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<h2 class="hd hd-2 unit-title">1. Motivation</h2>
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<h3 class="hd hd-2">Mean Value Theorem</h3>
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<h2 class="hd hd-2 unit-title">2. The Mean Value Theorem and some applications</h2>
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<p><b class="bfseries">Objectives</b></p><ul class="itemize"><li><p>
Know the hypothesis and conclusion of the <span style="color:#27408C"><b class="bf">Mean Value Theorem</b></span>. </p></li><li><p>
Use <span style="color:#27408C"><b class="bf">upper bounds</b></span> and <span style="color:#27408C"><b class="bf">lower bounds</b></span> on the derivative to establish inequalities between functions. </p></li></ul><p><b class="bfseries">Contents: 19 pages</b></p><p>
9 videos (33 minutes 1x speed) 35 questions </p>
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<h2 class="hd hd-2 unit-title">3. Exploration: Average vs instantaneous rate of change</h2>
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Review of the average rate of change
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<p>
Recall the definition of the average rate of change of a function [mathjaxinline]x(t)[/mathjaxinline] over an interval [mathjaxinline][a,b][/mathjaxinline]. </p>
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<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \text {Average rate of change } \ = \ \frac{x(b)-x(a)}{b-a}.[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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Geometrically, the average rate of change over [mathjaxinline][a,b][/mathjaxinline] is the slope of <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_antider_1-tab3-problem1_2_1">
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Review of instantaneous rate of change
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Recall the instantaneous rate of the change of the function [mathjaxinline]x(t)[/mathjaxinline] is the derivative: </p>
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<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle x'(t) \ = \ \lim _{\Delta t\rightarrow 0} \frac{ x(t+ \Delta t)- x(t)}{\Delta t}[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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Geometrically, the instantaneous rate of change at [mathjaxinline]t[/mathjaxinline] is the slope of <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_antider_1-tab3-problem3_2_1">
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Draw your answer, instantaneous rate of change
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Draw your answer from the previous problem on the graph below. </p>
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Comparing average and instantaneous rates of change
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On the graph below, the secant line through [mathjaxinline](a,x(a))[/mathjaxinline], [mathjaxinline](b,x(b))[/mathjaxinline] has the same slope as the tangent line(s) at which of the following point(s)? (Check all that apply.) </p>
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<text>[mathjaxinline]t_2[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">4. Identifying necessary hypotheses</h2>
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How the Mean Value Theorem can go wrong
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The MVT conclusion: <br/>"There is a point [mathjaxinline]c[/mathjaxinline], such that [mathjaxinline]a&lt;c&lt;b[/mathjaxinline], at which the tangent line is parallel to the secant line through [mathjaxinline](a,x(a))[/mathjaxinline] and [mathjaxinline](b,x(b))[/mathjaxinline]." </p>
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We may abbreviate "such that" with "s.t." from now on.<br/></p>
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For which of the graphs below is the MVT conclusion <b class="bfseries">false</b>? </p>
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Note that solid points are the end points. Dotted lines are asymptotes. </p>
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<h2 class="hd hd-2 unit-title">5. Statement of the Mean Value Theorem</h2>
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If [mathjaxinline]x(t)[/mathjaxinline] is continuous on [mathjaxinline]a\leq t \leq b[/mathjaxinline], and differentiable on [mathjaxinline]a<t<b[/mathjaxinline], that is, [mathjaxinline]x'(t)[/mathjaxinline] is defined for all [mathjaxinline]t[/mathjaxinline], [mathjaxinline]\, \, a<t<b[/mathjaxinline], then <br/></p><table id="a0000000005" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \frac{x(b)-x(a)}{b-a} \, = x'(c) \qquad \text {for some }c,\, \, \text {with } a<c<b.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Equivalently, in geometric terms, there is at least one point [mathjaxinline]c[/mathjaxinline], with [mathjaxinline]a<c<b[/mathjaxinline], at which the tangent line is parallel to the secant line through [mathjaxinline](a, x(a))[/mathjaxinline] and [mathjaxinline](b, x(b))[/mathjaxinline]: </p><center><img src="/assets/courseware/v1/ff1709fc4f125c8fffb302d051212c2d/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_xtgraph_MVT_statement.svg" width="280px" alt="An increasing, concave down curve is plotted in the x versus t plane. The points on the horizontal axis are indicated by a, c, and b where a is less c and c is less than b. The tangent line to the graph is drawn at c comma x of c." style="margin: 10px 25px 25px 25px"/></center>
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The logic of the MVT
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The following graph has a discontinuity within the interval [mathjaxinline][a,b][/mathjaxinline]. </p>
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Is there a point [mathjaxinline]c[/mathjaxinline] with [mathjaxinline]a&lt;c&lt;b[/mathjaxinline] at which the tangent line is parallel to the secant line through [mathjaxinline](a,x(a))[/mathjaxinline] and [mathjaxinline](b,x(b))[/mathjaxinline]? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="inputtype option-input ">
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What does this say about the MVT? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_antider_1-tab5-problem1_3_1">
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<input type="radio" name="input_antider_1-tab5-problem1_3_1" id="input_antider_1-tab5-problem1_3_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="antider_1-tab5-problem1_3_1-choice_2-label" for="input_antider_1-tab5-problem1_3_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab5-problem1_3_1"> <text> This graph does not satisfy the MVT hypothesis but satisfies the MVT conclusion. This example does not contradict the MVT.</text>
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<input type="radio" name="input_antider_1-tab5-problem1_3_1" id="input_antider_1-tab5-problem1_3_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="antider_1-tab5-problem1_3_1-choice_3-label" for="input_antider_1-tab5-problem1_3_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab5-problem1_3_1"> <text> This graph shows that the MVT statement above must have the wrong hypothesis.</text>
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<input type="radio" name="input_antider_1-tab5-problem1_3_1" id="input_antider_1-tab5-problem1_3_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="antider_1-tab5-problem1_3_1-choice_4-label" for="input_antider_1-tab5-problem1_3_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab5-problem1_3_1"> <text> This graph shows that the MVT statement above must have the wrong conclusion.</text>
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<p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Proof of MVT<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
To prove the Mean Value Theorem, we will start by proving a special case in which the function has the same values at the two end points, and then use this special case to prove the full theorem.<br/></p><p>
