{"id":332,"date":"2021-02-04T01:13:06","date_gmt":"2021-02-04T01:13:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=332"},"modified":"2022-03-16T05:28:23","modified_gmt":"2022-03-16T05:28:23","slug":"graphing-a-derivative","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/graphing-a-derivative\/","title":{"raw":"Graphing a Derivative","rendered":"Graphing a Derivative"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Graph a derivative function from the graph of a given function<\/li>\r\n \t<li>State the connection between derivatives and continuity<\/li>\r\n \t<li>Describe three conditions for when a function does not have a derivative<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Derivative Function Graphs<\/h2>\r\n<p id=\"fs-id1169737952664\">We have already discussed how to graph a function, so given the equation of a function or the equation of a derivative function, we could graph it. Given both, we would expect to see a correspondence between the graphs of these two functions, since [latex]f^{\\prime}(x)[\/latex] gives the rate of change of a function [latex]f(x)[\/latex] (or slope of the tangent line to [latex]f(x)[\/latex]). It may be useful here to recall the characteristics of lines with certain slopes.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Recall: slopes of lines and their defining characteristics<\/h3>\r\nA line has a\u00a0<strong>positive<\/strong> slope if it is <em>increasing<\/em> from left to right.\r\n\r\nA line has a\u00a0<strong>negative\u00a0<\/strong>slope if it is <em>decreasing<\/em> from left to right.\r\n\r\nA <em>horizontal<\/em> line has a slope of\u00a0<strong>0.\u00a0<\/strong>\r\n\r\nA <em>vertical<\/em> line has an\u00a0<strong>undefined<\/strong> slope.\r\n\r\n<\/div>\r\n<p id=\"fs-id1169738190458\">In the first example we found that for [latex]f(x)=\\sqrt{x}, \\, f^{\\prime}(x)=\\frac{1}{2}\\sqrt{x}[\/latex]. If we graph these functions on the same axes, as in Figure 2, we can use the graphs to understand the relationship between these two functions. First, we notice that [latex]f(x)[\/latex] is increasing over its entire domain, which means that the slopes of its tangent lines at all points are positive. Consequently, we expect [latex]f^{\\prime}(x)&gt;0[\/latex] for all values of [latex]x[\/latex] in its domain. Furthermore, as [latex]x[\/latex] increases, the slopes of the tangent lines to [latex]f(x)[\/latex] are decreasing and we expect to see a corresponding decrease in [latex]f^{\\prime}(x)[\/latex]. We also observe that [latex]f(0)[\/latex] is undefined and that [latex]\\underset{x\\to 0^+}{\\lim}f^{\\prime}(x)=+\\infty[\/latex], corresponding to a vertical tangent to [latex]f(x)[\/latex] at 0.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205223\/CNX_Calc_Figure_03_02_002.jpg\" alt=\"The function f(x) = the square root of x is graphed as is its derivative f\u2019(x) = 1\/(2 times the square root of x).\" width=\"487\" height=\"322\" \/> Figure 2. The derivative [latex]f^{\\prime}(x)[\/latex] is positive everywhere because the function [latex]f(x)[\/latex] is increasing.[\/caption]\r\n<p id=\"fs-id1169738183946\">In the second example we found that for [latex]f(x)=x^2-2x, \\, f^{\\prime}(x)=2x-2[\/latex]. The graphs of these functions are shown in Figure 3. Observe that [latex]f(x)[\/latex] is decreasing for [latex]x&lt;1[\/latex]. For these same values of [latex]x, \\, f^{\\prime}(x)&lt;0[\/latex]. For values of [latex]x&gt;1, \\, f(x)[\/latex] is increasing and [latex]f^{\\prime}(x)&gt;0[\/latex]. Also, [latex]f(x)[\/latex] has a horizontal tangent at [latex]x=1[\/latex] and [latex]f^{\\prime}(1)=0[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205225\/CNX_Calc_Figure_03_02_003.jpg\" alt=\"The function f(x) = x squared \u2013 2x is graphed as is its derivative f\u2019(x) = 2x \u2212 2.\" width=\"487\" height=\"358\" \/> Figure 3. The derivative [latex]f^{\\prime}(x)&lt;0[\/latex] where the function [latex]f(x)[\/latex] is decreasing and [latex]f^{\\prime}(x)&gt;0[\/latex] where [latex]f(x)[\/latex] is increasing. The derivative is zero where the function has a horizontal tangent.[\/caption]\r\n<div id=\"fs-id1169737966982\" class=\"textbook exercises\">\r\n<h3>Example: Sketching a Derivative Using a Function<\/h3>\r\n<p id=\"fs-id1169738187865\">Use the following graph of [latex]f(x)[\/latex] to sketch a graph of [latex]f^{\\prime}(x)[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205228\/CNX_Calc_Figure_03_02_004.jpg\" alt=\"The function f(x) is roughly sinusoidal, starting at (\u22124, 3), decreasing to a local minimum at (\u22122, 2), then increasing to a local maximum at (3, 6), and getting cut off at (7, 2).\" width=\"487\" height=\"471\" \/> Figure 4. Graph of [latex]f(x)[\/latex].