{"id":3774,"date":"2021-05-12T20:24:07","date_gmt":"2021-05-12T20:24:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-section-2-4-continuity\/"},"modified":"2021-05-12T20:24:07","modified_gmt":"2021-05-12T20:24:07","slug":"review-for-section-2-4-continuity","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-section-2-4-continuity\/","title":{"raw":"Skills Review for Continuity","rendered":"Skills Review for Continuity"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n \t<li>Find the domain of a function algebraically<\/li>\n \t<li>Determine the input values for which a function is discontinuous<\/li>\n \t<li>Graph a piecewise function using domain<\/li>\n<\/ul>\n<\/div>\nThe Continuity section will explore the relationship between limits and continuity. To prepare you for this, we will review some of the various types of discontinuities you have previously encountered on your mathematical journey.\n<h2>Classify Discontinuities of Rational Functions<\/h2>\nIn general, to find the domain of a rational function, we need to determine which inputs would cause division by zero.\n<div class=\"textbox\">\n<h3>A General Note: Domain of a Rational Function<\/h3>\nThe domain of a rational function includes all real numbers except those that cause the denominator to equal zero.\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a rational function, find the domain.<\/h3>\n<ol>\n \t<li>Set the denominator equal to zero.<\/li>\n \t<li>Solve to find the values of the variable that cause the denominator to equal zero.<\/li>\n \t<li>The domain contains all real numbers except those found in Step 2.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Domain of a Rational Function<\/h3>\nFind the domain of [latex]f\\left(x\\right)=\\dfrac{x+3}{{x}^{2}-9}[\/latex].\n\n[reveal-answer q=\"860212\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"860212\"]\n\nBegin by setting the denominator equal to zero and solving.\n<p style=\"text-align: center;\">[latex]\\begin{align} {x}^{2}-9&amp;=0 \\\\ {x}^{2}&amp;=9 \\\\ x&amp;=\\pm 3 \\end{align}[\/latex]<\/p>\nThe denominator is equal to zero when [latex]x=\\pm 3[\/latex]. The domain of the function is all real numbers except [latex]x=\\pm 3[\/latex].\n<h4>Analysis of the Solution<\/h4>\nA graph of this function confirms that the function is not defined when [latex]x=\\pm 3[\/latex].\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213925\/CNX_Precalc_Figure_03_07_0092.jpg\" alt=\"Graph of f(x)=1\/(x-3) with its vertical asymptote at x=3 and its horizontal asymptote at y=0.\" width=\"487\" height=\"364\">\n\nThere is a vertical asymptote at [latex]x=3[\/latex] and a hole in the graph at [latex]x=-3[\/latex]. We will discuss these types of holes in greater detail later in this section.\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFind the domain of [latex]f\\left(x\\right)=\\dfrac{4x}{5\\left(x - 1\\right)\\left(x - 5\\right)}[\/latex].\n\n[reveal-answer q=\"553731\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"553731\"]\n\nThe domain is all real numbers except [latex]x=1[\/latex] and [latex]x=5[\/latex].\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129068&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\n\n<\/div>\nValues that are restricted from the domain of a rational function are values where the rational function has either a vertical asymptote or a hole (removable discontinuity).\n\nA <strong>vertical asymptote<\/strong> of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors. In other words, once you factor and simplify the rational function, vertical asymptotes occur at values that cause the denominator of the <em>simplified<\/em> rational function to equal 0.\n<div class=\"textbox\">\n<h3>How To: Given a rational function, identify any vertical asymptotes of its graph.<\/h3>\n<ol>\n \t<li>Factor the numerator and denominator.<\/li>\n \t<li>Simplify the expression by canceling common factors in the numerator and the denominator.<\/li>\n \t<li>Note any values that cause the denominator to be zero in this simplified version. These are the values where the vertical asymptotes occur.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Vertical Asymptotes<\/h3>\nFind the vertical asymptotes of the graph of [latex]k\\left(x\\right)=\\dfrac{5+2{x}^{2}}{2-x-{x}^{2}}[\/latex].