see that there must be a real number [latex]c[\/latex] in [latex]\\left[0,\\pi \\text{\/}2\\right][\/latex] that satisfies [latex]f(c)=0.[\/latex] Therefore, [latex]f(x)=x- \\cos x[\/latex] has at least one zero.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573439386\" class=\"textbox examples\">\n<h3>When Can You Apply the Intermediate Value Theorem?<\/h3>\n<div id=\"fs-id1170573439388\" class=\"exercise\">\n<div id=\"fs-id1170573413593\" class=\"textbox\">\n<p id=\"fs-id1170573413598\">If [latex]f(x)[\/latex] is continuous over [latex]\\left[0,2\\right],f(0)>0[\/latex] and [latex]f(2)>0,[\/latex] can we use the Intermediate Value Theorem to conclude that [latex]f(x)[\/latex] has no zeros in the interval [latex]\\left[0,2\\right]\\text{?}[\/latex] Explain.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571262087\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571262087\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571262087\">No. The Intermediate Value Theorem only allows us to conclude that we can find a value between [latex]f(0)[\/latex] and [latex]f(2);[\/latex] it doesn\u2019t allow us to conclude that we can\u2019t find other values. To see this more clearly, consider the function [latex]f(x)={(x-1)}^{2}.[\/latex] It satisfies [latex]f(0)=1>0,f(2)=1>0,[\/latex] and [latex]f(1)=0.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170570996536\" class=\"textbox examples\">\n<h3>When Can You Apply the Intermediate Value Theorem?<\/h3>\n<div id=\"fs-id1170573586870\" class=\"exercise\">\n<div id=\"fs-id1170573586872\" class=\"textbox\">\n<p id=\"fs-id1170573586819\">For [latex]f(x)=1\\text{\/}x,f(-1)=-1<0[\/latex] and [latex]f(1)=1>0.[\/latex] Can we conclude that [latex]f(x)[\/latex] has a zero in the interval [latex]\\left[-1,1\\right]?[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571138880\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571138880\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571138880\">No. The function is not continuous over [latex]\\left[-1,1\\right].[\/latex] The Intermediate Value Theorem does not apply here.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571100300\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1170571100303\" class=\"exercise\">\n<div id=\"fs-id1170571048764\" class=\"textbox\">\n<p id=\"fs-id1170571048766\">Show that [latex]f(x)={x}^{3}-{x}^{2}-3x+1[\/latex] has a zero over the interval [latex]\\left[0,1\\right].[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573382710\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573382710\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573382710\">[latex]f(0)=1>0,f(1)=-2<0;f(x)[\/latex] is continuous over [latex]\\left[0,1\\right].[\/latex] It must have a zero on this interval.<\/p>\n<\/div>\n<div id=\"fs-id1170572203444\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1170573569887\">Find [latex]f(0)[\/latex] and [latex]f(1).[\/latex] Apply the Intermediate Value Theorem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573391172\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1170573574290\">\n<li>For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point.<\/li>\n<li>Discontinuities may be classified as removable, jump, or infinite.<\/li>\n<li>A function is continuous over an open interval if it is continuous at every point in the interval. It is continuous over a closed interval if it is continuous at every point in its interior and is continuous at its endpoints.<\/li>\n<li>The composite function theorem states: If [latex]f(x)[\/latex] is continuous at <em>L<\/em> and [latex]\\underset{x\\to a}{\\text{lim}}g(x)=L,[\/latex] then [latex]\\underset{x\\to a}{\\text{lim}}f(g(x))=f(\\underset{x\\to a}{\\text{lim}}g(x))=f(L).[\/latex]<\/li>\n<li>The Intermediate Value Theorem guarantees that if a function is continuous over a closed interval, then the function takes on every value between the values at its endpoints.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1170573397457\" class=\"textbox exercises\">\n<p id=\"fs-id1170573397460\">For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.<\/p>\n<div id=\"fs-id1170571246283\" class=\"exercise\">\n<div id=\"fs-id1170571246285\" class=\"textbox\">\n<p id=\"fs-id1170571246287\">[latex]f(x)=\\frac{1}{\\sqrt{x}}[\/latex]<\/p>\n<\/div>\n<div class=\"solution\">\n<p id=\"fs-id1170571301576\">The function is defined for all [latex]x[\/latex] in the interval [latex](0,\\infty ).