{"id":2429,"date":"2024-05-22T14:52:09","date_gmt":"2024-05-22T14:52:09","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=2429"},"modified":"2024-08-05T12:39:48","modified_gmt":"2024-08-05T12:39:48","slug":"continuity-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/continuity-learn-it-4\/","title":{"raw":"Continuity: Learn It 4","rendered":"Continuity: Learn It 4"},"content":{"raw":"<h2>Composite Function Theorem<\/h2>\r\n<p>The Composite Function Theorem helps expand our ability to compute limits, particularly demonstrating the continuity of trigonometric functions over their domains.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>composite function theorem<\/h3>\r\n<p id=\"fs-id1170573762934\">If [latex]f(x)[\/latex] is continuous at [latex]L[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=L[\/latex], then<\/p>\r\n<div id=\"fs-id1170573400410\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}f(g(x))=f(\\underset{x\\to a}{\\lim}g(x))=f(L)[\/latex].<\/div>\r\n<\/section>\r\n<p>This theorem allows us to demonstrate that the composition of functions is continuous if the inner function approaches a limit where the outer function is continuous.<\/p>\r\n<section class=\"textbox recall\">\r\n<p>Before we move on to the next example, recall that earlier, in the section on limit laws, we showed [latex]\\underset{x\\to 0}{\\lim} \\cos x=1= \\cos (0)[\/latex]. Consequently, we know that [latex]f(x)= \\cos x[\/latex] is continuous at [latex]0[\/latex]. In the next example we see how to combine this result with the composite function theorem.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Evaluate [latex]\\underset{x\\to \\pi\/2}{\\lim}\\cos(x-\\dfrac{\\pi }{2})[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1170573408578\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170573408578\"]<\/p>\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\/2}{\\lim}(x-\\frac{\\pi}{2})=0[\/latex] and [latex]\\cos x[\/latex] is continuous at [latex]0[\/latex], we may apply the composite function theorem. Thus,<\/p>\r\n<div id=\"fs-id1170573570975\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pi\/2}{\\lim}\\cos(x-\\frac{\\pi}{2})= \\cos (\\underset{x\\to \\pi\/2}{\\lim}(x-\\frac{\\pi}{2}))= \\cos (0)=1[\/latex].<br \/>\r\n[\/hidden-answer]<\/div>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170571131881\">Evaluate [latex]\\underset{x\\to \\pi}{\\lim}\\sin(x-\\pi)[\/latex].<\/p>\r\n<p>[reveal-answer q=\"790361\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"790361\"]<\/p>\r\n<p id=\"fs-id1170573359398\">[latex]f(x)= \\sin x[\/latex] is continuous at [latex]0[\/latex].\u00a0<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1170573362583\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170573362583\"]<\/p>\r\n<p id=\"fs-id1170573362583\">[latex]0[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox watchIt\">\r\n<p>[caption]Watch the following video to see examples of solving limits of composite functions. [\/caption]<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/RgfKNIkpFWc?controls=0\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]<br \/>\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.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/LimitsOfCompositeFunctions_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Limits of composite functions | Limits and continuity | AP Calculus AB | Khan Academy\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p id=\"fs-id1170571101566\">The proof of the next theorem uses the composite function theorem and the continuity of [latex]f(x)= \\sin x[\/latex] and [latex]g(x)= \\cos x[\/latex] at the point [latex]0[\/latex] to demonstrate that trigonometric functions are continuous over their entire domains.<\/p>\r\n<section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n<hr \/>\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}{\\lim}\\cos x = \\cos a[\/latex] for all values of [latex]a[\/latex].<\/p>\r\n<p id=\"fs-id1170571099777\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\underset{x\\to a}{\\lim}\\cos x &amp; =\\underset{x\\to a}{\\lim}\\cos((x-a)+a) &amp; &amp; &amp; \\text{rewrite} \\, x \\, \\text{as} \\, x-a+a \\, \\text{and group} \\, (x-a) \\\\ &amp; =\\underset{x\\to a}{\\lim}(\\cos(x-a)\\cos a - \\sin(x-a)\\sin a) &amp; &amp; &amp; \\text{apply the identity for the cosine of the sum of two angles} \\\\ &amp; = \\cos(\\underset{x\\to a}{\\lim}(x-a)) \\cos a - \\sin(\\underset{x\\to a}{\\lim}(x-a))\\sin a &amp; &amp; &amp; \\underset{x\\to a}{\\lim}(x-a)=0, \\, \\text{and} \\, \\sin x \\, \\text{and} \\, \\cos x \\, \\text{are continuous at 0} \\\\ &amp; = \\cos(0)\\cos a - \\sin(0)\\sin a &amp; &amp; &amp; \\text{evaluate cos(0) and sin(0) and simplify} \\\\ &amp; =1 \\cdot \\cos a - 0 \\cdot \\sin a = \\cos a \\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>[latex]_\\blacksquare[\/latex]<\/p>\r\n<\/section>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>continuity of trigonometric functions<\/h3>\r\n<p>Trigonometric functions are continuous over their entire domains.<\/p>\r\n<\/section>\r\n<h2 id=\"fs-id1170573370285\">The Intermediate Value Theorem<\/h2>\r\n<p id=\"fs-id1170573717714\">Functions that are continuous over intervals of the form [latex][a,b][\/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<section class=\"textbox keyTakeaway\">\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][a,b][\/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][a,b][\/latex] satisfying [latex]f(c)=z[\/latex].