\n\n[reveal-answer q=\"787718\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"787718\"]\n\nFirst, factor the numerator and denominator.\n<p style=\"text-align: center;\">[latex]\\begin{align}k\\left(x\\right)&amp;=\\dfrac{5+2{x}^{2}}{2-x-{x}^{2}} \\\\[1mm] &amp;=\\dfrac{5+2{x}^{2}}{\\left(2+x\\right)\\left(1-x\\right)} \\end{align}[\/latex]<\/p>\nTo find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:\n<p style=\"text-align: center;\">[latex]\\left(2+x\\right)\\left(1-x\\right)=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=-2,1[\/latex]<\/p>\nNeither [latex]x=-2[\/latex] nor [latex]x=1[\/latex] are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph below confirms the location of the two vertical asymptotes.\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213927\/CNX_Precalc_Figure_03_07_0102.jpg\" alt=\"Graph of k(x)=(5+2x)^2\/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.\" width=\"487\" height=\"514\">\n\n[\/hidden-answer]\n\n<\/div>\nOccasionally, a graph will contain a <strong>hole<\/strong>: a single point where the graph is not defined, indicated by an open circle. We call such a hole a <strong>removable discontinuity<\/strong>.\n\nFor example, the function [latex]f\\left(x\\right)=\\dfrac{{x}^{2}-1}{{x}^{2}-2x - 3}[\/latex] may be re-written by factoring the numerator and the denominator.\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{\\left(x+1\\right)\\left(x - 1\\right)}{\\left(x+1\\right)\\left(x - 3\\right)}[\/latex]<\/p>\nNotice that [latex]x+1[\/latex] is a common factor to the numerator and the denominator. The zero of this factor, [latex]x=-1[\/latex], is the location of the removable discontinuity. Notice also that [latex]x - 3[\/latex] is not a factor in both the numerator and denominator. The zero of this factor, [latex]x=3[\/latex], is the vertical asymptote.\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213929\/CNX_Precalc_Figure_03_07_0112.jpg\" alt=\"Graph of f(x)=(x^2-1)\/(x^2-2x-3) with its vertical asymptote at x=3 and a removable discontinuity at x=-1.\" width=\"487\" height=\"326\">\n<div class=\"textbox\">\n<h3>A General Note: Removable Discontinuities of Rational Functions<\/h3>\nA <strong>removable discontinuity<\/strong> occurs in the graph of a rational function at [latex]x=a[\/latex] if <em>a<\/em>&nbsp;is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Vertical Asymptotes and Removable Discontinuities for a Graph<\/h3>\nFind the vertical asymptotes and removable discontinuities of the graph of [latex]k\\left(x\\right)=\\dfrac{x - 2}{{x}^{2}-4}[\/latex].\n\n[reveal-answer q=\"519186\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"519186\"]\n\nFactor the numerator and the denominator.\n<p style=\"text-align: center;\">[latex]k\\left(x\\right)=\\dfrac{x - 2}{\\left(x - 2\\right)\\left(x+2\\right)}=\\dfrac{1}{x+2}[\/latex]<\/p>\nNotice that there is a common factor in the numerator and the denominator, [latex]x - 2[\/latex]. The zero for this factor is [latex]x=2[\/latex]. This is the location of the removable discontinuity.\n\nNotice that there is a factor in the denominator that is not in the numerator, [latex]x+2[\/latex]. The zero for this factor is [latex]x=-2[\/latex]. The vertical asymptote is [latex]x=-2[\/latex].\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213931\/CNX_Precalc_Figure_03_07_0122.jpg\" alt=\"Graph of k(x)=(x-2)\/(x-2)(x+2) with its vertical asymptote at x=-2 and a removable discontinuity at x=2.\" width=\"487\" height=\"364\">\n\nThe graph of this function will have a vertical asymptote at [latex]x=-2[\/latex], but at [latex]x=2[\/latex] the graph will have a hole.\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n[ohm_question]74565[\/ohm_question]\n\n<\/div>\n<h2>Determine Whether a Piecewise Function is Continuous at Its Boundary Value<\/h2>\nRecall that a&nbsp;<strong>piecewise function<\/strong>, also called a&nbsp;<strong>piecewise-defined<\/strong><strong> function<\/strong>,&nbsp;is a function in which more than one formula is used to define the output over different pieces of the domain.