[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571103211\" class=\"exercise\">\n<div id=\"fs-id1170570973770\" class=\"textbox\">\n<p id=\"fs-id1170570973773\">[latex]f(x)=\\frac{2}{{x}^{2}+1}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573367933\" class=\"exercise\">\n<div id=\"fs-id1170573367935\" class=\"textbox\">\n<p id=\"fs-id1170573367937\">[latex]f(x)=\\frac{x}{{x}^{2}-x}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573590405\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573590405\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573590405\">Removable discontinuity at [latex]x=0;[\/latex] infinite discontinuity at [latex]x=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573734927\" class=\"exercise\">\n<div id=\"fs-id1170573734930\" class=\"textbox\">\n<p id=\"fs-id1170573404382\">[latex]g(t)={t}^{-1}+1[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573581616\" class=\"exercise\">\n<div id=\"fs-id1170573581618\" class=\"textbox\">\n<p id=\"fs-id1170573581620\">[latex]f(x)=\\frac{5}{{e}^{x}-2}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573586367\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573586367\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573586367\">Infinite discontinuity at [latex]x=\\text{ln}2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573403108\" class=\"exercise\">\n<div id=\"fs-id1170573403110\" class=\"textbox\">\n<p id=\"fs-id1170573403112\">[latex]f(x)=\\frac{|x-2|}{x-2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573361738\" class=\"exercise\">\n<div id=\"fs-id1170573361740\" class=\"textbox\">\n<p id=\"fs-id1170573361742\">[latex]H(x)= \\tan 2x[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571047553\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571047553\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571047553\">Infinite discontinuities at [latex]x=\\frac{(2k+1)\\pi }{4},[\/latex] for [latex]k=0,\u00b11,\u00b12,\u00b13\\text{,\u2026}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573581931\" class=\"exercise\">\n<div id=\"fs-id1170573581933\" class=\"textbox\">\n<p id=\"fs-id1170573408756\">[latex]f(t)=\\frac{t+3}{{t}^{2}+5t+6}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573750406\">For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?<\/p>\n<div id=\"fs-id1170573750411\" class=\"exercise\">\n<div id=\"fs-id1170573750413\" class=\"textbox\">\n<p id=\"fs-id1170573593159\">[latex]\\frac{2{x}^{2}-5x+3}{x-1}[\/latex] at [latex]x=1[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573331459\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573331459\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573331459\">No. It is a removable discontinuity.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170570999796\" class=\"exercise\">\n<div id=\"fs-id1170570999798\" class=\"textbox\">\n<p id=\"fs-id1170570999800\">[latex]h(\\theta )=\\frac{ \\sin \\theta - \\cos \\theta }{ \\tan \\theta }[\/latex] at [latex]\\theta =\\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573580625\" class=\"exercise\">\n<div id=\"fs-id1170573580627\" class=\"textbox\">\n<p id=\"fs-id1170573580629\">[latex]g(u)=\\bigg\\{\\begin{array}{ll}\\frac{6{u}^{2}+u-2}{2u-1}& \\text{ if }u\\ne \\frac{1}{2}\\hfill \\\\ \\frac{7}{2}\\hfill & \\text{ if }u=\\frac{1}{2}\\hfill \\end{array},[\/latex] at [latex]u=\\frac{1}{2}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571120270\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571120270\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571120270\">Yes. It is continuous.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571120275\" class=\"exercise\">\n<div id=\"fs-id1170573590166\" class=\"textbox\">\n<p id=\"fs-id1170573590169\">[latex]f(y)=\\frac{ \\sin (\\pi y)}{ \\tan (\\pi y)},[\/latex] at [latex]y=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571120812\" class=\"exercise\">\n<div id=\"fs-id1170571130844\" class=\"textbox\">\n<p id=\"fs-id1170571130846\">[latex]f(x)=\\bigg\\{\\begin{array}{ll}{x}^{2}-{e}^{x}& \\text{ if }x<0\\hfill \\\\ x-1& \\text{ if }x\\ge 0\\hfill \\end{array},[\/latex] at [latex]x=0[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573381211\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573381211\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573381211\">Yes. It is continuous.