[\/caption][reveal-answer q=\"fs-id1169738226318\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738226318\"]\r\n<p id=\"fs-id1169738226318\">The solution is shown in the following graph. Observe that [latex]f(x)[\/latex] is increasing and [latex]f^{\\prime}(x)&gt;0[\/latex] on [latex](\u20132,3)[\/latex]. Also, [latex]f(x)[\/latex] is decreasing and [latex]f^{\\prime}(x)&lt;0[\/latex] on [latex](\u2212\\infty ,-2)[\/latex] and on [latex](3,+\\infty)[\/latex]. Also note that [latex]f(x)[\/latex] has horizontal tangents at -2 and 3, and [latex]f^{\\prime}(-2)=0[\/latex] and [latex]f^{\\prime}(3)=0[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205232\/CNX_Calc_Figure_03_02_005.jpg\" alt=\"Two functions are graphed here: f(x) and f\u2019(x). The function f(x) is the same as the above graph, that is, roughly sinusoidal, starting at (\u22124, 3), decreasing to a local minimum at (\u22122, 2), then increasing to a local maximum at (3, 6), and getting cut off at (7, 2). The function f\u2019(x) is an downward-facing parabola with vertex near (0.5, 1.75), y-intercept (0, 1.5), and x-intercepts (\u22121.9, 0) and (3, 0).\" width=\"487\" height=\"471\" \/> Figure 5. Graph of\u00a0[latex]f^{\\prime}(x)[\/latex].[\/caption]<span id=\"fs-id1169738042183\"><\/span>[\/hidden-answer]<\/div>\r\nWatch the following video to see the worked solution to Example: Sketching a Derivative Using a Function.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/w9m4h4i47-M?controls=0&amp;start=415&amp;end=535&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.2TheDerivativeAsAFunction415to535_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.2 The Derivative as a Function\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1169738226881\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169738223964\">Sketch the graph of [latex]f(x)=x^2-4[\/latex]. On what interval is the graph of [latex]f^{\\prime}(x)[\/latex] above the [latex]x[\/latex]-axis?<\/p>\r\n[reveal-answer q=\"3005289\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3005289\"]\r\n<p id=\"fs-id1169738101726\">The graph of [latex]f^{\\prime}(x)[\/latex] is positive where [latex]f(x)[\/latex] is increasing.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169737951946\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737951946\"]\r\n<p id=\"fs-id1169737951946\">[latex](0,+\\infty)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]17245[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Derivatives and Continuity<\/h2>\r\n<p id=\"fs-id1169738188294\">Now that we can graph a derivative, let\u2019s examine the behavior of the graphs. First, we consider the relationship between differentiability and continuity. We will see that if a function is differentiable at a point, it must be continuous there; however, a function that is continuous at a point need not be differentiable at that point. In fact, a function may be continuous at a point and fail to be differentiable at the point for one of several reasons.<\/p>\r\n\r\n<div id=\"fs-id1169737985034\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Differentiability Implies Continuity<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169737985040\">Let [latex]f(x)[\/latex] be a function and [latex]a[\/latex] be in its domain. If [latex]f(x)[\/latex] is differentiable at [latex]a[\/latex], then [latex]f[\/latex] is continuous at [latex]a[\/latex].<\/p>\r\n\r\n<\/div>\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1169738184909\">If [latex]f(x)[\/latex] is differentiable at [latex]a[\/latex], then [latex]f^{\\prime}(a)[\/latex] exists and<\/p>\r\n\r\n<div id=\"fs-id1169738220653\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(a)=\\underset{x\\to a}{\\lim}\\dfrac{f(x)-f(a)}{x-a}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737140882\">We want to show that [latex]f(x)[\/latex] is continuous at [latex]a[\/latex] by showing that [latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1169738221978\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} \\underset{x\\to a}{\\lim}f(x) &amp; =\\underset{x\\to a}{\\lim}(f(x)-f(a)+f(a)) &amp; &amp; &amp; \\\\ &amp; =\\underset{x\\to a}{\\lim}(\\frac{f(x)-f(a)}{x-a}\\cdot (x-a)+f(a)) &amp; &amp; &amp; \\text{Multiply and divide} \\, f(x)-f(a) \\, \\text{by} \\, x-a. \\\\ &amp; =(\\underset{x\\to a}{\\lim}\\frac{f(x)-f(a)}{x-a}) \\cdot (\\underset{x\\to a}{\\lim}(x-a))+\\underset{x\\to a}{\\lim}f(a) &amp; &amp; &amp; \\\\ &amp; =f^{\\prime}(a) \\cdot 0+f(a) &amp; &amp; &amp; \\\\ &amp; =f(a). &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738226778\">Therefore, since [latex]f(a)[\/latex] is defined and [latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex], we conclude that [latex]f[\/latex] is continuous at [latex]a[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n<p id=\"fs-id1169737978576\">We have just proven that differentiability implies continuity, but now we consider whether continuity implies differentiability. To