\n<div class=\"textbox\">\n<h3>A General Note: Piecewise Functions<\/h3>\nA piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this:\n<p style=\"text-align: center;\">[latex] f\\left(x\\right)=\\begin{cases}\\text{formula 1 if x is in domain 1}\\\\ \\text{formula 2 if x is in domain 2}\\\\ \\text{formula 3 if x is in domain 3}\\end{cases} [\/latex]<\/p>\n\n<\/div>\nA number in the domain where a piecewise function switches from one piece to the next is called a boundary value. Sometimes, a piecewise function is not continuous at its boundary value(s) and thus can have a discontinuity at a boundary value. In the example below, 2 is a boundary value.\n<p style=\"text-align: center;\">[latex] f\\left(x\\right)=\\begin{cases}\\text{formula 1 if } x&lt;2\\\\ \\text{formula 2 if } x\\geq2\\\\ \\end{cases} [\/latex]<\/p>\n\n<div class=\"textbox\">\n<h3>How To:&nbsp;Given a piecewise function, Determine if it is discontinuous at its boundary value(S)<strong>\n<\/strong><\/h3>\n<ol>\n \t<li>Identify the boundary value(s) of the piecewise function. For every formula (piece) the boundary value is listed by, plug the boundary value into that piece.<\/li>\n \t<li>To be continuous at a boundary value, all pieces evaluated in step 1 must produce the same function value. If the same function value is not produced, the piecewise-defined function is discontinuous at its boundary value.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Determining Whether a Piecewise function is continuous at its boundary value(S)<\/h3>\nDetermine whether the function is continuous at its boundary value.\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\begin{cases}\\begin{align}{ x }^{2}-4 \\hspace{2mm}&amp;\\text{ if }\\hspace{2mm}{ x }\\le{ -3 }\\\\ { x+8 } \\hspace{2mm}&amp;\\text{ if }\\hspace{2mm}{ x }&amp;gt{ -3 }\\end{align}\\end{cases}[\/latex]<\/p>\n[reveal-answer q=\"375071\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"375071\"]\n\nThe function has a boundary value at [latex]-3[\/latex].\n\nThe boundary value of [latex]-3[\/latex] is listed by pieces 1 and 2, so evaluate each of these pieces at [latex]x=-3[\/latex].\n\nPiece 1: [latex]f(-3)=(-3)^2-4=9-4=5[\/latex]\n\nPiece 2: [latex]f(-3)=-3+8=5[\/latex]\n\nSince you get the same function value when plugging [latex]-3[\/latex] into each piece it is listed by, the function is continuous at [latex]x=-3[\/latex].\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Determining Whether a Piecewise function is continuous at its boundary value(S)<\/h3>\nDetermine whether the function is continuous at each of its boundary values.\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\begin{cases}\\begin{align}{ x }^{2} \\hspace{2mm}&amp;\\text{ if }\\hspace{2mm}{ x }\\le{ 1 }\\\\ { 3 } \\hspace{2mm}&amp;\\text{ if }\\hspace{2mm} { 1 }&amp;lt{ x }\\le 2\\\\ { x } \\hspace{2mm}&amp;\\text{ if }\\hspace{2mm}{ x }&amp;gt{ 2 }\\end{align}\\end{cases}[\/latex]<\/p>\n[reveal-answer q=\"375072\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"375072\"]\n\nThe function has two boundary values: 1 and 2.\n\nThe boundary value of 1 is listed by pieces 1 and 2, so evaluate each of these pieces at [latex]x=1[\/latex].\n\nPiece 1: [latex]f(1)=(1)^2=1[\/latex]\n\nPiece 2: [latex]f(1)=3[\/latex]\n\nSince you do not get the same function value when plugging 1 into each piece it is listed by, the function is not continuous at [latex]x=1[\/latex].\n\nThe boundary value of 2 is listed by pieces 2 and 3, so evaluate each of these pieces at&nbsp;[latex]x=2[\/latex].\n\nPiece 2: [latex]f(2)=3[\/latex]\n\nPiece 3: [latex]f(2)=2[\/latex]\n\nSince you do not get the same function value when plugging 2 into each piece it is listed by, the function is not continuous at [latex]x=2[\/latex].\n\nBy looking at the graph of the piecewise function below, you can confirm that it is not continuous at either of its boundary values.\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18193635\/CNX_Precalc_Figure_01_02_0262.jpg\" alt=\"Graph of the entire function.\" width=\"487\" height=\"333\">\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try IT<\/h3>\n[ohm_question]219357[\/ohm_question]\n\n<\/div>\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the domain of a function algebraically<\/li>\n<li>Determine the input values for which a function is discontinuous<\/li>\n<li>Graph a piecewise function using domain<\/li>\n<\/ul>\n<\/div>\n<p>The Continuity section will explore the relationship between limits and continuity. To prepare you for this, we will review some of the various types of discontinuities you have previously encountered on your mathematical journey.