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573381217\" class=\"exercise\">\n<div id=\"fs-id1170573381219\" class=\"textbox\">\n<p id=\"fs-id1170571285463\">[latex]f(x)=\\bigg\\{\\begin{array}{l}x \\sin (x)\\text{ if }x\\le \\pi \\\\ x \\tan (x)\\text{ if }x>\\pi \\end{array},[\/latex] at [latex]x=\\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571123088\">In the following exercises, find the value(s) of [latex]k[\/latex] that makes each function continuous over the given interval.<\/p>\n<div id=\"fs-id1170573413991\" class=\"exercise\">\n<div id=\"fs-id1170573413993\" class=\"textbox\">\n<p id=\"fs-id1170571137944\">[latex]f(x)=\\bigg\\{\\begin{array}{cc}3x+2,\\hfill & x<k\\hfill \\\\ 2x-3,\\hfill & k\\le x\\le 8\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573440208\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573440208\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573440208\">[latex]k=-5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571068186\" class=\"exercise\">\n<div id=\"fs-id1170571068188\" class=\"textbox\">\n<p id=\"fs-id1170573760712\">[latex]f(\\theta )=\\bigg\\{\\begin{array}{ll}\\hfill \\sin \\theta ,& 0\\le \\theta <\\frac{\\pi }{2}\\hfill \\\\ \\cos (\\theta +k),\\hfill & \\frac{\\pi }{2}\\le \\theta \\le \\pi \\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573575226\" class=\"exercise\">\n<div id=\"fs-id1170573575228\" class=\"textbox\">\n<p id=\"fs-id1170573575230\">[latex]f(x)=\\bigg\\{\\begin{array}{cc}\\frac{{x}^{2}+3x+2}{x+2},\\hfill & x\\ne -2\\hfill \\\\ \\hfill k,& x=-2\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573502717\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573502717\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573502717\">[latex]k=-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"exercise\">\n<div class=\"textbox\">\n<p id=\"fs-id1170573589735\">[latex]f(x)=\\bigg\\{\\begin{array}{cc}\\hfill {e}^{kx},& 0\\le x<4\\hfill \\\\ x+3,\\hfill & 4\\le x\\le 8\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573449551\" class=\"exercise\">\n<div id=\"fs-id1170573449554\" class=\"textbox\">\n<p id=\"fs-id1170573413580\">[latex]f(x)=\\bigg\\{\\begin{array}{cc}\\hfill \\sqrt{kx},& 0\\le x\\le 3\\hfill \\\\ x+1,\\hfill & 3<x\\le 10\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571050054\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571050054\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571050054\">[latex]k=\\frac{16}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571121702\">In the following exercises, use the Intermediate Value Theorem (IVT).<\/p>\n<div id=\"fs-id1170571121705\" class=\"exercise\">\n<div id=\"fs-id1170573541393\" class=\"textbox\">\n<p id=\"fs-id1170573541395\">Let [latex]h(x)=\\bigg\\{\\begin{array}{ll}3{x}^{2}-4,& x\\le 2\\hfill \\\\ 5+4x,& x>2\\hfill \\end{array}[\/latex] Over the interval [latex]\\left[0,4\\right],[\/latex] there is no value of [latex]x[\/latex] such that [latex]h(x)=10,[\/latex] although [latex]h(0)<10[\/latex] and [latex]h(4)>10.[\/latex] Explain why this does not contradict the IVT.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571087121\" class=\"exercise\">\n<div id=\"fs-id1170571087123\" class=\"textbox\">\n<p id=\"fs-id1170571087125\">A particle moving along a line has at each time [latex]t[\/latex] a position function [latex]s(t),[\/latex] which is continuous. Assume [latex]s(2)=5[\/latex] and [latex]s(5)=2.[\/latex] Another particle moves such that its position is given by [latex]h(t)=s(t)-t.[\/latex] Explain why there must be a value [latex]c[\/latex] for [latex]2<c<5[\/latex] such that [latex]h(c)=0.[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170570998991\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170570998991\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170570998991\">Since both [latex]s[\/latex] and [latex]y=t[\/latex] are continuous everywhere, then [latex]h(t)=s(t)-t[\/latex] is continuous everywhere and, in particular, it is continuous over the closed interval [latex]\\left[2,5\\right].[\/latex] Also, [latex]h(2)=3>0[\/latex] and [latex]h(5)=-3<0.[\/latex] Therefore, by the IVT, there is a value [latex]x=c[\/latex] such that [latex]h(c)=0.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573590374\" class=\"exercise\">\n<div id=\"fs-id1170573590376\" class=\"textbox\">\n<p id=\"fs-id1170573590378\"><strong>[T]<\/strong> Use the statement \u201cThe cosine of [latex]t[\/latex] is equal to [latex]t[\/latex] cubed.