determine an answer to this question, we examine the function [latex]f(x)=|x|[\/latex]. This function is continuous everywhere; however, [latex]f^{\\prime}(0)[\/latex] is undefined. This observation leads us to believe that continuity does not imply differentiability. Let\u2019s explore further. For [latex]f(x)=|x|[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1169738223612\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\dfrac{f(x)-f(0)}{x-0}=\\underset{x\\to 0}{\\lim}\\dfrac{|x|-|0|}{x-0}=\\underset{x\\to 0}{\\lim}\\dfrac{|x|}{x}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738216093\">This limit does not exist because<\/p>\r\n\r\n<div id=\"fs-id1169738216096\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^-}{\\lim}\\dfrac{|x|}{x}=-1 \\, \\text{and} \\, \\underset{x\\to 0^+}{\\lim}\\dfrac{|x|}{x}=1[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738063963\">See Figure 6.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205235\/CNX_Calc_Figure_03_02_006.jpg\" alt=\"The function f(x) = the absolute value of x is graphed. It consists of two straight line segments: the first follows the equation y = \u2212x and ends at the origin; the second follows the equation y = x and starts at the origin.\" width=\"487\" height=\"471\" \/> Figure 6. The function [latex]f(x)=|x|[\/latex] is continuous at 0 but is not differentiable at 0.[\/caption]\r\n<p id=\"fs-id1169738228170\">Let\u2019s consider some additional situations in which a continuous function fails to be differentiable. Consider the function [latex]f(x)=\\sqrt[3]{x}[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1169738228211\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\dfrac{\\sqrt[3]{x}-0}{x-0}=\\underset{x\\to 0}{\\lim}\\dfrac{1}{\\sqrt[3]{x^2}}=+\\infty[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738217227\">Thus [latex]f^{\\prime}(0)[\/latex] does not exist. A quick look at the graph of [latex]f(x)=\\sqrt[3]{x}[\/latex] clarifies the situation. The function has a vertical tangent line at 0 (Figure 7).<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205237\/CNX_Calc_Figure_03_02_007.jpg\" alt=\"The function f(x) = the cube root of x is graphed. It has a vertical tangent at x = 0.\" width=\"487\" height=\"275\" \/> Figure 7. The function [latex]f(x)=\\sqrt[3]{x}[\/latex] has a vertical tangent at [latex]x=0[\/latex]. It is continuous at 0 but is not differentiable at 0.[\/caption]\r\n<p id=\"fs-id1169738217510\">The function [latex]f(x)=\\begin{cases} x \\sin\\left(\\frac{1}{x}\\right) &amp; \\text{ if } \\, x \\ne 0 \\\\ 0 &amp; \\text{ if } \\, x = 0 \\end{cases}[\/latex] also has a derivative that exhibits interesting behavior at 0. We see that<\/p>\r\n\r\n<div id=\"fs-id1169738225665\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\dfrac{x \\sin\\left(\\frac{1}{x}\\right)-0}{x-0}=\\underset{x\\to 0}{\\lim} \\sin\\left(\\dfrac{1}{x}\\right)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737946726\">This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approach zero (Figure 8).<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"426\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205240\/CNX_Calc_Figure_03_02_008.jpg\" alt=\"The function f(x) = x sin (1\/2) if x does not equal 0 and f(x) = 0 if x = 0 is graphed. It looks like a rapidly oscillating sinusoidal function with amplitude decreasing to 0 at the origin.\" width=\"426\" height=\"391\" \/> Figure 8. The function [latex]f(x)=\\begin{cases} x \\sin(\\frac{1}{x}) &amp; \\text{ if } \\, x \\ne 0 \\\\ 0 &amp; \\text{ if } \\, x = 0 \\end{cases}[\/latex] is not differentiable at 0.[\/caption]&nbsp;\r\n<p id=\"fs-id1169738042233\">In summary:<\/p>\r\n\r\n<ol id=\"fs-id1169738042236\">\r\n \t<li>We observe that if a function is not continuous, it cannot be differentiable, since every differentiable function must be continuous. However, if a function is continuous, it may still fail to be differentiable.<\/li>\r\n \t<li>We saw that [latex]f(x)=|x|[\/latex] failed to be differentiable at 0 because the limit of the slopes of the tangent lines on the left and right were not the same. Visually, this resulted in a sharp corner on the graph of the function at 0. From this we conclude that in order to be differentiable at a point, a function must be \u201csmooth\u201d at that point.<\/li>\r\n \t<li>As we saw in the example of [latex]f(x)=\\sqrt[3]{x}[\/latex], a function fails to be differentiable at a point where there is a vertical tangent line.<\/li>\r\n \t<li>As we saw with [latex]f(x)=\\begin{cases} x \\sin(\\frac{1}{x}) &amp; \\text{ if } \\, x \\ne 0 \\\\ 0 &amp; \\text{ if } \\, x = 0 \\end{cases}[\/latex] a function may fail to be differentiable at a point in more complicated ways as well.