{"id":86380,"date":"2021-07-19T16:44:08","date_gmt":"2021-07-19T21:44:08","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=86380"},"modified":"2026-04-23T11:05:16","modified_gmt":"2026-04-23T16:05:16","slug":"mean-value-theorem","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/mean-value-theorem\/","title":{"rendered":"Mean Value Theorem"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_6i09KylL-Nc\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_6i09KylL-Nc\" data-source-videoID=\"6i09KylL-Nc\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Mean Value Theorem Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Mean Value Theorem\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_6i09KylL-Nc:hover {cursor:pointer;} img#videoThumbnailImage_6i09KylL-Nc {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/1763-thumb-final-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_6i09KylL-Nc\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_6i09KylL-Nc\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Mean Value Theorem\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_6i09KylL-Nc\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_6i09KylL-Nc\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_6i09KylL-Nc\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction TLt_Function() {\n  var x = document.getElementById(\"TLt\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"TLt_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"TLt\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Introduction_to_the_Mean_Value_Theorem\" class=\"smooth-scroll\">Introduction to the Mean Value Theorem<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Graphing_the_Mean_Value_Theorem\" class=\"smooth-scroll\">Graphing the Mean Value Theorem<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Rolle%E2%80%99s_Theorem\" class=\"smooth-scroll\">Rolle\u2019s Theorem<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Applying_the_Mean_Value_Theorem\" class=\"smooth-scroll\">Applying the Mean Value Theorem<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<h2><span id=\"Introduction_to_the_Mean_Value_Theorem\" class=\"m-toc-anchor\"><\/span>Introduction to the Mean Value Theorem<\/h2>\n<p>\nPicture this: You drive through a tollbooth at the start of a 20-mile stretch of road at 12:30 pm. The speed limit over the entire section is 60 miles per hour. You exit through the tollbooth on the other side at 12:48 pm only to find a police officer waiting to give you a speeding ticket.<\/p>\n<p>How did the officer know you were speeding without shooting your car with their radar gun? Calculus, of course!<\/p>\n<p>Since the speed limit is 60 miles per hour, the fastest you can legally go is one mile per minute (60 miles in 60 minutes). So, it should have taken at least 20 minutes to make it to the other side (you <em>can<\/em> go slower). But you arrive in 18 minutes, so at some point you must have gone faster than 60 miles per hour. No radar gun needed.<\/p>\n<p>The police officer applied the <strong>mean value theorem<\/strong>, which says for a continuous, differentiable function over a closed interval, at some point the instantaneous rate of change equals the average rate of change over the interval.<\/p>\n<p>Formally stated, the theorem says:<\/p>\n<div class=\"yellow-quote\" style=\"text-style: italic; font-weight: normal\">\n<p style=\"line-height: 40px\">If \\(f\\) is continuous over \\([a,b]\\) and differentiable over \\((a,b)\\), then there is a \\(c\\) in \\((a,b)\\) such that \\(f^{&#8216;}\\left ( c \\right )=\\large{\\frac{f\\left ( b \\right )-f\\left ( a \\right )}{b-a}}\\).<\/p>\n<\/div>\n<p>The police officer knew that if your average rate of change was 20 miles per 18 minutes, or 1.1 miles per minute, at some instant on your trip, you must have traveled 1.1 miles per minute (speeding).