\u201d<\/p>\n<ol id=\"fs-id1170571132593\" style=\"list-style-type: lower-alpha\">\n<li>Write a mathematical equation of the statement.<\/li>\n<li>Prove that the equation in part a. has at least one real solution.<\/li>\n<li>Use a calculator to find an interval of length 0.01 that contains a solution.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"fs-id1170570982565\" class=\"exercise\">\n<div id=\"fs-id1170570982567\" class=\"textbox\">\n<p id=\"fs-id1170570982569\">Apply the IVT to determine whether [latex]{2}^{x}={x}^{3}[\/latex] has a solution in one of the intervals [latex]\\left[1.25,1.375\\right][\/latex] or [latex]\\left[1.375,1.5\\right].[\/latex] Briefly explain your response for each interval.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170570998210\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170570998210\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170570998210\">The function [latex]f(x)={2}^{x}-{x}^{3}[\/latex] is continuous over the interval [latex]\\left[1.25,1.375\\right][\/latex] and has opposite signs at the endpoints.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573423919\" class=\"exercise\">\n<div id=\"fs-id1170573423921\" class=\"textbox\">\n<p id=\"fs-id1170573423923\">Consider the graph of the function [latex]y=f(x)[\/latex] shown in the following graph.<\/p>\n<p><span id=\"fs-id1170573581162\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203520\/CNX_Calc_Figure_02_04_201.jpg\" alt=\"A diagram illustrating the intermediate value theorem. There is a generic continuous curved function shown over the interval [a,b]. The points fa. and fb. are marked, and dotted lines are drawn from a, b, fa., and fb. to the points (a, fa.) and (b, fb.). A third point, c, is plotted between a and b. Since the function is continuous, there is a value for fc. along the curve, and a line is drawn from c to (c, fc.) and from (c, fc.) to fc., which is labeled as z on the y axis.\" \/><\/span><\/p>\n<ol id=\"fs-id1170571053591\" style=\"list-style-type: lower-alpha\">\n<li>Find all values for which the function is discontinuous.<\/li>\n<li>For each value in part a., state why the formal definition of continuity does not apply.<\/li>\n<li>Classify each discontinuity as either jump, removable, or infinite.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571290233\" class=\"exercise\">\n<div id=\"fs-id1170571290235\" class=\"textbox\">\n<p id=\"fs-id1170571290237\">Let [latex]f(x)=\\bigg\\{\\begin{array}{c}3x,x>1\\\\ {x}^{3},x<1\\end{array}.[\/latex]<\/p>\n<ol id=\"fs-id1170571101024\" style=\"list-style-type: lower-alpha\">\n<li>Sketch the graph of [latex]f[\/latex].<\/li>\n<li>Is it possible to find a value [latex]k[\/latex] such that [latex]f(1)=k,[\/latex] which makes [latex]f(x)[\/latex] continuous for all real numbers? Briefly explain.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571100273\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571100273\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571100273\">a.<\/p>\n<p><span id=\"fs-id1170571100281\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203523\/CNX_Calc_Figure_02_04_202.jpg\" alt=\"A graph of the given piecewise function containing two segments. The first, x^3, exists for x &lt; 1 and ends with an open circle at (1,1). The second, 3x, exists for x &gt; 1. It beings with an open circle at (1,3).\" \/><\/span><br \/>\nb. It is not possible to redefine [latex]f(1)[\/latex] since the discontinuity is a jump discontinuity.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573359564\" class=\"exercise\">\n<div id=\"fs-id1170573359566\" class=\"textbox\">\n<p id=\"fs-id1170573359568\">Let [latex]f(x)=\\frac{{x}^{4}-1}{{x}^{2}-1}[\/latex] for [latex]x\\ne -1,1.[\/latex]<\/p>\n<ol id=\"fs-id1170573633982\" style=\"list-style-type: lower-alpha\">\n<li>Sketch the graph of [latex]f[\/latex].<\/li>\n<li>Is it possible to find values [latex]{k}_{1}[\/latex] and [latex]{k}_{2}[\/latex] such that [latex]f(-1)=k[\/latex] and [latex]f(1)={k}_{2},[\/latex] and that makes [latex]f(x)[\/latex] continuous for all real numbers? Briefly explain.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573512268\" class=\"exercise\">\n<div id=\"fs-id1170573439551\" class=\"textbox\">\n<p id=\"fs-id1170573439554\">Sketch the graph of the function [latex]y=f(x)[\/latex] with properties i. through vii.<\/p>\n<ol id=\"fs-id1170573397517\">\n<li>The domain of [latex]f[\/latex] is [latex](\\text{\u2212}\\infty ,\\text{+}\\infty ).