<\/li>\r\n<\/ol>\r\n&nbsp;\r\n<div id=\"fs-id1169738218186\" class=\"textbook exercises\">\r\n<h3>Example: A Piecewise Function that is Continuous and Differentiable<\/h3>\r\n<p id=\"fs-id1169738218195\">A toy company wants to design a track for a toy car that starts out along a parabolic curve and then converts to a straight line (Figure 9). The function that describes the track is to have the form [latex]f(x)=\\begin{cases} \\frac{1}{10}x^2 + bx + c &amp; \\text{ if } \\, x &lt; -10 \\\\ -\\frac{1}{4}x + \\frac{5}{2} &amp; \\text{ if } \\, x \\ge -10 \\end{cases}[\/latex], where [latex]x[\/latex] and [latex]f(x)[\/latex] are in inches. For the car to move smoothly along the track, the function [latex]f(x)[\/latex] must be both continuous and differentiable at -10. Find values of [latex]b[\/latex] and [latex]c[\/latex] that make [latex]f(x)[\/latex] both continuous and differentiable.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205243\/CNX_Calc_Figure_03_02_009.jpg\" alt=\"A cart is drawn on a line that curves through (\u221210, 5) to (10, 0) with y-intercept roughly (0, 2).\" width=\"487\" height=\"237\" \/> Figure 9. For the car to move smoothly along the track, the function must be both continuous and differentiable.[\/caption]\r\n\r\n[reveal-answer q=\"fs-id1169738222036\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738222036\"]\r\n<p id=\"fs-id1169738222036\">For the function to be continuous at [latex]x=-10, \\, \\underset{x\\to 10^-}{\\lim}f(x)=f(-10)[\/latex]. Thus, since<\/p>\r\n\r\n<div id=\"fs-id1169737950192\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \u221210^-}{\\lim}f(x)=\\frac{1}{10}(-10)^2-10b+c=10-10b+c[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738216001\">and [latex]f(-10)=5[\/latex], we must have [latex]10-10b+c=5[\/latex]. Equivalently, we have [latex]c=10b-5[\/latex].<\/p>\r\n<p id=\"fs-id1169738219145\">For the function to be differentiable at -10,<\/p>\r\n\r\n<div id=\"fs-id1169738219154\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(10)=\\underset{x\\to \u221210}{\\lim}\\frac{f(x)-f(-10)}{x+10}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738224010\">must exist. Since [latex]f(x)[\/latex] is defined using different rules on the right and the left, we must evaluate this limit from the right and the left and then set them equal to each other:<\/p>\r\n\r\n<div id=\"fs-id1169738183518\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} \\underset{x\\to \u221210^-}{\\lim}\\frac{f(x)-f(-10)}{x+10} &amp; =\\underset{x\\to \u221210^-}{\\lim}\\frac{\\frac{1}{10}x^2+bx+c-5}{x+10} &amp; &amp; &amp; \\\\ &amp; =\\underset{x\\to \u221210^-}{\\lim}\\frac{\\frac{1}{10}x^2+bx+(10b-5)-5}{x+10} &amp; &amp; &amp; \\text{Substitute} \\, c=10b-5. \\\\ &amp; =\\underset{x\\to \u221210^-}{\\lim}\\frac{x^2-100+10bx+100b}{10(x+10)} &amp; &amp; &amp; \\\\ &amp; =\\underset{x\\to \u221210^-}{\\lim}\\frac{(x+10)(x-10+10b)}{10(x+10)} &amp; &amp; &amp; \\text{Factor by grouping.} \\\\ &amp; =b-2 &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738183170\">We also have<\/p>\r\n\r\n<div id=\"fs-id1169738183173\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to \u221210^+}{\\lim}\\frac{f(x)-f(-10)}{x+10} &amp; =\\underset{x\\to \u221210^+}{\\lim}\\frac{-\\frac{1}{4}x+\\frac{5}{2}-5}{x+10} \\\\ &amp; =\\underset{x\\to \u221210^+}{\\lim}\\frac{\u2212(x+10)}{4(x+10)} \\\\ &amp; =-\\frac{1}{4} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738221364\">This gives us [latex]b-2=-\\frac{1}{4}[\/latex].<\/p>\r\nThus [latex]b=\\frac{7}{4}[\/latex] and [latex]c=10(\\frac{7}{4})-5=\\frac{25}{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: A Piecewise Function that is Continuous and Differentiable.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/w9m4h4i47-M?controls=0&amp;start=904&amp;end=1275&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.2TheDerivativeAsAFunction904to1275_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.2 The Derivative as a Function\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1169738185315\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169738185323\">Find values of [latex]a[\/latex] and [latex]b[\/latex] that make [latex]f(x)=\\begin{cases} ax+b &amp; \\text{ if } \\, x &lt; 3 \\\\ x^2 &amp; \\text{ if } \\, x \\ge 3 \\end{cases}[\/latex] both continuous and differentiable at 3.<\/p>\r\n[reveal-answer q=\"fs-id1169738225596\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738225596\"]\r\n<p id=\"fs-id1169738225596\">[latex]a=6[\/latex] and [latex]b=-9[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169738225624\">Use the previous example as a guide.