To prove the special case, we will rely on the <span style="color:#27408C"><b class="bf">Extreme Value Theorem</b></span>, which says that any function which is continuous on a closed interval must attain both its maximum and minimum values in that closed interval. <br/>This theorem requires deeper analysis of the real numbers and we will not prove it here. The point is that we need continuity to guarantee that the function attains both its maximum and minimum.<br/></p><p><b class="bfseries">Proof of the special case</b></p><p>
Suppose a function [mathjaxinline]x_0(t)[/mathjaxinline] satisfies the hypothesis of the MVT, that is, [mathjaxinline]x_0(t)[/mathjaxinline] is continuous on [mathjaxinline][a,b][/mathjaxinline], and differentiable on [mathjaxinline](a,b)[/mathjaxinline].<br/></p><p>
In this special case, suppose also that [mathjaxinline]x_0(a)=x_0(b)[/mathjaxinline]. By the Extreme Value Theorem, [mathjaxinline]x_0(t)[/mathjaxinline] attains both its maximum and minimum in [mathjaxinline][a,b][/mathjaxinline]. In other words, there is at least one point [mathjaxinline]t_1[/mathjaxinline] in [mathjaxinline][a,b][/mathjaxinline] such that [mathjaxinline]\displaystyle \ x_0(t_1)\, =\, \min _{a\leq t\leq b} x_0(t)[/mathjaxinline], and at least one point [mathjaxinline]t_2[/mathjaxinline] in [mathjaxinline][a,b][/mathjaxinline] such that [mathjaxinline]\displaystyle \ x_0(t_2)\, =\, \max _{a\leq t\leq b} x_0(t)[/mathjaxinline]. <br/></p><p>
There are only two possibilities. The maximum and minimum are either equal or not.<br/></p><dl class="description"><dt>Case 1: [mathjaxinline]\displaystyle \ \max _{a\leq t\leq b} x_0(t)\, =\, \min _{a\leq t\leq b} x_0(t)[/mathjaxinline]</dt><dd><p>
[mathjaxinline]x_0(t)[/mathjaxinline] must be constant over [mathjaxinline][a,b][/mathjaxinline], so [mathjaxinline]x'_0(t)=0[/mathjaxinline] for all [mathjaxinline]a<t<b[/mathjaxinline]. In particular, there is at least one point [mathjaxinline]c[/mathjaxinline], with [mathjaxinline]a<c<b[/mathjaxinline] at which [mathjaxinline]x'_0(c)=0[/mathjaxinline]. </p></dd><dt>Case 2: [mathjaxinline]\displaystyle \ \max _{a\leq t\leq b} x_0(t)\, \neq \min _{a\leq t\leq b} x_0(t)[/mathjaxinline]</dt><dd><p>
Since [mathjaxinline]x_0(a)=x_0(b)[/mathjaxinline], they cannot both be at the end points. Hence at least one of [mathjaxinline]\displaystyle \max _{a\leq t\leq b} x_0(t)[/mathjaxinline] and [mathjaxinline]\displaystyle \, \min _{a\leq t\leq b} x_0(t)[/mathjaxinline] must be achieved in [mathjaxinline](a,b)[/mathjaxinline]. Hence, there must be a [mathjaxinline]c[/mathjaxinline], with [mathjaxinline]a<c<b[/mathjaxinline] such that [mathjaxinline]x_0(c)=\max _{a\leq t\leq b} x_0(t)[/mathjaxinline] or [mathjaxinline]x_0(c)=\min _{a\leq t\leq b} x_0(t)[/mathjaxinline]. Now, recall the derivative of a differentiable function at a local maximum or minimum. By the hypothesis, [mathjaxinline]x_0(t)[/mathjaxinline] is differentiable in [mathjaxinline](a,b)[/mathjaxinline], so [mathjaxinline]x'_0(c)=0[/mathjaxinline] since [mathjaxinline]c[/mathjaxinline] is either a local maximum or a minimum. </p></dd></dl><p>
In both cases, there is a point [mathjaxinline]c[/mathjaxinline], with [mathjaxinline]a<c<b[/mathjaxinline], such that </p><table id="a0000000006" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle x'_0(c)\, =\, 0 \, =\, \frac{0}{b-a} \, =\, \frac{x_0(b)-x_0(a)}{b-a}\qquad \textrm{since}\, x_0(a)=x_0(b).[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
This special case of the MVT is called <span style="color:#27408C"><b class="bf">Rolle's Theorem.</b></span><br/></p><p>
Let us now use the special case above to prove the MVT for functions with possibly different endpoint values. </p><p>
Suppose a function [mathjaxinline]x(t)[/mathjaxinline] satisfies the hypothesis of the MVT, that is, [mathjaxinline]x(t)[/mathjaxinline] is continuous on [mathjaxinline][a,b][/mathjaxinline], and differentiable on [mathjaxinline](a,b)[/mathjaxinline]. Let </p><table id="a0000000007" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle x_0(t)\, =\, x(t)-\left(x(a)+ \frac{x(b)-x(a)}{b-a}(t-a)\right).[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
That is, construct a function [mathjaxinline]x_0(t)[/mathjaxinline] by subtracting from [mathjaxinline]x(t)[/mathjaxinline] the line that goes through [mathjaxinline](a,x(a))[/mathjaxinline], [mathjaxinline](b,x(b))[/mathjaxinline]. Then [mathjaxinline]x_0(t)[/mathjaxinline] also satisfies the hypothesis of the MVT, and [mathjaxinline]x_0(a)=x_0(b)=0[/mathjaxinline]. So we can apply Rolle's Theorem to [mathjaxinline]x_0(t)[/mathjaxinline], and know that there is a [mathjaxinline]c[/mathjaxinline] in [mathjaxinline](a, b)[/mathjaxinline], such that [mathjaxinline]x_0'(c)=0[/mathjaxinline].<br/>Now we can rearrange the equation above and get [mathjaxinline]x(t)[/mathjaxinline] in terms of [mathjaxinline]x_0(t)[/mathjaxinline]. </p><table id="a0000000008" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000009"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle x(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle x_0(t)+ \left(x(a)+ \frac{x(b)-x(a)}{b-a}(t-a)\right)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
Taking the derivative on both sides. we get </p><table id="a0000000010" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000011"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle x'(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle x_0'(t)\, + \, \frac{x(b)-x(a)}{b-a}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
And at the point [mathjaxinline]c[/mathjaxinline] in [mathjaxinline](a,b)[/mathjaxinline] at which [mathjaxinline]x_0'(c)=0[/mathjaxinline], the equation above reduces to </p><table id="a0000000012" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000013"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle x'(c)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle x_0'(c)+ \frac{x(b)-x(a)}{b-a}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000014"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{x(b)-x(a)}{b-a}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
Thus, we have found a [mathjaxinline]c[/mathjaxinline] we need for the conclusion of the MVT. </p></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><p><b class="bf">Supplementary link:</b> There is a nice visualization tool @EmilyMcKee made; the tool lets you explore what the existence of [mathjaxinline]c[/mathjaxinline], explained in this proof, actually looks like when visualized in a graphic. The tool is at <a href="https://www.desmos.com/calculator/vdhfhltxcr" target="_blank">this link</a>, it was made using Desmos graphing calculator and enables you to play with it changing the parameters as you wish. </p><SCRIPT src="/assets/courseware/v1/631e447105fca1b243137b21b9ed6f90/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/latex2edx.js" type="text/javascript"/><LINK href="/assets/courseware/v1/daf81af0af57b85a105e0ed27b7873a0/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/latex2edx.css" rel="stylesheet" type="text/css"/>
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Time-position graph of the speeding car
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<p>
Recall in the example of the speeding car, the only information the police had was that our car was at the [mathjaxinline]50[/mathjaxinline] mile marker at [mathjaxinline]8[/mathjaxinline] a.m., and [mathjaxinline]220[/mathjaxinline] mile marker at [mathjaxinline]10[/mathjaxinline] a.m. </p>
<p>
Let [mathjaxinline]x(t)[/mathjaxinline] be the position of the car at time [mathjaxinline]t[/mathjaxinline] and let the units of [mathjaxinline]x[/mathjaxinline] be miles and [mathjaxinline]t[/mathjaxinline] be hours, so that [mathjaxinline]x(8)=50[/mathjaxinline], and [mathjaxinline]x(10)=220[/mathjaxinline]. </p>
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Which of the following graphs can be the graph of [mathjaxinline]x(t)[/mathjaxinline]? Check all that applies. </p>
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When 85mph?