<\/p>\n<h2>Classify Discontinuities of Rational Functions<\/h2>\n<p>In general, to find the domain of a rational function, we need to determine which inputs would cause division by zero.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Domain of a Rational Function<\/h3>\n<p>The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a rational function, find the domain.<\/h3>\n<ol>\n<li>Set the denominator equal to zero.<\/li>\n<li>Solve to find the values of the variable that cause the denominator to equal zero.<\/li>\n<li>The domain contains all real numbers except those found in Step 2.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Domain of a Rational Function<\/h3>\n<p>Find the domain of [latex]f\\left(x\\right)=\\dfrac{x+3}{{x}^{2}-9}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q860212\">Show Solution<\/span><\/p>\n<div id=\"q860212\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by setting the denominator equal to zero and solving.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} {x}^{2}-9&=0 \\\\ {x}^{2}&=9 \\\\ x&=\\pm 3 \\end{align}[\/latex]<\/p>\n<p>The denominator is equal to zero when [latex]x=\\pm 3[\/latex]. The domain of the function is all real numbers except [latex]x=\\pm 3[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>A graph of this function confirms that the function is not defined when [latex]x=\\pm 3[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213925\/CNX_Precalc_Figure_03_07_0092.jpg\" alt=\"Graph of f(x)=1\/(x-3) with its vertical asymptote at x=3 and its horizontal asymptote at y=0.\" width=\"487\" height=\"364\" \/><\/p>\n<p>There is a vertical asymptote at [latex]x=3[\/latex] and a hole in the graph at [latex]x=-3[\/latex]. We will discuss these types of holes in greater detail later in this section.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the domain of [latex]f\\left(x\\right)=\\dfrac{4x}{5\\left(x - 1\\right)\\left(x - 5\\right)}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q553731\">Show Solution<\/span><\/p>\n<div id=\"q553731\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain is all real numbers except [latex]x=1[\/latex] and [latex]x=5[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm129068\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129068&#38;theme=oea&#38;iframe_resize_id=ohm129068&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Values that are restricted from the domain of a rational function are values where the rational function has either a vertical asymptote or a hole (removable discontinuity).<\/p>\n<p>A <strong>vertical asymptote<\/strong> of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors. In other words, once you factor and simplify the rational function, vertical asymptotes occur at values that cause the denominator of the <em>simplified<\/em> rational function to equal 0.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a rational function, identify any vertical asymptotes of its graph.<\/h3>\n<ol>\n<li>Factor the numerator and denominator.<\/li>\n<li>Simplify the expression by canceling common factors in the numerator and the denominator.<\/li>\n<li>Note any values that cause the denominator to be zero in this simplified version. These are the values where the vertical asymptotes occur.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Vertical Asymptotes<\/h3>\n<p>Find the vertical asymptotes of the graph of [latex]k\\left(x\\right)=\\dfrac{5+2{x}^{2}}{2-x-{x}^{2}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q787718\">Show Solution<\/span><\/p>\n<div id=\"q787718\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, factor the numerator and denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k\\left(x\\right)&=\\dfrac{5+2{x}^{2}}{2-x-{x}^{2}} \\\\[1mm] &=\\dfrac{5+2{x}^{2}}{\\left(2+x\\right)\\left(1-x\\right)} \\end{align}[\/latex]<\/p>\n<p>To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(2+x\\right)\\left(1-x\\right)=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=-2,1[\/latex]<\/p>\n<p>Neither [latex]x=-2[\/latex] nor [latex]x=1[\/latex] are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph below confirms the location of the two vertical asymptotes.