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738225636\" class=\"bc-section section\"><\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Graph a derivative function from the graph of a given function<\/li>\n<li>State the connection between derivatives and continuity<\/li>\n<li>Describe three conditions for when a function does not have a derivative<\/li>\n<\/ul>\n<\/div>\n<h2>Derivative Function Graphs<\/h2>\n<p id=\"fs-id1169737952664\">We have already discussed how to graph a function, so given the equation of a function or the equation of a derivative function, we could graph it. Given both, we would expect to see a correspondence between the graphs of these two functions, since [latex]f^{\\prime}(x)[\/latex] gives the rate of change of a function [latex]f(x)[\/latex] (or slope of the tangent line to [latex]f(x)[\/latex]). It may be useful here to recall the characteristics of lines with certain slopes.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: slopes of lines and their defining characteristics<\/h3>\n<p>A line has a\u00a0<strong>positive<\/strong> slope if it is <em>increasing<\/em> from left to right.<\/p>\n<p>A line has a\u00a0<strong>negative\u00a0<\/strong>slope if it is <em>decreasing<\/em> from left to right.<\/p>\n<p>A <em>horizontal<\/em> line has a slope of\u00a0<strong>0.\u00a0<\/strong><\/p>\n<p>A <em>vertical<\/em> line has an\u00a0<strong>undefined<\/strong> slope.<\/p>\n<\/div>\n<p id=\"fs-id1169738190458\">In the first example we found that for [latex]f(x)=\\sqrt{x}, \\, f^{\\prime}(x)=\\frac{1}{2}\\sqrt{x}[\/latex]. If we graph these functions on the same axes, as in Figure 2, we can use the graphs to understand the relationship between these two functions. First, we notice that [latex]f(x)[\/latex] is increasing over its entire domain, which means that the slopes of its tangent lines at all points are positive. Consequently, we expect [latex]f^{\\prime}(x)>0[\/latex] for all values of [latex]x[\/latex] in its domain. Furthermore, as [latex]x[\/latex] increases, the slopes of the tangent lines to [latex]f(x)[\/latex] are decreasing and we expect to see a corresponding decrease in [latex]f^{\\prime}(x)[\/latex]. We also observe that [latex]f(0)[\/latex] is undefined and that [latex]\\underset{x\\to 0^+}{\\lim}f^{\\prime}(x)=+\\infty[\/latex], corresponding to a vertical tangent to [latex]f(x)[\/latex] at 0.<\/p>\n<div style=\"width: 497px\" 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\/11205223\/CNX_Calc_Figure_03_02_002.jpg\" alt=\"The function f(x) = the square root of x is graphed as is its derivative f\u2019(x) = 1\/(2 times the square root of x).\" width=\"487\" height=\"322\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The derivative [latex]f^{\\prime}(x)[\/latex] is positive everywhere because the function [latex]f(x)[\/latex] is increasing.<\/p>\n<\/div>\n<p id=\"fs-id1169738183946\">In the second example we found that for [latex]f(x)=x^2-2x, \\, f^{\\prime}(x)=2x-2[\/latex]. The graphs of these functions are shown in Figure 3. Observe that [latex]f(x)[\/latex] is decreasing for [latex]x<1[\/latex]. For these same values of [latex]x, \\, f^{\\prime}(x)<0[\/latex]. For values of [latex]x>1, \\, f(x)[\/latex] is increasing and [latex]f^{\\prime}(x)>0[\/latex]. Also, [latex]f(x)[\/latex] has a horizontal tangent at [latex]x=1[\/latex] and [latex]f^{\\prime}(1)=0[\/latex].<\/p>\n<div style=\"width: 497px\" 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\/11205225\/CNX_Calc_Figure_03_02_003.jpg\" alt=\"The function f(x) = x squared \u2013 2x is graphed as is its derivative f\u2019(x) = 2x \u2212 2.\" width=\"487\" height=\"358\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The derivative [latex]f^{\\prime}(x)&lt;0[\/latex] where the function [latex]f(x)[\/latex] is decreasing and [latex]f^{\\prime}(x)&gt;0[\/latex] where [latex]f(x)[\/latex] is increasing. The derivative is zero where the function has a horizontal tangent.<\/p>\n<\/div>\n<div id=\"fs-id1169737966982\" class=\"textbook exercises\">\n<h3>Example: Sketching a Derivative Using a Function<\/h3>\n<p id=\"fs-id1169738187865\">Use the following graph of [latex]f(x)[\/latex] to sketch a graph of [latex]f^{\\prime}(x)[\/latex].<\/p>\n<div style=\"width: 497px\" 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\/11205228\/CNX_Calc_Figure_03_02_004.jpg\" alt=\"The function f(x) is roughly sinusoidal, starting at (\u22124, 3), decreasing to a local minimum at (\u22122, 2), then increasing to a local maximum at (3, 6), and getting cut off at (7, 2).\" width=\"487\" height=\"471\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. Graph of [latex]f(x)[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738226318\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738226318\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738226318\">The solution is shown in the following graph. Observe that [latex]f(x)[\/latex] is increasing and [latex]f^{\\prime}(x)>0[\/latex] on [latex](\u20132,3)[\/latex]. Also, [latex]f(x)[\/latex] is decreasing and [latex]f^{\\prime}(x)<0[\/latex] on [latex](\u2212\\infty ,-2)[\/latex] and on [latex](3,+\\infty)[\/latex]. Also note that [latex]f(x)[\/latex] has horizontal tangents at -2 and 3, and [latex]f^{\\prime}(-2)=0[\/latex] and [latex]f^{\\prime}(3)=0[\/latex].