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<p>
Recall the average velocity of the car between 8 and 10 am is [mathjaxinline]85[/mathjaxinline] mph.<br/></p>
<p>
According to the MVT, the strongest conclusion the police officer could make about when the car is traveling at exactly [mathjaxinline]85[/mathjaxinline] mph is </p>
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<p style="display:inline">There is/are </p>
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<h2 class="hd hd-2 unit-title">7. Application to simultaneous rates</h2>
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Let us use the MVT to obtain a qualitative result on simultaneous rates of change. </p><p>
Suppose that we have two tanks each of volume [mathjaxinline]6000[/mathjaxinline] liters. Initially, both are empty. They start getting filled at 1:00, although at variable rates that may be different. Both tanks become completely filled at exactly 1:30 (30 minutes later). Note that the rate of filling of each tank is a continuous function of time.<br/></p><p>
We want to determine whether there was some moment when both tanks were being filled at the same rate. </p>
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Applying the MVT
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<p>
The MVT tells us that at some [mathjaxinline]t_1[/mathjaxinline] in the 30 minutes between 1:00 and 1:30, the instantaneous rate of filling of tank 1 is <div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_antider_1-tab7-problem1_2_1" class=" capa_inputtype inline textline">
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<span class="trailing_text" id="trailing_text_antider_1-tab7-problem1_2_1">L/min.</span>
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<p>
The MVT tells us that at some [mathjaxinline]t_2[/mathjaxinline] between 1:00 and 1:30, the instantaneous rate of filling of tank 2 is <div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_antider_1-tab7-problem1_3_1" class=" capa_inputtype inline textline">
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<span class="trailing_text" id="trailing_text_antider_1-tab7-problem1_3_1">L/min.</span>
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<p>
<p style="display:inline">Are [mathjaxinline]t_1[/mathjaxinline] and [mathjaxinline]t_2[/mathjaxinline] the same?</p>
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<text> Yes.</text>
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Applying the MVT to a new function
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Let us use the MVT in a different way to see if there is a moment at which the rate of filling is the same for the two tanks.<br/></p>
<p>
Suppose that [mathjaxinline]v_1(t)[/mathjaxinline] is the volume of water in the tank 1 at time [mathjaxinline]t[/mathjaxinline], and [mathjaxinline]v_2(t)[/mathjaxinline] is the volume of water in tank 2 at time [mathjaxinline]t[/mathjaxinline].<br/></p>
<p>
Set [mathjaxinline]h(t) = v_1(t)-v_2(t)[/mathjaxinline]. Then </p>
<p>
<p style="display:inline">[mathjaxinline]h(\text {1:00})=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]h(\text {1:30})=[/mathjaxinline]</p>
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<br/>
<p style="display:inline">The average rate of change of [mathjaxinline]h[/mathjaxinline] between 1:00 and 1:30 is </p>
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The MVT states that <p style="display:inline">[mathjaxinline]h'=[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 4" role="group"><div id="inputtype_antider_1-tab7-problem2_5_1" class=" capa_inputtype inline textline">
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<p>
If at time [mathjaxinline]t[/mathjaxinline], [mathjaxinline]h'(t) = 0[/mathjaxinline]. Then what can we say about [mathjaxinline]v_1'(t)[/mathjaxinline] and [mathjaxinline]v_2'(t)[/mathjaxinline]? </p>
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<text> [mathjaxinline]v_1'(t) = v_2'(t) = 200[/mathjaxinline] L/min</text>
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<text> [mathjaxinline]v_1'(t) = v_2'(t),[/mathjaxinline] but we don't know what value they take at [mathjaxinline]t[/mathjaxinline]</text>
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<text> [mathjaxinline]v_1'(t) = v_2'(t)[/mathjaxinline] at all times [mathjaxinline]t[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">8. Upper and lower bounds</h2>
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<p>
We will be using the notions of upper and lower bounds so we will discuss these here.<br/></p><p>
A number [mathjaxinline]M[/mathjaxinline] is an <span style="color:#27408C"><b class="bf">upper bound</b></span> on a function [mathjaxinline]\ f(x)[/mathjaxinline] if </p><table id="a0000000021" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000022"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle f(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle M[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {for all } \, \, x\, \,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
and a number [mathjaxinline]m[/mathjaxinline] is a <span style="color:#27408C"><b class="bf">lower bound</b></span> on a function [mathjaxinline]\ f(x)[/mathjaxinline] if </p><table id="a0000000023" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000024"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle m[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle f(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {for all } \, \, x\, \,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
We can consider upper and lower bounds over the entire real number line, or over an interval. </p><center><img src="/assets/courseware/v1/a7e90678f7204ee64f42438d34e042bb/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_upperbounddef2.svg" width="450px" alt="A function y equals f of x is plotted in the x y plane. The function is bounded above by capital M and bounded below by lowercase m. The function's behavior fluctuates, but it never reaches heights above capital M or below lowercase m." style="margin: 10px 25px 25px 25px"/><table id="a0000000025" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle m\leq f(x)\leq M[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></center><p>
In other words, an upper bound on a function is a number that is larger than or equal to all values of the function. A lower bound on a function is a number which is smaller than or equal to all values of the function. </p>
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Upper and lower bounds
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Consider the following graph of a function [mathjaxinline]y=f(x)[/mathjaxinline]. </p>
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<img alt="A function y equals f of x is plotted in the x y plane. The horizontal lines y equals A, y equals B, y equals C, y equals zero, y equals D, y equals E, y equals F, and y equals G are indicated in increasing order from least to greatest. The function increases up to a height of F, decreases to a height of B, increases to a height of E, then decreases to a horizontal asymptote at a height of 0. The limit of f of x as x approaches negative infinity is negative infinity. The limit of f of x as x approaches positive infinity is 0." src="/assets/courseware/v1/b8a29dd649dfa8a32a5f88c0b3c8494c/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_upperbound.svg" style="margin: 10px 25px 25px 25px" width="450px"/>
</center>
<p>
Among the numbers A,B,C,[mathjaxinline]0[/mathjaxinline],D,E,F,G, which are upper bounds and lower bounds on [mathjaxinline]\ f(x)[/mathjaxinline] ?<br/></p>
<p>
(Enter your answer separated by commas, e.g. "C, 0, D". Enter &#8220;None" if none of the above is an answer.)<br/></p>
<p>
<p style="display:inline">Which are the upper bounds on [mathjaxinline]\ f(x)[/mathjaxinline]?</p>
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<p style="display:inline">Which are lower bounds on [mathjaxinline]\ f(x)[/mathjaxinline]?</p>