\u00a0<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"400\"]<img class=\"\" 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=\"400\" height=\"370\" \/> Figure 7. There is a number [latex]c \\in [a,b][\/latex] that satisfies [latex]f(c)=z[\/latex].[\/caption]\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170570997458\">Show that [latex]f(x)=x- \\cos x[\/latex] has at least one zero.<\/p>\r\n<p>[reveal-answer q=\"fs-id1170571136342\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170571136342\"]<\/p>\r\n<p id=\"fs-id1170571136342\">Since [latex]f(x)=x-\\cos x[\/latex] is continuous over [latex](\u2212\\infty,+\\infty)[\/latex], it is continuous over any closed interval of the form [latex][a,b][\/latex]. If you can find an interval [latex][a,b][\/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<div id=\"fs-id1170571239008\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(0)=0 - \\cos (0)=-1&lt;0[\/latex]<\/div>\r\n<p id=\"fs-id1170573424328\">and<\/p>\r\n<div id=\"fs-id1170573424331\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(\\frac{\\pi}{2})=\\frac{\\pi}{2} - \\cos \\frac{\\pi}{2}=\\frac{\\pi}{2}&gt;0[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\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][0,\\pi\/2][\/latex] that satisfies [latex]f(c)=0[\/latex]. Therefore, [latex]f(x)=x- \\cos x[\/latex] has at least one zero.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170573413598\">If [latex]f(x)[\/latex] is continuous over [latex][0,2], \\, 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][0,2][\/latex]? Explain.<\/p>\r\n<p>[reveal-answer q=\"fs-id1170571262087\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170571262087\"]<\/p>\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<p>[caption]Watch the following video to see the worked solution to this example. [\/caption]<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/BXUu5bG1CXU?controls=0&amp;start=804&amp;end=860&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]<br \/>\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.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.4Continuity804to860_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.4 Continuity\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\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][0,1][\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1170573382710\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170573382710\"]<\/p>\r\n<p id=\"fs-id1170573382710\">[latex]f(0)=1&gt;0, \\, f(1)=-2&lt;0[\/latex]; [latex]f(x)[\/latex] is continuous over [latex][0,1][\/latex]. It must have a zero on this interval.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]288280[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<h2>Composite Function Theorem<\/h2>\n<p>The Composite Function Theorem helps expand our ability to compute limits, particularly demonstrating the continuity of trigonometric functions over their domains.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>composite function theorem<\/h3>\n<p id=\"fs-id1170573762934\">If [latex]f(x)[\/latex] is continuous at [latex]L[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=L[\/latex], then<\/p>\n<div id=\"fs-id1170573400410\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}f(g(x))=f(\\underset{x\\to a}{\\lim}g(x))=f(L)[\/latex].<\/div>\n<\/section>\n<p>This theorem allows us to demonstrate that the composition of functions is continuous if the inner function approaches a limit where the outer function is continuous.<\/p>\n<section class=\"textbox recall\">\n<p>Before we move on to the next example, recall that earlier, in the section on limit laws, we showed [latex]\\underset{x\\to 0}{\\lim} \\cos x=1= \\cos (0)[\/latex]. Consequently, we know that [latex]f(x)= \\cos x[\/latex] is continuous at [latex]0[\/latex]. In the next example we see how to combine this result with the composite function theorem.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Evaluate [latex]\\underset{x\\to \\pi\/2}{\\lim}\\cos(x-\\dfrac{\\pi }{2})[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170573408578\">Show Solution<\/button><\/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\/2}{\\lim}(x-\\frac{\\pi}{2})=0[\/latex] and [latex]\\cos x[\/latex] is continuous at [latex]0[\/latex], we may apply the composite function theorem. Thus,<\/p>\n<div id=\"fs-id1170573570975\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pi\/2}{\\lim}\\cos(x-\\frac{\\pi}{2})= \\cos (\\underset{x\\to \\pi\/2}{\\lim}(x-\\frac{\\pi}{2}))= \\cos (0)=1[\/latex].\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170571131881\">Evaluate [latex]\\underset{x\\to \\pi}{\\lim}\\sin(x-\\pi)[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q790361\">Hint<\/button><\/p>\n<div id=\"q790361\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573359398\">[latex]f(x)= \\sin x[\/latex] is continuous at [latex]0[\/latex].\u00a0<\/p>\n<\/div>\n<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170573362583\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170573362583\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573362583\">[latex]0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\">\nWatch the following video to see examples of solving limits of composite functions. <\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/RgfKNIkpFWc?controls=0\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/button><\/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\/LimitsOfCompositeFunctions_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Limits of composite functions | Limits and continuity | AP Calculus AB | Khan Academy&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1170571101566\">The proof of the next theorem uses the composite function theorem and the continuity of [latex]f(x)= \\sin x[\/latex] and [latex]g(x)= \\cos x[\/latex] at the point [latex]0[\/latex] to demonstrate that trigonometric functions are continuous over their entire domains.