[\/latex]<\/li>\n<li>[latex]f[\/latex] has an infinite discontinuity at [latex]x=-6.[\/latex]<\/li>\n<li>[latex]f(-6)=3[\/latex]<\/li>\n<li>[latex]\\underset{x\\to {-3}^{-}}{\\text{lim}}f(x)=\\underset{x\\to {-3}^{+}}{\\text{lim}}f(x)=2[\/latex]<\/li>\n<li>[latex]f(-3)=3[\/latex]<\/li>\n<li>[latex]f[\/latex] is left continuous but not right continuous at [latex]x=3.[\/latex]<\/li>\n<li>[latex]\\underset{x\\to -\\infty }{\\text{lim}}f(x)=\\text{\u2212}\\infty[\/latex] and [latex]\\underset{x\\to +\\infty }{\\text{lim}}f(x)=\\text{+}\\infty[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170570982616\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170570982616\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170570982616\">Answers may vary; see the following example:<\/p>\n<p><span id=\"fs-id1170571130672\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203525\/CNX_Calc_Figure_02_04_207.jpg\" alt=\"A graph of a piecewise function with several segments. The first is an increasing line that exists for x &lt; -8. It ends at an open circle at (-8,-8). The second is an increasing curve that exists from -8 &lt;= x &lt; -6. It begins with a closed circle at (-8, 0 ) and goes to infinity as x goes to -6 from the left. The third is a closed circle at the point (-6, 3). The fourth is a line that exists from -6 &lt; x &lt;= 3. It begins with an open circle at (-6, 2) and ends with a closed circle at (3,2). The fifth is an increasing line starting with an open circle at (3,3). It exists for x &gt; 3.\" \/><\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571130669\" class=\"exercise\">\n<div id=\"fs-id1170571073840\" class=\"textbox\">\n<p id=\"fs-id1170571073842\">Sketch the graph of the function [latex]y=f(x)[\/latex] with properties i. through iv.<\/p>\n<ol id=\"fs-id1170573388516\">\n<li>The domain of [latex]f[\/latex] is [latex]\\left[0,5\\right].[\/latex]<\/li>\n<li>[latex]\\underset{x\\to {1}^{+}}{\\text{lim}}f(x)[\/latex] and [latex]\\underset{x\\to {1}^{-}}{\\text{lim}}f(x)[\/latex] exist and are equal.<\/li>\n<li>[latex]f(x)[\/latex] is left continuous but not continuous at [latex]x=2,[\/latex] and right continuous but not continuous at [latex]x=3.[\/latex]<\/li>\n<li>[latex]f(x)[\/latex] has a removable discontinuity at [latex]x=1,[\/latex] a jump discontinuity at [latex]x=2,[\/latex] and the following limits hold: [latex]\\underset{x\\to {3}^{-}}{\\text{lim}}f(x)=\\text{\u2212}\\infty[\/latex] and [latex]\\underset{x\\to {3}^{+}}{\\text{lim}}f(x)=2.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573425311\">In the following exercises, suppose [latex]y=f(x)[\/latex] is defined for all [latex]x[\/latex]. For each description, sketch a graph with the indicated property.<\/p>\n<div id=\"fs-id1170571096173\" class=\"exercise\">\n<div id=\"fs-id1170571096175\" class=\"textbox\">\n<p id=\"fs-id1170571096177\">Discontinuous at [latex]x=1[\/latex] with [latex]\\underset{x\\to -1}{\\text{lim}}f(x)=-1[\/latex] and [latex]\\underset{x\\to 2}{\\text{lim}}f(x)=4[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571123483\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571123483\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571123483\">Answers may vary; see the following example:<\/p>\n<p><span id=\"fs-id1170571123491\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11203529\/CNX_Calc_Figure_02_04_205.jpg\" alt=\"The graph of a piecewise function with two parts. The first part is an increasing curve that exists for x &lt; 1. It ends at (1,1). The second part is an increasing line that exists for x &gt; 1. It begins at (1,3).\" \/><\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571123500\" class=\"exercise\">\n<div id=\"fs-id1170571123502\" class=\"textbox\">\n<p id=\"fs-id1170573401181\">Discontinuous at [latex]x=2[\/latex] but continuous elsewhere with [latex]\\underset{x\\to 0}{\\text{lim}}f(x)=\\frac{1}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573422225\">Determine whether each of the given statements is true. Justify your response with an explanation or counterexample.<\/p>\n<div id=\"fs-id1170573422229\" class=\"exercise\">\n<div id=\"fs-id1170573422231\" class=\"textbox\">\n<p id=\"fs-id1170573422234\">[latex]f(t)=\\frac{2}{{e}^{t}-{e}^{-t}}[\/latex] is continuous everywhere.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573404350\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573404350\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573404350\">False. It is continuous over [latex](\\text{\u2212}\\infty ,0)\\cup (0,\\infty ).