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Upper and lower bounds on an interval
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<p>
We also consider <span style="color:#27408C"><b class="bf">upper and lower bounds over an interval</b></span>. That is, [mathjaxinline]M[/mathjaxinline] is an upper bound and [mathjaxinline]m[/mathjaxinline] is a lower bound on [mathjaxinline]\ f(x)[/mathjaxinline] over [mathjaxinline][a,b][/mathjaxinline] if </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000000026" style="table-layout:auto" width="100%">
<tr id="a0000000027">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle m[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle f(x)[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle M[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {for all } \, \, x\, \, \text {in}\, \, [a,b][/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Consider the same graph as above. </p>
<center>
<img alt="A function y equals f of x is plotted in the x y plane. The horizontal lines y equals A, y equals B, y equals C, y equals zero, y equals D, y equals E, y equals F, and y equals G are indicated in increasing order from least to greatest. The function increases up to a height of F, decreases to a height of B, increases to a height of E, then decreases to a horizontal asymptote at a height of 0. The limit of f of x as x approaches negative infinity is negative infinity. The limit of f of x as x approaches positive infinity is 0. The interval from x equals lowercase a to x equals lowercase b is indicated. On this interval, the function f starts at a height of C at the point x equals lowercase a, increases to a height of E, then decreases to a height of D at the point x equals lowercase b." src="/assets/courseware/v1/45b4da03138501398e9cedb7dd98b762/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_upperboundwithab.svg" style="margin: 10px 25px 25px 25px" width="450px"/>
</center>
<p>
Among the numbers A,B,C,[mathjaxinline]0[/mathjaxinline],D,E,F,G:<br/></p>
<p>
(Enter your answer separated by commas, e.g. "C, 0, D". Enter &#8220;None" if none of the above is an answer.)<br/></p>
<p>
<p style="display:inline">Which are the upper bounds on [mathjaxinline]\ f(x)[/mathjaxinline] over [mathjaxinline][a,b][/mathjaxinline]?</p>
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<p style="display:inline">Which are lower bounds on [mathjaxinline]\ f(x)[/mathjaxinline] over [mathjaxinline][a,b][/mathjaxinline] ?</p>
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Best upper and lower bounds
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<p>
You may have noticed that the smaller the upper bound, the more it tells you where the function is. Let us consider the "best" upper and lower bounds. <br/></p>
<p>
The best upper bound is the smallest number that is an upper bound, and is called the <span style="color:#27408C"><b class="bf">least upper bound</b></span>. The best lower bound is the biggest number which is a lower bound, and is called the <span style="color:#27408C"><b class="bf">greatest lower bound</b></span>. <br/></p>
<p>
Consider the same graph as above. </p>
<center>
<img alt="A function y equals f of x is plotted in the x y plane. The horizontal lines y equals A, y equals B, y equals C, y equals zero, y equals D, y equals E, y equals F, and y equals G are indicated in increasing order from least to greatest. The function increases up to a height of F, decreases to a height of B, increases to a height of E, then decreases to a horizontal asymptote at a height of 0. The limit of f of x as x approaches negative infinity is negative infinity. The limit of f of x as x approaches positive infinity is 0. The interval from x equals lowercase a to x equals lowercase b is indicated. On this interval, the function f starts at a height of C at the point x equals lowercase a, increases to a height of E, then decreases to a height of D at the point x equals lowercase b." src="/assets/courseware/v1/45b4da03138501398e9cedb7dd98b762/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_upperboundwithab.svg" style="margin: 10px 25px 25px 25px" width="450px"/>
</center>
<p>
Among the numbers A,B,C,[mathjaxinline]0[/mathjaxinline],D,E,F,G:<br/>(Enter your answer separated by commas, e.g. "C, 0, D". Enter &#8220;None" if none of the above is an answer.)<br/></p>
<p>
<p style="display:inline">Which is the least upper bound on [mathjaxinline]\ f(x)[/mathjaxinline] over [mathjaxinline][a,b][/mathjaxinline]?</p>
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<p style="display:inline">Which is the greatest lower bound on [mathjaxinline]\ f(x)[/mathjaxinline] over [mathjaxinline][a,b][/mathjaxinline] ?</p>
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Greatest lower bound which is not a minimum
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Consider the same graph as above. </p>
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<img alt="A function y equals f of x is plotted in the x y plane. The horizontal lines y equals A, y equals B, y equals C, y equals zero, y equals D, y equals E, y equals F, and y equals G are indicated in increasing order from least to greatest. The function increases up to a height of F, decreases to a height of B, increases to a height of E, then decreases to a horizontal asymptote at a height of 0. The limit of f of x as x approaches negative infinity is negative infinity. The limit of f of x as x approaches positive infinity is 0. The interval from x equals lowercase a to x equals lowercase b is indicated. On this interval, the function f starts at a height of C at the point x equals lowercase a, increases to a height of E, then decreases to a height of D at the point x equals lowercase b." src="/assets/courseware/v1/45b4da03138501398e9cedb7dd98b762/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_upperboundwithab.svg" style="margin: 10px 25px 25px 25px" width="450px"/>
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Among the numbers A, B, C, [mathjaxinline]0[/mathjaxinline], D, E, F, G:<br/></p>
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(Enter your answer separated by commas, e.g. "C, 0, D". Enter &#8220;None" if none of the above is an answer.)<br/></p>
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<p style="display:inline">Which are lower bounds on [mathjaxinline]\ f(x)[/mathjaxinline] over [mathjaxinline][b,+\infty )[/mathjaxinline]?</p>
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<p style="display:inline">Which is the greatest lower bound on [mathjaxinline]\ f(x)[/mathjaxinline] over [mathjaxinline][b,+\infty )[/mathjaxinline] ?</p>
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Lower bound of another function
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Let </p>
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<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle f(x)=\left(\frac{xe^{x^2}+2}{x\cos (x)+7\sqrt {x}}\right)^{2}.[/mathjax]</td>
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Find a lower bound on [mathjaxinline]\ f(x)[/mathjaxinline] as quickly as you can (without any calculations). <br/> <style> .xmodule_display.xmodule_CapaModule div.problem section div span.MathJax { display: inline-block !important; } .xmodule_display.xmodule_CapaModule div.problem section div span.MathJax_Preview { display: inline-block !important; } </style> </p>
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[mathjaxinline]\displaystyle \leq \, \, f(x)[/mathjaxinline] </td>
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Upper bound of absolute values
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Suppose you know a lower bound and an upper bound of the velocity [mathjaxinline]v(t)[/mathjaxinline] of your car to be </p>
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[mathjaxinline]\displaystyle -85[/mathjaxinline]