<\/p>\n<div style=\"width: 497px\" 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\/11205232\/CNX_Calc_Figure_03_02_005.jpg\" alt=\"Two functions are graphed here: f(x) and f\u2019(x). The function f(x) is the same as the above graph, that is, roughly sinusoidal, starting at (\u22124, 3), decreasing to a local minimum at (\u22122, 2), then increasing to a local maximum at (3, 6), and getting cut off at (7, 2). The function f\u2019(x) is an downward-facing parabola with vertex near (0.5, 1.75), y-intercept (0, 1.5), and x-intercepts (\u22121.9, 0) and (3, 0).\" width=\"487\" height=\"471\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. Graph of\u00a0[latex]f^{\\prime}(x)[\/latex].<\/p>\n<\/div>\n<p><span id=\"fs-id1169738042183\"><\/span><\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Sketching a Derivative Using a Function.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/w9m4h4i47-M?controls=0&amp;start=415&amp;end=535&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.2TheDerivativeAsAFunction415to535_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.2 The Derivative as a Function&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1169738226881\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169738223964\">Sketch the graph of [latex]f(x)=x^2-4[\/latex]. On what interval is the graph of [latex]f^{\\prime}(x)[\/latex] above the [latex]x[\/latex]-axis?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3005289\">Hint<\/span><\/p>\n<div id=\"q3005289\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738101726\">The graph of [latex]f^{\\prime}(x)[\/latex] is positive where [latex]f(x)[\/latex] is increasing.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737951946\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737951946\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737951946\">[latex](0,+\\infty)[\/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=\"ohm17245\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=17245&theme=oea&iframe_resize_id=ohm17245&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Derivatives and Continuity<\/h2>\n<p id=\"fs-id1169738188294\">Now that we can graph a derivative, let\u2019s examine the behavior of the graphs. First, we consider the relationship between differentiability and continuity. We will see that if a function is differentiable at a point, it must be continuous there; however, a function that is continuous at a point need not be differentiable at that point. In fact, a function may be continuous at a point and fail to be differentiable at the point for one of several reasons.<\/p>\n<div id=\"fs-id1169737985034\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Differentiability Implies Continuity<\/h3>\n<hr \/>\n<p id=\"fs-id1169737985040\">Let [latex]f(x)[\/latex] be a function and [latex]a[\/latex] be in its domain. If [latex]f(x)[\/latex] is differentiable at [latex]a[\/latex], then [latex]f[\/latex] is continuous at [latex]a[\/latex].<\/p>\n<\/div>\n<h3>Proof<\/h3>\n<p id=\"fs-id1169738184909\">If [latex]f(x)[\/latex] is differentiable at [latex]a[\/latex], then [latex]f^{\\prime}(a)[\/latex] exists and<\/p>\n<div id=\"fs-id1169738220653\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(a)=\\underset{x\\to a}{\\lim}\\dfrac{f(x)-f(a)}{x-a}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737140882\">We want to show that [latex]f(x)[\/latex] is continuous at [latex]a[\/latex] by showing that [latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex]. Thus,<\/p>\n<div id=\"fs-id1169738221978\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} \\underset{x\\to a}{\\lim}f(x) & =\\underset{x\\to a}{\\lim}(f(x)-f(a)+f(a)) & & & \\\\ & =\\underset{x\\to a}{\\lim}(\\frac{f(x)-f(a)}{x-a}\\cdot (x-a)+f(a)) & & & \\text{Multiply and divide} \\, f(x)-f(a) \\, \\text{by} \\, x-a. \\\\ & =(\\underset{x\\to a}{\\lim}\\frac{f(x)-f(a)}{x-a}) \\cdot (\\underset{x\\to a}{\\lim}(x-a))+\\underset{x\\to a}{\\lim}f(a) & & & \\\\ & =f^{\\prime}(a) \\cdot 0+f(a) & & & \\\\ & =f(a). & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738226778\">Therefore, since [latex]f(a)[\/latex] is defined and [latex]\\underset{x\\to a}{\\lim}f(x)=f(a)[\/latex], we conclude that [latex]f[\/latex] is continuous at [latex]a[\/latex].