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573624587\" class=\"exercise\">\n<div id=\"fs-id1170573624589\" class=\"textbox\">\n<p id=\"fs-id1170573624591\">If the left- and right-hand limits of [latex]f(x)[\/latex] as [latex]x\\to a[\/latex] exist and are equal, then [latex]f[\/latex] cannot be discontinuous at [latex]x=a.[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571287009\" class=\"exercise\">\n<div id=\"fs-id1170571287012\" class=\"textbox\">\n<p id=\"fs-id1170571287014\">If a function is not continuous at a point, then it is not defined at that point.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571287020\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571287020\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571287020\">False. Consider [latex]f(x)=\\bigg\\{\\begin{array}{l}x\\text{ if }x\\ne 0\\\\ 4\\text{ if }x=0\\end{array}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571131867\" class=\"exercise\">\n<div id=\"fs-id1170571131869\" class=\"textbox\">\n<p id=\"fs-id1170571131872\">According to the IVT, [latex]\\cos x- \\sin x-x=2[\/latex] has a solution over the interval [latex]\\left[-1,1\\right].[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571131236\" class=\"exercise\">\n<div id=\"fs-id1170571131238\" class=\"textbox\">\n<p id=\"fs-id1170571131241\">If [latex]f(x)[\/latex] is continuous such that [latex]f(a)[\/latex] and [latex]f(b)[\/latex] have opposite signs, then [latex]f(x)=0[\/latex] has exactly one solution in [latex]\\left[a,b\\right].[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573519289\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573519289\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573519289\">False. Consider [latex]f(x)= \\cos (x)[\/latex] on [latex]\\left[-\\pi ,2\\pi \\right].[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571099130\" class=\"exercise\">\n<div id=\"fs-id1170571099132\" class=\"textbox\">\n<p id=\"fs-id1170571099134\">The function [latex]f(x)=\\frac{{x}^{2}-4x+3}{{x}^{2}-1}[\/latex] is continuous over the interval [latex]\\left[0,3\\right].[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170570998957\" class=\"exercise\">\n<div id=\"fs-id1170570998960\" class=\"textbox\">\n<p id=\"fs-id1170570998962\">If [latex]f(x)[\/latex] is continuous everywhere and [latex]f(a),f(b)>0,[\/latex] then there is no root of [latex]f(x)[\/latex] in the interval [latex]\\left[a,b\\right].[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571197414\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571197414\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571197414\">False. The IVT does <em>not<\/em> work in reverse! Consider [latex]{(x-1)}^{2}[\/latex] over the interval [latex]\\left[-2,2\\right].[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573506388\"><strong>[T]<\/strong> The following problems consider the scalar form of Coulomb\u2019s law, which describes the electrostatic force between two point charges, such as electrons. It is given by the equation [latex]F(r)={k}_{e}\\frac{|{q}_{1}{q}_{2}|}{{r}^{2}},[\/latex] where [latex]{k}_{e}[\/latex] is Coulomb\u2019s constant, [latex]{q}_{i}[\/latex] are the magnitudes of the charges of the two particles, and [latex]r[\/latex] is the distance between the two particles.<\/p>\n<div id=\"fs-id1170573393357\" class=\"exercise\">\n<div id=\"fs-id1170573393359\" class=\"textbox\">\n<p id=\"fs-id1170573393361\">To simplify the calculation of a model with many interacting particles, after some threshold value [latex]r=R,[\/latex] we approximate <em>F<\/em> as zero.<\/p>\n<ol id=\"fs-id1170573750444\" style=\"list-style-type: lower-alpha\">\n<li>Explain the physical reasoning behind this assumption.<\/li>\n<li>What is the force equation?<\/li>\n<li>Evaluate the force <em>F<\/em> using both Coulomb\u2019s law and our approximation, assuming two protons with a charge magnitude of [latex]1.6022\u00d7{10}^{-19}\\text{coulombs (C)},[\/latex] and the Coulomb constant [latex]{k}_{e}=8.988\u00d7{10}^{9}{\\text{Nm}}^{2}\\text{\/}{\\text{C}}^{2}[\/latex] are 1 m apart. Also, assume [latex]R<1\\text{m}.[\/latex] How much inaccuracy does our approximation generate? Is our approximation reasonable?<\/li>\n<li>Is there any finite value of <em>R<\/em> for which this system remains continuous at <em>R<\/em>?