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[mathjaxinline]\displaystyle \leq[/mathjaxinline]
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[mathjaxinline]\displaystyle v(t)[/mathjaxinline]
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[mathjaxinline]\displaystyle \leq[/mathjaxinline]
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[mathjaxinline]\displaystyle 65[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.3)</td>
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Then what is the strongest statement you can make about the speed of your car? In other words, <style> .xmodule_display.xmodule_CapaModule .problem .capa_inputtype.textline input { min-width: 0 !important; } </style> </p>
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[mathjaxinline]\displaystyle \leq \, \, \left|v(t)\right|\, \, \leq[/mathjaxinline] </td>
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Lower bounds for the average rate of change
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<p>
We will now go through the following three steps to derive from the MVT the fundamental fact that if the derivative of a function is non-negative, then the function is increasing or staying at the same value. </p>
<p>
Suppose [mathjaxinline]x(t)[/mathjaxinline] satisfies the hypothesis of the MVT, that is, <br/>[mathjaxinline]x(t)[/mathjaxinline] is continuous on [mathjaxinline]A\leq t \leq B[/mathjaxinline], and differentiable on [mathjaxinline]A&lt;t &lt;B[/mathjaxinline]. </p>
<p>
Let </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000000036" style="table-layout:auto" width="100%">
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \displaystyle \Delta x[/mathjaxinline]
</td>
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[mathjaxinline]\displaystyle = \ x(B)-x(A)[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.4)</td>
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[mathjaxinline]\displaystyle \Delta t[/mathjaxinline]
</td>
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[mathjaxinline]\displaystyle = \ B-A[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.5)</td>
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<p>
Suppose we know a <span style="color:#99182C"><b class="bf">lower bound</b></span> [mathjaxinline]m[/mathjaxinline] for the derivative over the interval. In other words, </p>
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[mathjaxinline]\displaystyle \displaystyle m\leq x'(t) \qquad[/mathjaxinline]
</td>
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[mathjaxinline]\displaystyle \text { for all } \ t\ \text { such that } \ A&lt;t&lt;B.[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.6)</td>
</tr>
</table>
<p>
Using the MVT, find a lower bound for the average rate of change and the total change.<br/>(Enter your answer in terms of [mathjaxinline]m[/mathjaxinline], [mathjaxinline]A[/mathjaxinline], and [mathjaxinline]B[/mathjaxinline].) </p>
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[mathjaxinline]\displaystyle \leq \, \, \frac{\Delta x}{\Delta t}\qquad \left(\text {Average rate of change over}\, \, [A,B]\right)[/mathjaxinline] </td>
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[mathjaxinline]\displaystyle \leq \, \, \Delta x\qquad \left(\text {Total change over}\, \, [A,B]\right)[/mathjaxinline] </td>
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If we know a lower bound for [mathjaxinline]x'(t)[/mathjaxinline] over the interval [mathjaxinline](A,B)[/mathjaxinline], we can conclude more than the above.<br/></p>
<p>
As above, suppose [mathjaxinline]x(t)[/mathjaxinline] satisfies the hypothesis of the MVT, that is, <br/>[mathjaxinline]x(t)[/mathjaxinline] is continuous on [mathjaxinline]A\leq t \leq B[/mathjaxinline], and differentiable on [mathjaxinline]A&lt;t &lt;B[/mathjaxinline], and suppose </p>
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<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle m\leq x'(t) \qquad \text { for all } \ t \ \text { such that } \ A&lt;t&lt;B.[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.10)</td>
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<p>
let us consider the MVT on shorter intervals. The strongest conclusion we can draw is <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_antider_1-tab9-problem2_2_1">
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<input type="radio" name="input_antider_1-tab9-problem2_2_1" id="input_antider_1-tab9-problem2_2_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="antider_1-tab9-problem2_2_1-choice_1-label" for="input_antider_1-tab9-problem2_2_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab9-problem2_2_1"> <text> [mathjaxinline]\displaystyle m\leq \frac{x(B)-x(A)}{B-A}[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab9-problem2_2_1" id="input_antider_1-tab9-problem2_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="antider_1-tab9-problem2_2_1-choice_2-label" for="input_antider_1-tab9-problem2_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab9-problem2_2_1"> <text> [mathjaxinline]\displaystyle m\leq \frac{x(b)-x(a)}{b-a} \, \,[/mathjaxinline] for SOME [mathjaxinline]\, a,b\,[/mathjaxinline] such that [mathjaxinline]\, A\leq a&lt;b\leq B[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab9-problem2_2_1" id="input_antider_1-tab9-problem2_2_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="antider_1-tab9-problem2_2_1-choice_3-label" for="input_antider_1-tab9-problem2_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab9-problem2_2_1"> <text> [mathjaxinline]\displaystyle m\leq \frac{x(b)-x(a)}{b-a} \, \,[/mathjaxinline] for ALL [mathjaxinline]\, a,b \,[/mathjaxinline] such that [mathjaxinline]\, A&lt;a&lt;b&lt;B[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab9-problem2_2_1" id="input_antider_1-tab9-problem2_2_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="antider_1-tab9-problem2_2_1-choice_4-label" for="input_antider_1-tab9-problem2_2_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab9-problem2_2_1"> <text> [mathjaxinline]\displaystyle m\leq \frac{x(b)-x(a)}{b-a} \, \,[/mathjaxinline] for ALL [mathjaxinline]\, a,b\,[/mathjaxinline] such that [mathjaxinline]\, A\leq a&lt;b\leq B[/mathjaxinline]</text>
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When the lower bound is zero
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<p>
As above, suppose [mathjaxinline]x(t)[/mathjaxinline] satisfies the hypothesis of the MVT, that is, [mathjaxinline]x(t)[/mathjaxinline] is continuous on [mathjaxinline]A\leq t \leq B[/mathjaxinline], and differentiable on [mathjaxinline]A&lt;t &lt;B[/mathjaxinline]. </p>
<p>
Now, let the lower bound of [mathjaxinline]x'(t)[/mathjaxinline] be zero. In other words, suppose </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000000050" style="table-layout:auto" width="100%">
<tr id="a0000000051">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle 0\leq x'(t) \qquad \text { for all } \ t \ \text { in } \ (A,B),[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.12)</td>
</tr>
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<p>
then using the MVT, the strongest conclusion we can draw is <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_antider_1-tab9-problem3_2_1">
<fieldset aria-describedby="status_antider_1-tab9-problem3_2_1">
<div class="field">