<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1169737978576\">We have just proven that differentiability implies continuity, but now we consider whether continuity implies differentiability. To determine an answer to this question, we examine the function [latex]f(x)=|x|[\/latex]. This function is continuous everywhere; however, [latex]f^{\\prime}(0)[\/latex] is undefined. This observation leads us to believe that continuity does not imply differentiability. Let\u2019s explore further. For [latex]f(x)=|x|[\/latex],<\/p>\n<div id=\"fs-id1169738223612\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\dfrac{f(x)-f(0)}{x-0}=\\underset{x\\to 0}{\\lim}\\dfrac{|x|-|0|}{x-0}=\\underset{x\\to 0}{\\lim}\\dfrac{|x|}{x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738216093\">This limit does not exist because<\/p>\n<div id=\"fs-id1169738216096\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 0^-}{\\lim}\\dfrac{|x|}{x}=-1 \\, \\text{and} \\, \\underset{x\\to 0^+}{\\lim}\\dfrac{|x|}{x}=1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738063963\">See Figure 6.<\/p>\n<div style=\"width: 497px\" 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\/11205235\/CNX_Calc_Figure_03_02_006.jpg\" alt=\"The function f(x) = the absolute value of x is graphed. It consists of two straight line segments: the first follows the equation y = \u2212x and ends at the origin; the second follows the equation y = x and starts at the origin.\" width=\"487\" height=\"471\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. The function [latex]f(x)=|x|[\/latex] is continuous at 0 but is not differentiable at 0.<\/p>\n<\/div>\n<p id=\"fs-id1169738228170\">Let\u2019s consider some additional situations in which a continuous function fails to be differentiable. Consider the function [latex]f(x)=\\sqrt[3]{x}[\/latex]:<\/p>\n<div id=\"fs-id1169738228211\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\dfrac{\\sqrt[3]{x}-0}{x-0}=\\underset{x\\to 0}{\\lim}\\dfrac{1}{\\sqrt[3]{x^2}}=+\\infty[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738217227\">Thus [latex]f^{\\prime}(0)[\/latex] does not exist. A quick look at the graph of [latex]f(x)=\\sqrt[3]{x}[\/latex] clarifies the situation. The function has a vertical tangent line at 0 (Figure 7).<\/p>\n<div style=\"width: 497px\" 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\/11205237\/CNX_Calc_Figure_03_02_007.jpg\" alt=\"The function f(x) = the cube root of x is graphed. It has a vertical tangent at x = 0.\" width=\"487\" height=\"275\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. The function [latex]f(x)=\\sqrt[3]{x}[\/latex] has a vertical tangent at [latex]x=0[\/latex]. It is continuous at 0 but is not differentiable at 0.<\/p>\n<\/div>\n<p id=\"fs-id1169738217510\">The function [latex]f(x)=\\begin{cases} x \\sin\\left(\\frac{1}{x}\\right) & \\text{ if } \\, x \\ne 0 \\\\ 0 & \\text{ if } \\, x = 0 \\end{cases}[\/latex] also has a derivative that exhibits interesting behavior at 0. We see that<\/p>\n<div id=\"fs-id1169738225665\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\dfrac{x \\sin\\left(\\frac{1}{x}\\right)-0}{x-0}=\\underset{x\\to 0}{\\lim} \\sin\\left(\\dfrac{1}{x}\\right)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737946726\">This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approach zero (Figure 8).<\/p>\n<div style=\"width: 436px\" 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\/11205240\/CNX_Calc_Figure_03_02_008.jpg\" alt=\"The function f(x) = x sin (1\/2) if x does not equal 0 and f(x) = 0 if x = 0 is graphed. It looks like a rapidly oscillating sinusoidal function with amplitude decreasing to 0 at the origin.\" width=\"426\" height=\"391\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8. The function [latex]f(x)=\\begin{cases} x \\sin(\\frac{1}{x}) &amp; \\text{ if } \\, x \\ne 0 \\\\ 0 &amp; \\text{ if } \\, x = 0 \\end{cases}[\/latex] is not differentiable at 0.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738042233\">In summary:<\/p>\n<ol id=\"fs-id1169738042236\">\n<li>We observe that if a function is not continuous, it cannot be differentiable, since every differentiable function must be continuous. However, if a function is continuous, it may still fail to be differentiable.<\/li>\n<li>We saw that [latex]f(x)=|x|[\/latex] failed to be differentiable at 0 because the limit of the slopes of the tangent lines on the left and right were not the same. Visually, this resulted in a sharp corner on the graph of the function at 0. From this we conclude that in order to be differentiable at a point, a function must be \u201csmooth\u201d at that point.<\/li>\n<li>As we saw in the example of [latex]f(x)=\\sqrt[3]{x}[\/latex], a function fails to be differentiable at a point where there is a vertical tangent line.<\/li>\n<li>As we saw with [latex]f(x)=\\begin{cases} x \\sin(\\frac{1}{x}) & \\text{ if } \\, x \\ne 0 \\\\ 0 & \\text{ if } \\, x = 0 \\end{cases}[\/latex] a function may fail to be differentiable at a point in more complicated ways as well.