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573438919\" class=\"exercise\">\n<div id=\"fs-id1170573438922\" class=\"textbox\">\n<p id=\"fs-id1170573438924\">Instead of making the force 0 at <em>R<\/em>, instead we let the force be 10<sup>\u221220<\/sup> for [latex]r\\ge R.[\/latex] Assume two protons, which have a magnitude of charge [latex]1.6022\u00d7{10}^{-19}\\text{C},[\/latex] and the Coulomb constant [latex]{k}_{e}=8.988\u00d7{10}^{9}{\\text{Nm}}^{2}\\text{\/}{\\text{C}}^{2}.[\/latex] Is there a value <em>R<\/em> that can make this system continuous? If so, find it.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573420081\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573420081\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573420081\">[latex]R=0.0001519\\text{m}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571038053\">Recall the discussion on spacecraft from the chapter opener. The following problems consider a rocket launch from Earth\u2019s surface. The force of gravity on the rocket is given by [latex]F(d)=-mk\\text{\/}{d}^{2},[\/latex] where [latex]m[\/latex] is the mass of the rocket, [latex]d[\/latex] is the distance of the rocket from the center of Earth, and [latex]k[\/latex] is a constant.<\/p>\n<div id=\"fs-id1170573587225\" class=\"exercise\">\n<div id=\"fs-id1170570991016\" class=\"textbox\">\n<p id=\"fs-id1170570991018\"><strong>[T]<\/strong> Determine the value and units of [latex]k[\/latex] given that the mass of the rocket on Earth is 3 million kg. (<em>Hint<\/em>: The distance from the center of Earth to its surface is 6378 km.)<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573571185\" class=\"exercise\">\n<div id=\"fs-id1170573571187\" class=\"textbox\">\n<p id=\"fs-id1170573571189\"><strong>[T]<\/strong> After a certain distance <em>D<\/em> has passed, the gravitational effect of Earth becomes quite negligible, so we can approximate the force function by [latex]F(d)=\\bigg\\{\\begin{array}{ll}-\\frac{mk}{{d}^{2}}\\hfill & \\text{ if }d<D\\hfill \\\\ 10,000\\hfill & \\text{ if }d\\ge D\\hfill \\end{array}.[\/latex] Find the necessary condition <em>D<\/em> such that the force function remains continuous.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170570974552\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170570974552\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170570974552\">[latex]D=63.78\\text{km}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571124644\" class=\"exercise\">\n<div id=\"fs-id1170571124646\" class=\"textbox\">\n<p id=\"fs-id1170571124648\">As the rocket travels away from Earth\u2019s surface, there is a distance <em>D<\/em> where the rocket sheds some of its mass, since it no longer needs the excess fuel storage. We can write this function as [latex]F(d)=\\bigg\\{\\begin{array}{l}-\\frac{{m}_{1}k}{{d}^{2}}\\text{ if }d<D\\hfill \\\\ -\\frac{{m}_{2}k}{{d}^{2}}\\text{ if }d\\ge D\\hfill \\end{array}.[\/latex] Is there a <em>D<\/em> value such that this function is continuous, assuming [latex]{m}_{1}\\ne {m}_{2}?[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571136676\">Prove the following functions are continuous everywhere<\/p>\n<div id=\"fs-id1170571136680\" class=\"exercise\">\n<div id=\"fs-id1170571276945\" class=\"textbox\">\n<p id=\"fs-id1170571276947\">[latex]f(\\theta )= \\sin \\theta[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571276973\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571276973\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571276973\">For all values of [latex]a,f(a)[\/latex] is defined, [latex]\\underset{\\theta \\to a}{\\text{lim}}f(\\theta )[\/latex] exists, and [latex]\\underset{\\theta \\to a}{\\text{lim}}f(\\theta )=f(a).[\/latex] Therefore, [latex]f(\\theta )[\/latex] is continuous everywhere.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573413789\" class=\"exercise\">\n<div id=\"fs-id1170573413791\" class=\"textbox\">\n<p id=\"fs-id1170573413793\">[latex]g(x)=|x|[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573414086\" class=\"exercise\">\n<div id=\"fs-id1170573414089\" class=\"textbox\">\n<p id=\"fs-id1170573414091\">Where is [latex]f(x)=\\bigg\\{\\begin{array}{l}0\\text{ if }x\\text{is irrational}\\\\ 1\\text{ if }x\\text{is rational}\\end{array}[\/latex] continuous?