<input type="radio" name="input_antider_1-tab9-problem3_2_1" id="input_antider_1-tab9-problem3_2_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="antider_1-tab9-problem3_2_1-choice_1-label" for="input_antider_1-tab9-problem3_2_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab9-problem3_2_1"> <text> [mathjaxinline]x(A)\leq x(B)[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab9-problem3_2_1" id="input_antider_1-tab9-problem3_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="antider_1-tab9-problem3_2_1-choice_2-label" for="input_antider_1-tab9-problem3_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab9-problem3_2_1"> <text> [mathjaxinline]x(a)\leq x(b)\,[/mathjaxinline] for SOME [mathjaxinline]\, a,b\,[/mathjaxinline] such that [mathjaxinline]\, A\leq a&lt;b \leq B[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab9-problem3_2_1" id="input_antider_1-tab9-problem3_2_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="antider_1-tab9-problem3_2_1-choice_3-label" for="input_antider_1-tab9-problem3_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab9-problem3_2_1"> <text> [mathjaxinline]x(a)\leq x(b)\,[/mathjaxinline] for ALL [mathjaxinline]\, a,b\,[/mathjaxinline] such that [mathjaxinline]\, A&lt;a&lt;b&lt;B[/mathjaxinline]</text>
</label>
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<div class="field">
<input type="radio" name="input_antider_1-tab9-problem3_2_1" id="input_antider_1-tab9-problem3_2_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="antider_1-tab9-problem3_2_1-choice_4-label" for="input_antider_1-tab9-problem3_2_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab9-problem3_2_1"> <text> [mathjaxinline]x(a)\leq x(b) \,[/mathjaxinline] for ALL [mathjaxinline]\, a,b\,[/mathjaxinline] such that [mathjaxinline]\, A\leq a&lt;b \leq B[/mathjaxinline]</text>
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<p>
<p style="display:inline">This implies that over the big interval [mathjaxinline][A,B][/mathjaxinline], [mathjaxinline]x(t)[/mathjaxinline] is </p>
<div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div class="inputtype option-input inline">
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When the upper bound is zero
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<p>
As above, suppose [mathjaxinline]x(t)[/mathjaxinline] satisfies the hypothesis of the MVT, that is, <br/>[mathjaxinline]x(t)[/mathjaxinline] is continuous on [mathjaxinline]A\leq t \leq B[/mathjaxinline], and differentiable on [mathjaxinline]A&lt;t &lt;B[/mathjaxinline]. </p>
<p>
Now, suppose the UPPER bound on [mathjaxinline]x'(t)[/mathjaxinline] is zero. In other words, </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000000056" style="table-layout:auto" width="100%">
<tr id="a0000000057">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle x'(t)\leq 0 \qquad \text { for all } \ t \ \text { in } \ (A,B),[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.15)</td>
</tr>
</table>
<p>
then using the MVT, the strongest conclusion we can draw is <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_antider_1-tab9-problem4_2_1">
<fieldset aria-describedby="status_antider_1-tab9-problem4_2_1">
<div class="field">
<input type="radio" name="input_antider_1-tab9-problem4_2_1" id="input_antider_1-tab9-problem4_2_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="antider_1-tab9-problem4_2_1-choice_1-label" for="input_antider_1-tab9-problem4_2_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab9-problem4_2_1"> <text> [mathjaxinline]x(A)\geq x(B)[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab9-problem4_2_1" id="input_antider_1-tab9-problem4_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="antider_1-tab9-problem4_2_1-choice_2-label" for="input_antider_1-tab9-problem4_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab9-problem4_2_1"> <text> [mathjaxinline]x(a)\geq x(b)\,[/mathjaxinline] for SOME [mathjaxinline]\, a,b\,[/mathjaxinline] such that [mathjaxinline]\, A\leq a&lt;b \leq B[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab9-problem4_2_1" id="input_antider_1-tab9-problem4_2_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="antider_1-tab9-problem4_2_1-choice_4-label" for="input_antider_1-tab9-problem4_2_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab9-problem4_2_1"> <text> [mathjaxinline]x(a)\geq x(b) \,[/mathjaxinline] for ALL [mathjaxinline]\, a,b\,[/mathjaxinline] such that [mathjaxinline]\, A\leq a&lt;b \leq B[/mathjaxinline]</text>
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</div></div> <p style="display:inline">Hence, if [mathjaxinline]x'(t)\leq 0[/mathjaxinline] for all [mathjaxinline]t[/mathjaxinline] in [mathjaxinline](A,B)[/mathjaxinline], then over [mathjaxinline][A,B][/mathjaxinline], [mathjaxinline]x(t)[/mathjaxinline] is </p> <div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div class="inputtype option-input inline">
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<option value="decreasing or staying at the same value."> decreasing or staying at the same value.</option>
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<p>
Using both of these facts, which are both consequences of the MVT, we can conclude that <p style="display:inline">If [mathjaxinline]x'(t)=0[/mathjaxinline] for all [mathjaxinline]t[/mathjaxinline] in [mathjaxinline](A,B)[/mathjaxinline], then over [mathjaxinline][A,B][/mathjaxinline], [mathjaxinline]x(t)[/mathjaxinline] is </p><div class="inline" tabindex="-1" aria-label="Question 3" role="group"><div class="inputtype option-input inline">
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<option value="option_antider_1-tab9-problem4_4_1_dummy_default">Select an option</option>
<option value="increasing or staying at the same value."> increasing or staying at the same value.</option>
<option value="decreasing or staying at the same value."> decreasing or staying at the same value.</option>
<option value="staying at the same value."> staying at the same value.</option>
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<h2 class="hd hd-2 unit-title">10. Old news</h2>
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Recall that a function [mathjaxinline]f(x)[/mathjaxinline] is <b class="bf">increasing</b> if whenever [mathjaxinline]a < b[/mathjaxinline], we have [mathjaxinline]f(a) \leq f(b)[/mathjaxinline].<br/>Similarly, a function [mathjaxinline]f(x)[/mathjaxinline] is <b class="bf">decreasing</b> if whenever [mathjaxinline]a < b[/mathjaxinline], we have [mathjaxinline]f(a) \geq f(b)[/mathjaxinline]. </p><p>
The conclusions we just made using the MVT on all subintervals of [mathjaxinline][A,B][/mathjaxinline] are fundamental facts we have been relying on: </p><ul class="itemize"><li><p>
If [mathjaxinline]x'(t) \geq 0[/mathjaxinline] for all [mathjaxinline]t[/mathjaxinline] in [mathjaxinline](A,B)[/mathjaxinline], then [mathjaxinline]x(t)[/mathjaxinline] is <b class="bfseries">increasing or staying the same</b> over [mathjaxinline][A,B][/mathjaxinline]. </p></li><li><p>
If [mathjaxinline]x'(t) \leq 0[/mathjaxinline] for all [mathjaxinline]t[/mathjaxinline] in [mathjaxinline](A,B)[/mathjaxinline], then [mathjaxinline]x(t)[/mathjaxinline] is <b class="bfseries">decreasing or staying the same</b> over [mathjaxinline][A,B][/mathjaxinline]. </p></li><li><p>
If [mathjaxinline]x'(t)=0[/mathjaxinline] for all [mathjaxinline]t[/mathjaxinline] in [mathjaxinline](A,B)[/mathjaxinline], then [mathjaxinline]x(t)[/mathjaxinline] is <b class="bfseries">constant</b> over [mathjaxinline][A,B][/mathjaxinline]. </p></li></ul><p>
And the MVT gives the following consequences with strict inequalities as well: </p><ul class="itemize"><li><p>
If [mathjaxinline]x'(t) >0[/mathjaxinline] for all [mathjaxinline]t[/mathjaxinline] in [mathjaxinline](A,B)[/mathjaxinline], then [mathjaxinline]x(t)[/mathjaxinline] is <b class="bfseries">strictly increasing</b> over [mathjaxinline][A,B][/mathjaxinline]. </p></li><li><p>
If [mathjaxinline]x'(t) <0[/mathjaxinline] for all [mathjaxinline]t[/mathjaxinline] in [mathjaxinline](A,B)[/mathjaxinline], then [mathjaxinline]x(t)[/mathjaxinline] is <b class="bfseries">strictly decreasing</b> over [mathjaxinline][A,B][/mathjaxinline]. </p></li></ul><p>
These facts need proofs and their proofs are based on the MVT as seen in the exercises you have done. The subtlety is that the MVT relates the infinitesimal behavior of the function, the derivative, which is defined at each point, with the macroscopic behavior, the total change over an interval. </p>
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Fire! Fire!