<\/p>\n<h2><span id=\"Graphing_the_Mean_Value_Theorem\" class=\"m-toc-anchor\"><\/span>Graphing the Mean Value Theorem<\/h2>\n<p>\nLet\u2019s explore the mean value theorem graphically using the points \\((-2, 1)\\) and \\((4, 4)\\). Connecting the points with a linear function clearly demonstrates the mean value theorem, since all points have an instantaneous rate of change of \\(\\frac{1}{2}\\), which is the average rate of change over the interval:<\/p>\n<div class=\"examplesentence\">\\(\\dfrac{4-1}{4-\\left ( -2 \\right )}=\\dfrac{3}{6}=\\dfrac{1}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\n<img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-1.png\" alt=\"A red line passes through two labeled points, (-2, 1) and (4, 4), on an x-y axis grid.\" width=\"421.44\" height=\"280.32\" class=\"aligncenter size-full wp-image-291962\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-1.png 878w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-1-300x200.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-1-768x511.png 768w\" sizes=\"(max-width: 878px) 100vw, 878px\" \/><\/p>\n<p>What if a curved function goes through those points? On the interval \\((-2, 4)\\), there must be at least one tangent line that is parallel to the secant line. Recall that a secant line is a straight line that intersects a curve in at least two different points.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-2.png\" alt=\"A graph showing a sine curve with a red secant line through points (\u20132,1) and (4,4), and two blue tangent lines at those points.\" width=\"502.32\" height=\"284.76\" class=\"aligncenter size-full wp-image-291965\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-2.png 1196w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-2-300x170.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-2-1024x580.png 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-2-768x435.png 768w\" sizes=\"(max-width: 1196px) 100vw, 1196px\" \/><\/p>\n<p>In this case, there are two tangent lines that are parallel to the secant line. You can imagine how the theorem would hold for functions of many different shapes.<\/p>\n<h2><span id=\"Rolle%E2%80%99s_Theorem\" class=\"m-toc-anchor\"><\/span>Rolle\u2019s Theorem<\/h2>\n<p>\nWhat if the endpoints of the interval are the same height? In the case of \\(f(x)=1\\) over the interval \\([-2, 2]\\), the average rate of change is zero, since there is no change in height. What about a non-linear function that passes through those same points?<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-3.png\" alt=\"A parabola with a blue tangent line at the vertex and a red secant line passing through points (-2,1) and (2,1); axes and grid in background.\" width=\"570.72\" height=\"294.72\" class=\"aligncenter size-full wp-image-291959\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-3.png 1189w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-3-300x155.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-3-1024x529.png 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/04\/Mean-value-theorem-graph-3-768x397.png 768w\" sizes=\"(max-width: 1189px) 100vw, 1189px\" \/><\/p>\n<p>The function \\(f(x)=-x^{2}+5\\) passes through those points and there is a \\(c\\) such that \\({f}'(c)=0\\).<\/p>\n<p>This \u201czero-slope secant line\u201d scenario actually demonstrates <strong>Rolle\u2019s theorem<\/strong>, which says if a continuous, differentiable function\u2019s average rate of change over a closed interval is zero, then at some point in the interval the instantaneous rate of change must be zero.<\/p>\n<p>Formally stated, here\u2019s Rolle\u2019s theorem:<\/p>\n<div class=\"yellow-quote\" style=\"text-style: italic; font-weight: normal\">If \\(f\\) is continuous over \\([a,b]\\) and differentiable over \\((a,b)\\), and \\(f(a)=f(b)\\) then there is a \\(c\\) in <span style=\"white-space:nowrap\">\\((a,b)\\)<\/span> such that \\({f}'(c)=0\\).<\/div>\n<h2><span id=\"Applying_the_Mean_Value_Theorem\" class=\"m-toc-anchor\"><\/span>Applying the Mean Value Theorem<\/h2>\n<p>\nHere\u2019s what these theorems often look like in practice.