<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1169738218186\" class=\"textbook exercises\">\n<h3>Example: A Piecewise Function that is Continuous and Differentiable<\/h3>\n<p id=\"fs-id1169738218195\">A toy company wants to design a track for a toy car that starts out along a parabolic curve and then converts to a straight line (Figure 9). The function that describes the track is to have the form [latex]f(x)=\\begin{cases} \\frac{1}{10}x^2 + bx + c & \\text{ if } \\, x < -10 \\\\ -\\frac{1}{4}x + \\frac{5}{2} & \\text{ if } \\, x \\ge -10 \\end{cases}[\/latex], where [latex]x[\/latex] and [latex]f(x)[\/latex] are in inches. For the car to move smoothly along the track, the function [latex]f(x)[\/latex] must be both continuous and differentiable at -10. Find values of [latex]b[\/latex] and [latex]c[\/latex] that make [latex]f(x)[\/latex] both continuous and differentiable.<\/p>\n<div style=\"width: 497px\" 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\/11205243\/CNX_Calc_Figure_03_02_009.jpg\" alt=\"A cart is drawn on a line that curves through (\u221210, 5) to (10, 0) with y-intercept roughly (0, 2).\" width=\"487\" height=\"237\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 9. For the car to move smoothly along the track, the function must be both continuous and differentiable.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738222036\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738222036\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738222036\">For the function to be continuous at [latex]x=-10, \\, \\underset{x\\to 10^-}{\\lim}f(x)=f(-10)[\/latex]. Thus, since<\/p>\n<div id=\"fs-id1169737950192\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \u221210^-}{\\lim}f(x)=\\frac{1}{10}(-10)^2-10b+c=10-10b+c[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738216001\">and [latex]f(-10)=5[\/latex], we must have [latex]10-10b+c=5[\/latex]. Equivalently, we have [latex]c=10b-5[\/latex].<\/p>\n<p id=\"fs-id1169738219145\">For the function to be differentiable at -10,<\/p>\n<div id=\"fs-id1169738219154\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(10)=\\underset{x\\to \u221210}{\\lim}\\frac{f(x)-f(-10)}{x+10}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738224010\">must exist. Since [latex]f(x)[\/latex] is defined using different rules on the right and the left, we must evaluate this limit from the right and the left and then set them equal to each other:<\/p>\n<div id=\"fs-id1169738183518\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} \\underset{x\\to \u221210^-}{\\lim}\\frac{f(x)-f(-10)}{x+10} & =\\underset{x\\to \u221210^-}{\\lim}\\frac{\\frac{1}{10}x^2+bx+c-5}{x+10} & & & \\\\ & =\\underset{x\\to \u221210^-}{\\lim}\\frac{\\frac{1}{10}x^2+bx+(10b-5)-5}{x+10} & & & \\text{Substitute} \\, c=10b-5. \\\\ & =\\underset{x\\to \u221210^-}{\\lim}\\frac{x^2-100+10bx+100b}{10(x+10)} & & & \\\\ & =\\underset{x\\to \u221210^-}{\\lim}\\frac{(x+10)(x-10+10b)}{10(x+10)} & & & \\text{Factor by grouping.} \\\\ & =b-2 & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738183170\">We also have<\/p>\n<div id=\"fs-id1169738183173\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\underset{x\\to \u221210^+}{\\lim}\\frac{f(x)-f(-10)}{x+10} & =\\underset{x\\to \u221210^+}{\\lim}\\frac{-\\frac{1}{4}x+\\frac{5}{2}-5}{x+10} \\\\ & =\\underset{x\\to \u221210^+}{\\lim}\\frac{\u2212(x+10)}{4(x+10)} \\\\ & =-\\frac{1}{4} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738221364\">This gives us [latex]b-2=-\\frac{1}{4}[\/latex].<\/p>\n<p>Thus [latex]b=\\frac{7}{4}[\/latex] and [latex]c=10(\\frac{7}{4})-5=\\frac{25}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: A Piecewise Function that is Continuous and Differentiable.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/w9m4h4i47-M?controls=0&amp;start=904&amp;end=1275&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.2TheDerivativeAsAFunction904to1275_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.2 The Derivative as a Function&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1169738185315\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169738185323\">Find values of [latex]a[\/latex] and [latex]b[\/latex] that make [latex]f(x)=\\begin{cases} ax+b & \\text{ if } \\, x < 3 \\\\ x^2 & \\text{ if } \\, x \\ge 3 \\end{cases}[\/latex] both continuous and differentiable at 3.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738225596\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738225596\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738225596\">[latex]a=6[\/latex] and [latex]b=-9[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1169738225624\">Use the previous example as a guide.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738225636\" class=\"bc-section section\"><\/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-332\">\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>3.2 The Derivative as a Function. <strong>Authored by<\/strong>: Ryan Melton. <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>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/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":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) 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