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At every moment in July, a wildfire is growing at a rate of at least 2 square kilometers/day. <p style="display:inline">On July 15, it was [mathjaxinline]50[/mathjaxinline] square kilometers in area. The strongest statement that we can make about its area on July 25 is that it had </p><div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div class="inputtype option-input inline">
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<span class="trailing_text" id="trailing_text_antider_1-tab11-problem1_3_1">square kilometers.</span>
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<p style="display:inline">The strongest statement that we can make about its area on July 5 is that it had </p>
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Sign of the function
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Let </p>
<table cellpadding="7" cellspacing="0" class="equation" id="a0000000080" style="table-layout:auto" width="100%">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle f(x) = -\frac{x^3}{6} - 3x - 2 \cos x[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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Without graphing the function, [mathjaxinline]\ f(x)[/mathjaxinline] is </p>
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<text> positive for all [mathjaxinline]x[/mathjaxinline]</text>
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<text> negative for all [mathjaxinline]x[/mathjaxinline]</text>
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Roots: Part 1
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Based solely on that information, what can we say about the number of solutions to the equation [mathjaxinline]\ f(x) = 0[/mathjaxinline]? </p>
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<text> There is at most one.</text>
</label>
</div>
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<input type="radio" name="input_antider_1-tab12-problem2_2_1" id="input_antider_1-tab12-problem2_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="antider_1-tab12-problem2_2_1-choice_2-label" for="input_antider_1-tab12-problem2_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab12-problem2_2_1">
<text> There is exactly one.</text>
</label>
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<input type="radio" name="input_antider_1-tab12-problem2_2_1" id="input_antider_1-tab12-problem2_2_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="antider_1-tab12-problem2_2_1-choice_3-label" for="input_antider_1-tab12-problem2_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab12-problem2_2_1">
<text> There is at least one.</text>
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<input type="radio" name="input_antider_1-tab12-problem2_2_1" id="input_antider_1-tab12-problem2_2_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="antider_1-tab12-problem2_2_1-choice_4-label" for="input_antider_1-tab12-problem2_2_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab12-problem2_2_1">
<text> None of the above.</text>
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Sign of the derivative
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Recall </p>
<table cellpadding="7" cellspacing="0" class="equation" id="a0000000081" style="table-layout:auto" width="100%">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle f(x) = -\frac{x^3}{6} - 3x - 2 \cos x[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
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[mathjaxinline]\ f'(x)[/mathjaxinline] is </p>
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<text> positive for all [mathjaxinline]x[/mathjaxinline]</text>
</label>
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<input type="radio" name="input_antider_1-tab12-problem3_2_1" id="input_antider_1-tab12-problem3_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="antider_1-tab12-problem3_2_1-choice_2-label" for="input_antider_1-tab12-problem3_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab12-problem3_2_1">
<text> negative for all [mathjaxinline]x[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab12-problem3_2_1" id="input_antider_1-tab12-problem3_2_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="antider_1-tab12-problem3_2_1-choice_3-label" for="input_antider_1-tab12-problem3_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab12-problem3_2_1">
<text> zero for all [mathjaxinline]x[/mathjaxinline]</text>
</label>
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<input type="radio" name="input_antider_1-tab12-problem3_2_1" id="input_antider_1-tab12-problem3_2_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="antider_1-tab12-problem3_2_1-choice_4-label" for="input_antider_1-tab12-problem3_2_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab12-problem3_2_1">
<text> sometimes positive and sometimes negative</text>
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Roots: Part 2
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<p>
Combining this new information with our facts above, what now can we say about the number of solutions to the equation [mathjaxinline]\ f(x) = 0[/mathjaxinline]? </p>
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<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_antider_1-tab12-problem4_2_1">
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<text> There is at most one.</text>
</label>
</div>
<div class="field">
<input type="radio" name="input_antider_1-tab12-problem4_2_1" id="input_antider_1-tab12-problem4_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="antider_1-tab12-problem4_2_1-choice_2-label" for="input_antider_1-tab12-problem4_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab12-problem4_2_1">
<text> There is exactly one.</text>
</label>
</div>
<div class="field">
<input type="radio" name="input_antider_1-tab12-problem4_2_1" id="input_antider_1-tab12-problem4_2_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="antider_1-tab12-problem4_2_1-choice_3-label" for="input_antider_1-tab12-problem4_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab12-problem4_2_1">
<text> There is at least one.</text>
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<input type="radio" name="input_antider_1-tab12-problem4_2_1" id="input_antider_1-tab12-problem4_2_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="antider_1-tab12-problem4_2_1-choice_4-label" for="input_antider_1-tab12-problem4_2_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab12-problem4_2_1">
<text> None of the above.</text>
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<h2 class="hd hd-2 unit-title">13. Applying MVT to inequalities</h2>
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Inequalities
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We will also use our old news to justify some inequalities between functions, such as [mathjaxinline]e^ x&gt;1+x[/mathjaxinline] for all [mathjaxinline]x&gt;0[/mathjaxinline]. </p>
<p>
Suppose our goal is to show [mathjaxinline]\ f(x)&gt;g(x)[/mathjaxinline] on an interval. The following three steps are an argument to confirm the inequality.<br/></p>
<p>
Let [mathjaxinline]h(x)=f(x)-g(x)[/mathjaxinline] for [mathjaxinline]x\geq A[/mathjaxinline], and suppose [mathjaxinline]h(x)[/mathjaxinline] is continous on [mathjaxinline]x\geq A[/mathjaxinline], and differentiable on [mathjaxinline]x&gt;A[/mathjaxinline]. </p>
<p>
<p style="display:inline">If [mathjaxinline]\ f(A)=g(A)[/mathjaxinline], then [mathjaxinline]h(A)[/mathjaxinline]=</p>
<div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_antider_1-tab13-problem1_2_1" class=" capa_inputtype inline textline">
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<p>
If [mathjaxinline]h(A)=0[/mathjaxinline], then which of the following guarantees that [mathjaxinline]h(x)&gt;0[/mathjaxinline] for all [mathjaxinline]x&gt;A[/mathjaxinline]?<br/>(Check all that apply.) </p>
<p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_antider_1-tab13-problem1_3_1">
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<text>[mathjaxinline]h'(x)&gt;0[/mathjaxinline] for ALL [mathjaxinline]x&gt;A[/mathjaxinline]</text>
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<text>[mathjaxinline]h'(x)\geq 0[/mathjaxinline] for ALL [mathjaxinline]x&gt;A[/mathjaxinline]</text>
</label>
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<text>[mathjaxinline]h'(x)&gt;0[/mathjaxinline] for SOME [mathjaxinline]x&gt;A[/mathjaxinline]</text>
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<text>[mathjaxinline]h(x)[/mathjaxinline] is strictly increasing for ALL [mathjaxinline]x&gt;A[/mathjaxinline]</text>
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<text>[mathjaxinline]h(x)[/mathjaxinline] is increasing or staying at the same value for ALL [mathjaxinline]x&gt;A[/mathjaxinline]</text>
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<text>[mathjaxinline]h(x)[/mathjaxinline] is strictly increasing for SOME [mathjaxinline]x&gt;A[/mathjaxinline]</text>
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<text>None of the above</text>
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<p>
What does the inequality [mathjaxinline]h(x)&gt;0[/mathjaxinline] for all [mathjaxinline]x&gt;A[/mathjaxinline] say about the relation between [mathjaxinline]\ f(x)[/mathjaxinline] and [mathjaxinline]g(x)\, \,[/mathjaxinline] for [mathjaxinline]x&gt;A[/mathjaxinline]? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 3" role="group"><div class="choicegroup capa_inputtype" id="inputtype_antider_1-tab13-problem1_4_1">
<fieldset aria-describedby="status_antider_1-tab13-problem1_4_1">
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<input type="radio" name="input_antider_1-tab13-problem1_4_1" id="input_antider_1-tab13-problem1_4_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="antider_1-tab13-problem1_4_1-choice_1-label" for="input_antider_1-tab13-problem1_4_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab13-problem1_4_1"> <text> [mathjaxinline]\ f(x)&lt;1+g(x)\, \,[/mathjaxinline] for [mathjaxinline]x&gt;A[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab13-problem1_4_1" id="input_antider_1-tab13-problem1_4_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="antider_1-tab13-problem1_4_1-choice_2-label" for="input_antider_1-tab13-problem1_4_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab13-problem1_4_1"> <text> [mathjaxinline]\ f(x)&lt;g(x)\, \,[/mathjaxinline] for [mathjaxinline]x&gt;A[/mathjaxinline]</text>
</label>
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<input type="radio" name="input_antider_1-tab13-problem1_4_1" id="input_antider_1-tab13-problem1_4_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="antider_1-tab13-problem1_4_1-choice_3-label" for="input_antider_1-tab13-problem1_4_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab13-problem1_4_1"> <text> [mathjaxinline]\ f(x)&gt;g(x)\, \,[/mathjaxinline] for [mathjaxinline]x&gt;A[/mathjaxinline]</text>
</label>