<\/p>\n<p>Let\u2019s take a look at this function:<\/p>\n<div class=\"examplesentence\">\\(f(x)=x^{2}-2x-6\\)<\/div>\n<p>\n&nbsp;<br \/>\nBefore we begin, first note that since \\(f(x)\\) is a polynomial function, it satisfies the conditions of continuity and differentiability in both theorems, since it is continuous and differentiable everywhere. There are times where functions will not satisfy these conditions, so be sure to verify them!<\/p>\n<h3><span id=\"Finding_with_Rolle%E2%80%99s_Theorem\" class=\"m-toc-anchor\"><\/span>Finding \\(c\\) with Rolle\u2019s Theorem<\/h3>\n<p>\nWhat value(s) of \\(c\\) satisfy Rolle\u2019s theorem over the interval \\([-1,3]\\)?<\/p>\n<p>To use Rolle\u2019s theorem, \\(f(a)\\) must equal \\(f(b)\\). So we\u2019ll take \\(f(-1)\\) and plug it into our equation:<\/p>\n<div class=\"examplesentence\">\\(f(-1)=(-1)^{2}-2(-1)-6\\)\\(\\:=1+2-6=-3\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd this must equal our other end, so now let\u2019s plug in \\(f(3)\\).<\/p>\n<div class=\"examplesentence\">\\(f(3)=3^{2}-2(3)-6\\)\\(\\:=9-6-6=-3\\)<\/div>\n<p>\n&nbsp;<br \/>\nSince this is true, there must be a point \\(c\\) where the derivative also equals 0. So we\u2019re gonna take the derivative of our function, we have:<\/p>\n<div class=\"examplesentence\">\\({f}'(x)=2x-2\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo now we want to find where it equals 0. So, remember, we\u2019re looking for the point \\(c\\), so we&#8217;re gonna put \\(c\\) in the place of \\(x\\).<\/p>\n<div class=\"examplesentence\">\\(0=2c-2\\)<\/div>\n<p>\n&nbsp;<br \/>\nAdd 2 to both sides.<\/p>\n<div class=\"examplesentence\">\\(2=2c\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then divide by 2 on both sides. So:<\/p>\n<div class=\"examplesentence\">\\(1=c\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn this case, Rolle\u2019s theorem and the mean value theorem can be interchanged.<\/p>\n<p>Here\u2019s the solution using the mean value theorem. So the mean value theorem says that \\(\\frac{f(b)-f(a)}{b-a}\\) will give us our point where the derivative equals 0.<\/p>\n<p>So \\(f(b)=f(3)\\).<\/p>\n<div class=\"examplesentence\">\\(\\dfrac{f(3)-f(-1)}{3-(-1)}=\\dfrac{(-3)-(-3)}{4}=0\\)<\/div>\n<p>\n&nbsp;<br \/>\nAccording to the mean value theorem, there must be a place where the derivative equals 0. So again, we\u2019re gonna take the derivative of this function.<\/p>\n<div class=\"examplesentence\">\\({f}'(x)=2x-2\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd since there must be a place where it equals 0, do:<\/p>\n<div class=\"examplesentence\">\\(0=2c-2\\)<\/div>\n<p>\n&nbsp;<br \/>\nAdd 2 to both sides.<\/p>\n<div class=\"examplesentence\">\\(2=2c\\)<\/div>\n<p>\n&nbsp;<br \/>\nDivide by 2 on both sides.<\/p>\n<div class=\"examplesentence\">\\(1=c\\)<\/div>\n<p>\n&nbsp;<br \/>\nSee, we get the same result either way, whether we use Rolle&#8217;s theorem or mean value theorem.<\/p>\n<h3><span id=\"Finding_with_the_Mean_Value_Theorem\" class=\"m-toc-anchor\"><\/span>Finding \\(c\\) with the Mean Value Theorem<\/h3>\n<p>\nWhat values of \\(c\\) satisfy the mean value theorem over the interval \\([-1,4]\\)?<\/p>\n<p>To use the mean value theorem, first find the slope of the secant line:<\/p>\n<div class=\"examplesentence\">\\(\\dfrac{f(b)-f(a)}{b-a}=\\dfrac{f(4)-f(-1)}{4-(-1)}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo first, let\u2019s find \\(f(4)\\).<\/p>\n<div class=\"examplesentence\">\\(f(4)=(4)^{2}-2(4)-6\\)\\(=16-8-6=2\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then:<\/p>\n<div class=\"examplesentence\">\\(f(-1)=(-1)^{2}-2(-1)-6\\)\\(=1+2-6=-3\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo now we can go back up here and plug these values in. So \\(f(4)=2\\), \\(f(1)=-3\\), so:<\/p>\n<div class=\"examplesentence\">\\(\\dfrac{2-(-3)}{5}=\\dfrac{5}{5}=1\\)<\/div>\n<p>\n&nbsp;<br \/>\nAt some point in the interval, there is a tangent line with a slope of 1. So we\u2019re gonna take the derivative of our function:<\/p>\n<div class=\"examplesentence\">\\({f}'(x)=2x-2\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd at some point it must equal 1, so:<\/p>\n<div class=\"examplesentence\">\\(1=2c-2\\)<\/div>\n<p>\n&nbsp;<br \/>\nAdd 2 to both sides.