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213927\/CNX_Precalc_Figure_03_07_0102.jpg\" alt=\"Graph of k(x)=(5+2x)^2\/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.\" width=\"487\" height=\"514\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Occasionally, a graph will contain a <strong>hole<\/strong>: a single point where the graph is not defined, indicated by an open circle. We call such a hole a <strong>removable discontinuity<\/strong>.<\/p>\n<p>For example, the function [latex]f\\left(x\\right)=\\dfrac{{x}^{2}-1}{{x}^{2}-2x - 3}[\/latex] may be re-written by factoring the numerator and the denominator.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\dfrac{\\left(x+1\\right)\\left(x - 1\\right)}{\\left(x+1\\right)\\left(x - 3\\right)}[\/latex]<\/p>\n<p>Notice that [latex]x+1[\/latex] is a common factor to the numerator and the denominator. The zero of this factor, [latex]x=-1[\/latex], is the location of the removable discontinuity. Notice also that [latex]x - 3[\/latex] is not a factor in both the numerator and denominator. The zero of this factor, [latex]x=3[\/latex], is the vertical asymptote.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213929\/CNX_Precalc_Figure_03_07_0112.jpg\" alt=\"Graph of f(x)=(x^2-1)\/(x^2-2x-3) with its vertical asymptote at x=3 and a removable discontinuity at x=-1.\" width=\"487\" height=\"326\" \/><\/p>\n<div class=\"textbox\">\n<h3>A General Note: Removable Discontinuities of Rational Functions<\/h3>\n<p>A <strong>removable discontinuity<\/strong> occurs in the graph of a rational function at [latex]x=a[\/latex] if <em>a<\/em>&nbsp;is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Vertical Asymptotes and Removable Discontinuities for a Graph<\/h3>\n<p>Find the vertical asymptotes and removable discontinuities of the graph of [latex]k\\left(x\\right)=\\dfrac{x - 2}{{x}^{2}-4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q519186\">Show Solution<\/span><\/p>\n<div id=\"q519186\" class=\"hidden-answer\" style=\"display: none\">\n<p>Factor the numerator and the denominator.<\/p>\n<p style=\"text-align: center;\">[latex]k\\left(x\\right)=\\dfrac{x - 2}{\\left(x - 2\\right)\\left(x+2\\right)}=\\dfrac{1}{x+2}[\/latex]<\/p>\n<p>Notice that there is a common factor in the numerator and the denominator, [latex]x - 2[\/latex]. The zero for this factor is [latex]x=2[\/latex]. This is the location of the removable discontinuity.<\/p>\n<p>Notice that there is a factor in the denominator that is not in the numerator, [latex]x+2[\/latex]. The zero for this factor is [latex]x=-2[\/latex]. The vertical asymptote is [latex]x=-2[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213931\/CNX_Precalc_Figure_03_07_0122.jpg\" alt=\"Graph of k(x)=(x-2)\/(x-2)(x+2) with its vertical asymptote at x=-2 and a removable discontinuity at x=2.\" width=\"487\" height=\"364\" \/><\/p>\n<p>The graph of this function will have a vertical asymptote at [latex]x=-2[\/latex], but at [latex]x=2[\/latex] the graph will have a hole.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm74565\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=74565&theme=oea&iframe_resize_id=ohm74565&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Determine Whether a Piecewise Function is Continuous at Its Boundary Value<\/h2>\n<p>Recall that a&nbsp;<strong>piecewise function<\/strong>, also called a&nbsp;<strong>piecewise-defined<\/strong><strong> function<\/strong>,&nbsp;is a function in which more than one formula is used to define the output over different pieces of the domain.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Piecewise Functions<\/h3>\n<p>A piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this:<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\begin{cases}\\text{formula 1 if x is in domain 1}\\\\ \\text{formula 2 if x is in domain 2}\\\\ \\text{formula 3 if x is in domain 3}\\end{cases}[\/latex]<\/p>\n<\/div>\n<p>A number in the domain where a piecewise function switches from one piece to the next is called a boundary value. Sometimes, a piecewise function is not continuous at its boundary value(s) and thus can have a discontinuity at a boundary value. In the example below, 2 is a boundary value.