<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571258410\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571258410\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571258410\">Nowhere<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1170571119822\" class=\"definition\">\n<dt>continuity at a point<\/dt>\n<dd id=\"fs-id1170571119828\">A function [latex]f(x)[\/latex] is continuous at a point [latex]a[\/latex] if and only if the following three conditions are satisfied: (1) [latex]f(a)[\/latex] is defined, (2) [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists, and (3) [latex]\\underset{x\\to a}{\\text{lim}}f(x)=f(a)[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1170573574137\" class=\"definition\">\n<dt>continuity from the left<\/dt>\n<dd id=\"fs-id1170573574142\">A function is continuous from the left at [latex]b[\/latex] if [latex]\\underset{x\\to {b}^{-}}{\\text{lim}}f(x)=f(b)[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1170571275835\" class=\"definition\">\n<dt>continuity from the right<\/dt>\n<dd id=\"fs-id1170571275840\">A function is continuous from the right at [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)=f(a)[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1170573363352\" class=\"definition\">\n<dt>continuity over an interval<\/dt>\n<dd id=\"fs-id1170573363357\">a function that can be traced with a pencil without lifting the pencil; a function is continuous over an open interval if it is continuous at every point in the interval; a function [latex]f(x)[\/latex] is continuous over a closed interval of the form [latex]\\left[a,b\\right][\/latex] if it is continuous at every point in [latex](a,b),[\/latex] and it is continuous from the right at [latex]a[\/latex] and from the left at [latex]b[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1170573573745\" class=\"definition\">\n<dt>discontinuity at a point<\/dt>\n<dd id=\"fs-id1170573573750\">A function is discontinuous at a point or has a discontinuity at a point if it is not continuous at the point<\/dd>\n<\/dl>\n<dl id=\"fs-id1170573573755\" class=\"definition\">\n<dt>infinite discontinuity<\/dt>\n<dd id=\"fs-id1170573573761\">An infinite discontinuity occurs at a point [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)=\\text{\u00b1}\\infty[\/latex] or [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)=\\text{\u00b1}\\infty[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1170571049355\" class=\"definition\">\n<dt>Intermediate Value Theorem<\/dt>\n<dd id=\"fs-id1170571049360\">Let [latex]f[\/latex] be continuous over a closed bounded interval [latex]\\left[\\text{a},\\text{b}\\right];[\/latex] if [latex]z[\/latex] is any real number between [latex]f(a)[\/latex] and [latex]f(b),[\/latex] then there is a number [latex]c[\/latex] in [latex]\\left[a,b\\right][\/latex] satisfying [latex]f(c)=z[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1170570997709\" class=\"definition\">\n<dt>jump discontinuity<\/dt>\n<dd id=\"fs-id1170570997714\">A jump discontinuity occurs at a point [latex]a[\/latex] if [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)[\/latex] and [latex]\\underset{x\\to {a}^{+}}{\\text{lim}}f(x)[\/latex] both exist, but [latex]\\underset{x\\to {a}^{-}}{\\text{lim}}f(x)\\ne \\underset{x\\to {a}^{+}}{\\text{lim}}f(x)[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1170571146508\" class=\"definition\">\n<dt>removable discontinuity<\/dt>\n<dd id=\"fs-id1170571146514\">A removable discontinuity occurs at a point [latex]a[\/latex] if [latex]f(x)[\/latex] is discontinuous at [latex]a[\/latex], but [latex]\\underset{x\\to a}{\\text{lim}}f(x)[\/latex] exists<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":311,"menu_order":5,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1669","chapter","type-chapter","status-publish","hentry"],"part":1589,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters\/1669","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters\/1669\/revisions"}],"predecessor-version":[{"id":2426,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters\/1669\/revisions\/2426"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/parts\/1589"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapters\/1669\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/media?parent=1669"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/pressbooks\/v2\/chapter-type?post=1669"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/contributor?post=1669"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-geneseo-openstax-calculus1-1\/wp-json\/wp\/v2\/license?post=1669"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}