<\/p>\n<div class=\"examplesentence\">\\(3=2c\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd divide by 2 on both sides. So:<\/p>\n<div class=\"examplesentence\">\\(\\dfrac{3}{2}=c\\)<\/div>\n<p>\n&nbsp;<br \/>\nFor which value of \\(b\\) in the interval \\([-2,b]\\) is the mean value theorem satisfied at \\(c=0\\)?<\/p>\n<p>If the mean value theorem is satisfied, then the slope of the tangent line at \\(c=0\\) is the same as the slope of the secant line connecting the endpoints of the interval.<\/p>\n<p>Let\u2019s find the slope of the tangent line. Take the derivative of our function.<\/p>\n<div class=\"examplesentence\">\\({f}'(x)=2x-2\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then plug in 0 to find our derivative at \\(x=0\\).<\/p>\n<div class=\"examplesentence\">\\({f}'(0)=2(0)-2=0-2=-2\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe slope of the tangent line equals the slope of the secant line. So we\u2019re gonna take this \u22122 and set that equal to:<\/p>\n<div class=\"examplesentence\">\n<p style=\"margin-bottom: 1.2em\">\\(-2=\\dfrac{f(b)-f(-2)}{b-(-2)}\\)<\/p>\n<p class=\"longmath\" style=\"margin-bottom: 1.2em\">\\(-2=\\dfrac{(b^{2}-2b-6)-((-2)^{2}-2(-2)-6)}{b+2}\\)<\/p>\n<p style=\"margin-bottom: 1.2em\">\\(-2=\\dfrac{b^{2}-2b-6-(2)}{b+2}\\)<\/p>\n\\(-2=\\dfrac{b^{2}-2b-8}{b+2}\\)\n<\/div>\n<p>\n&nbsp;<br \/>\nThen, if we factor out our numerator, we get (our \\(b\\)\u2019s multiply together to get \\(b^{2}\\), \u22128 comes from \u22124 and +2):<\/p>\n<div class=\"examplesentence\">\\(-2=\\dfrac{(b-4)(b+2)}{b+2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur \\(b+2\\)\u2019s cancel out and we\u2019re left with:<\/p>\n<div class=\"examplesentence\">\\(-2=b-4\\)<\/div>\n<p>\n&nbsp;<br \/>\nAdd 4 to both sides, and we get:<\/p>\n<div class=\"examplesentence\">\\(b=2\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow let\u2019s take a look at a place where we might have gotten a little tripped up. So let\u2019s take a look at this step right here:<\/p>\n<div class=\"examplesentence\">\\(-2=\\dfrac{b^{2}-2b-8}{b+2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhat if we multiplied \\((b+2)\\times 2\\) and solved from there?<\/p>\n<div class=\"examplesentence\">\\(-2(b+2)=b^{2}-2b-8\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo if we distribute our \u22122, we get:<\/p>\n<div class=\"examplesentence\">\\(-2b-4=b^{2}-2b-8\\)<\/div>\n<p>\n&nbsp;<br \/>\nAdd \\(2b\\) to both sides.<\/p>\n<div class=\"examplesentence\">\\(-4=b^{2}-8\\)<\/div>\n<p>\n&nbsp;<br \/>\nAdd 8 to both sides.<\/p>\n<div class=\"examplesentence\">\\(4=b^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd take the square root of both sides. That gives us:<\/p>\n<div class=\"examplesentence\">\\(b=\\pm 2\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, there\u2019s a slight difference between our two answers. This answer (\\(b=2\\)) only gives us positive 2, and this answer (\\(b=\\pm 2\\)) includes negative 2. We can\u2019t include \u22122 because, if we look back, we have \\(b+2\\) in the denominator, so if we plug in \u22122, we\u2019ll get a 0 in the denominator, which means that it\u2019s undefined.<\/p>\n<p>So even though we found two solutions this way, we know that there&#8217;s really only one solution that we&#8217;re looking for. There are a couple other reasons this is true. Negative 2 must lie in the <em>open interval<\/em> we\u2019re considering, but \\((-2,-2)\\) is actually empty, and another reason is that since \\(f(x)\\) is quadratic, there\u2019s only one turning point. There can be two \\(c\\) values with two or more turning points. It\u2019s the same reason only one \\(c\\)-value was located in parts \\(a\\) and \\(b\\).<\/p>\n<p>That\u2019s it for the mean value theorem! I hope that this video helped with your understanding of this oh-so-important finding. You\u2019ll see that it pops up in various places as you discover more calculus! I hope this video was helpful. 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