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\begin{cases}\\text{formula 1 if } x<2\\\\ \\text{formula 2 if } x\\geq2\\\\ \\end{cases}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>How To:&nbsp;Given a piecewise function, Determine if it is discontinuous at its boundary value(S)<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Identify the boundary value(s) of the piecewise function. For every formula (piece) the boundary value is listed by, plug the boundary value into that piece.<\/li>\n<li>To be continuous at a boundary value, all pieces evaluated in step 1 must produce the same function value. If the same function value is not produced, the piecewise-defined function is discontinuous at its boundary value.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Determining Whether a Piecewise function is continuous at its boundary value(S)<\/h3>\n<p>Determine whether the function is continuous at its boundary value.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\begin{cases}\\begin{align}{ x }^{2}-4 \\hspace{2mm}&\\text{ if }\\hspace{2mm}{ x }\\le{ -3 }\\\\ { x+8 } \\hspace{2mm}&\\text{ if }\\hspace{2mm}{ x }&gt{ -3 }\\end{align}\\end{cases}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q375071\">Show Solution<\/span><\/p>\n<div id=\"q375071\" class=\"hidden-answer\" style=\"display: none\">\n<p>The function has a boundary value at [latex]-3[\/latex].<\/p>\n<p>The boundary value of [latex]-3[\/latex] is listed by pieces 1 and 2, so evaluate each of these pieces at [latex]x=-3[\/latex].<\/p>\n<p>Piece 1: [latex]f(-3)=(-3)^2-4=9-4=5[\/latex]<\/p>\n<p>Piece 2: [latex]f(-3)=-3+8=5[\/latex]<\/p>\n<p>Since you get the same function value when plugging [latex]-3[\/latex] into each piece it is listed by, the function is continuous at [latex]x=-3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Determining Whether a Piecewise function is continuous at its boundary value(S)<\/h3>\n<p>Determine whether the function is continuous at each of its boundary values.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\begin{cases}\\begin{align}{ x }^{2} \\hspace{2mm}&\\text{ if }\\hspace{2mm}{ x }\\le{ 1 }\\\\ { 3 } \\hspace{2mm}&\\text{ if }\\hspace{2mm} { 1 }&lt{ x }\\le 2\\\\ { x } \\hspace{2mm}&\\text{ if }\\hspace{2mm}{ x }&gt{ 2 }\\end{align}\\end{cases}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q375072\">Show Solution<\/span><\/p>\n<div id=\"q375072\" class=\"hidden-answer\" style=\"display: none\">\n<p>The function has two boundary values: 1 and 2.<\/p>\n<p>The boundary value of 1 is listed by pieces 1 and 2, so evaluate each of these pieces at [latex]x=1[\/latex].<\/p>\n<p>Piece 1: [latex]f(1)=(1)^2=1[\/latex]<\/p>\n<p>Piece 2: [latex]f(1)=3[\/latex]<\/p>\n<p>Since you do not get the same function value when plugging 1 into each piece it is listed by, the function is not continuous at [latex]x=1[\/latex].<\/p>\n<p>The boundary value of 2 is listed by pieces 2 and 3, so evaluate each of these pieces at&nbsp;[latex]x=2[\/latex].<\/p>\n<p>Piece 2: [latex]f(2)=3[\/latex]<\/p>\n<p>Piece 3: [latex]f(2)=2[\/latex]<\/p>\n<p>Since you do not get the same function value when plugging 2 into each piece it is listed by, the function is not continuous at [latex]x=2[\/latex].<\/p>\n<p>By looking at the graph of the piecewise function below, you can confirm that it is not continuous at either of its boundary values.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18193635\/CNX_Precalc_Figure_01_02_0262.jpg\" alt=\"Graph of the entire function.\" width=\"487\" height=\"333\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm219357\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=219357&theme=oea&iframe_resize_id=ohm219357&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3774\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Modification and Revision. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/precalculus\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Modification and 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