<\/p>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\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}{\\lim}\\cos x = \\cos a[\/latex] for all values of [latex]a[\/latex].<\/p>\n<p id=\"fs-id1170571099777\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\underset{x\\to a}{\\lim}\\cos x & =\\underset{x\\to a}{\\lim}\\cos((x-a)+a) & & & \\text{rewrite} \\, x \\, \\text{as} \\, x-a+a \\, \\text{and group} \\, (x-a) \\\\ & =\\underset{x\\to a}{\\lim}(\\cos(x-a)\\cos a - \\sin(x-a)\\sin a) & & & \\text{apply the identity for the cosine of the sum of two angles} \\\\ & = \\cos(\\underset{x\\to a}{\\lim}(x-a)) \\cos a - \\sin(\\underset{x\\to a}{\\lim}(x-a))\\sin a & & & \\underset{x\\to a}{\\lim}(x-a)=0, \\, \\text{and} \\, \\sin x \\, \\text{and} \\, \\cos x \\, \\text{are continuous at 0} \\\\ & = \\cos(0)\\cos a - \\sin(0)\\sin a & & & \\text{evaluate cos(0) and sin(0) and simplify} \\\\ & =1 \\cdot \\cos a - 0 \\cdot \\sin a = \\cos a \\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>[latex]_\\blacksquare[\/latex]<br \/>\n<\/section>\n<section class=\"textbox keyTakeaway\">\n<h3>continuity of trigonometric functions<\/h3>\n<p>Trigonometric functions are continuous over their entire domains.<\/p>\n<\/section>\n<h2 id=\"fs-id1170573370285\">The Intermediate Value Theorem<\/h2>\n<p id=\"fs-id1170573717714\">Functions that are continuous over intervals of the form [latex][a,b][\/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<section class=\"textbox keyTakeaway\">\n<h3>the intermediate value theorem<\/h3>\n<p id=\"fs-id1170573534418\">Let [latex]f[\/latex] be continuous over a closed, bounded interval [latex][a,b][\/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][a,b][\/latex] satisfying [latex]f(c)=z[\/latex].\u00a0<\/p>\n<figure style=\"width: 400px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" 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=\"400\" height=\"370\" \/><figcaption class=\"wp-caption-text\">Figure 7. There is a number [latex]c \\in [a,b][\/latex] that satisfies [latex]f(c)=z[\/latex].<\/figcaption><\/figure>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170570997458\">Show that [latex]f(x)=x- \\cos x[\/latex] has at least one zero.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571136342\">Show Solution<\/button><\/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](\u2212\\infty,+\\infty)[\/latex], it is continuous over any closed interval of the form [latex][a,b][\/latex]. If you can find an interval [latex][a,b][\/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\" style=\"text-align: center;\">[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\" style=\"text-align: center;\">[latex]f(\\frac{\\pi}{2})=\\frac{\\pi}{2} - \\cos \\frac{\\pi}{2}=\\frac{\\pi}{2}>0[\/latex]<\/div>\n<p>&nbsp;<\/p>\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][0,\\pi\/2][\/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<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170573413598\">If [latex]f(x)[\/latex] is continuous over [latex][0,2], \\, 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][0,2][\/latex]? Explain.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571262087\">Show Solution<\/button><\/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<p>Watch the following video to see the worked solution to this example. <\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/BXUu5bG1CXU?controls=0&amp;start=804&amp;end=860&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/button><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\"><\/div>\n<\/div>\n<p>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\/2.4Continuity804to860_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.4 Continuity&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170571048766\">Show that [latex]f(x)=x^3-x^2-3x+1[\/latex] has a zero over the interval [latex][0,1][\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170573382710\">Show Solution<\/button><\/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[\/latex]; [latex]f(x)[\/latex] is continuous over [latex][0,1][\/latex]. It must have a zero on this interval.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288280\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288280&theme=lumen&iframe_resize_id=ohm288280&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":15,"menu_order":13,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":185,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2429"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":14,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2429\/revisions"}],"predecessor-version":[{"id":4656,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2429\/revisions\/4656"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/185"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2429\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=2429"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=2429"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=2429"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=2429"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}