</div>
<div class="field">
<input type="radio" name="input_antider_1-tab13-problem1_4_1" id="input_antider_1-tab13-problem1_4_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="antider_1-tab13-problem1_4_1-choice_4-label" for="input_antider_1-tab13-problem1_4_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab13-problem1_4_1"> <text> [mathjaxinline]\ f(x)-g(x)&lt;0\, \,[/mathjaxinline] for [mathjaxinline]x&gt;A[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab13-problem1_4_1" id="input_antider_1-tab13-problem1_4_1_choice_5" class="field-input input-radio" value="choice_5"/><label id="antider_1-tab13-problem1_4_1-choice_5-label" for="input_antider_1-tab13-problem1_4_1_choice_5" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab13-problem1_4_1"> <text> None of the above</text>
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The exponential function
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<p>
Based on the graphs of [mathjaxinline]e^ x[/mathjaxinline] and [mathjaxinline]1+x[/mathjaxinline], we may guess that </p>
<table cellpadding="7" cellspacing="0" class="equation" id="a0000000088" style="table-layout:auto" width="100%">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle e^ x&gt;1+x \text { for all } x&gt;0.[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Let us confirm this is true using the argument above. </p>
<p>
<p style="display:inline">To get started, is [mathjaxinline]e^ x&gt;1[/mathjaxinline] for all [mathjaxinline]x&gt;0[/mathjaxinline]?</p>
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<option value="yes"> yes</option>
<option value="no"> no</option>
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<p>
Let [mathjaxinline]h(x)=e^ x-(1+x)[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]h(0)[/mathjaxinline]=</p>
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<br/>
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<p>
<p style="display:inline">[mathjaxinline]h'(x)=[/mathjaxinline]</p>
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<p>
Which of the following would guarantee [mathjaxinline]e^ x&gt;1+x[/mathjaxinline] for all [mathjaxinline]x&gt;0[/mathjaxinline]? <br/>(Check all that apply.) </p>
<p>
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<text>[mathjaxinline]h'(x)\geq 0[/mathjaxinline] for ALL [mathjaxinline]x&gt;0[/mathjaxinline]</text>
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<text>[mathjaxinline]h'(x)&gt;0[/mathjaxinline] for SOME [mathjaxinline]x&gt;0[/mathjaxinline]</text>
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<text>[mathjaxinline]h(x)[/mathjaxinline] is strictly increasing for ALL [mathjaxinline]x&gt;0[/mathjaxinline]</text>
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<text>[mathjaxinline]h(x)[/mathjaxinline] is increasing or staying at the same value for ALL [mathjaxinline]x&gt;0[/mathjaxinline]</text>
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<text>[mathjaxinline]h(x)[/mathjaxinline] is strictly increasing for SOME [mathjaxinline]x&gt;0[/mathjaxinline]</text>
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<p style="display:inline">True or False: [mathjaxinline]e^ x&gt;1+x[/mathjaxinline] for all [mathjaxinline]x&gt;0[/mathjaxinline].</p>
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<th class="formulainput" scope="col">Allowable Entries</th>
<th class="formulainput" scope="col">Descriptions</th>
<th class="formulainput" scope="col">Example Entries</th>
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<tr class="formulainput">
<th class="formulainput" rowspan="3" scope="row">Numbers</th>
<td class="formulainput">Integers</td>
<td class="formulainput">
<font color="#0078b0">2520</font>
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<tr class="formulainput">
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<font color="#0078b0">2/3</font>
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<td class="formulainput">Decimals </td>
<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
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<tr class="formulainput">
<th class="formulainput" rowspan="4" scope="row">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
<tr class="formulainput">
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<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" rowspan="3" scope="row">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Differentials</th>
<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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<h2 class="hd hd-2 unit-title">14. More inequalities</h2>
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Review: the Log function
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<p>
<p style="display:inline">[mathjaxinline]\displaystyle \frac{d}{dx} \ln (1-x)\, =\,[/mathjaxinline]</p>
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<div id="display_antider_1-tab14-problem1_2_1" class="equation">`{::}`</div>
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<br/>
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<p>
Which of the following can be the graph of [mathjaxinline]y=\ln (1-x)[/mathjaxinline]? </p>
<p>
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<text>
<img alt="A decreasing, concave down function is plotted on the interval minus infinity to negative 1, where the limit of f of x as x approaches negative 1 from the left is negative infinity." src="/assets/courseware/v1/0f996138108a7281844fc03e47996577/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_lognegxminus1.svg" style="margin: 10px 25px 25px 25px" width="230px"/>
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<img alt="A decreasing, concave down function is plotted on the interval minus infinity to 0, where the limit of f of x as x approaches 0 from the left is negative infinity. The graph passes through the point negative 1 comma 1." src="/assets/courseware/v1/944f65f77f95a7ba2c131977a4de777e/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_1pluslognegx.svg" style="margin: 10px 25px 25px 25px" width="230px"/>
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<img alt="A decreasing, concave up function is plotted on the interval minus 1 to infinity where the limit of f of x as x approaches negative 1 from the right is infinity." src="/assets/courseware/v1/a171dc92fdbcb4770d8b90deac96493b/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_neglogxplus1.svg" style="margin: 10px 25px 25px 25px" width="230px"/>
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<img alt="A decreasing, concave down function is plotted on the interval minus infinity to 1 where the limit of f of x as x approaches 1 from the left is negative infinity." src="/assets/courseware/v1/5197b9c943f34c27ce9f330a20519259/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_lognegxplus1.svg" style="margin: 10px 25px 25px 25px" width="230px"/>
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<text>
<img alt="A decreasing, concave up function is plotted on the interval 0 to infinity where the limit of f of x as x approaches 0 from the right is infinity." src="/assets/courseware/v1/fed7141fb3e591220ea68ffe963e6afc/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_1plusneglog.svg" style="margin: 10px 25px 25px 25px" width="230px"/>
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<th class="formulainput" rowspan="3" scope="row">Numbers</th>
<td class="formulainput">Integers</td>
<td class="formulainput">
<font color="#0078b0">2520</font>
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<td class="formulainput">Fractions</td>
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<font color="#0078b0">2/3</font>
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<td class="formulainput">Decimals </td>
<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
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<th class="formulainput" rowspan="4" scope="row">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
</tr>
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<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
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<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
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<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
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<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
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<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
</td>
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<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
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<th class="formulainput" rowspan="3" scope="row">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
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<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
</tr>
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<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Differentials</th>
<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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Inequalities for the logarithm(*)
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Let us compare [mathjaxinline]\ln (1-x)[/mathjaxinline] with its linear and quadratic approximations near [mathjaxinline]x=0[/mathjaxinline].<br/></p>
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Which of the following inequalities are true?<br/>(Check all that apply.) <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_antider_1-tab14-problem2_2_1">
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<input type="checkbox" name="input_antider_1-tab14-problem2_2_1[]" id="input_antider_1-tab14-problem2_2_1_choice_1" class="field-input input-checkbox" value="choice_1"/><label id="antider_1-tab14-problem2_2_1-choice_1-label" for="input_antider_1-tab14-problem2_2_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab14-problem2_2_1"> <text>[mathjaxinline]\ln (1-x)&lt;-x \, \, \text {for all }\, \, 0&lt;x&lt;1[/mathjaxinline]</text>
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<input type="checkbox" name="input_antider_1-tab14-problem2_2_1[]" id="input_antider_1-tab14-problem2_2_1_choice_2" class="field-input input-checkbox" value="choice_2"/><label id="antider_1-tab14-problem2_2_1-choice_2-label" for="input_antider_1-tab14-problem2_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab14-problem2_2_1"> <text>[mathjaxinline]\ln (1-x)&lt;-x \, \, \text {for all }\, \, x&lt;0[/mathjaxinline]</text>
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<input type="checkbox" name="input_antider_1-tab14-problem2_2_1[]" id="input_antider_1-tab14-problem2_2_1_choice_3" class="field-input input-checkbox" value="choice_3"/><label id="antider_1-tab14-problem2_2_1-choice_3-label" for="input_antider_1-tab14-problem2_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab14-problem2_2_1"> <text>[mathjaxinline]\ln (1-x)&lt;-x-\frac{x^2}{2} \, \, \text {for all } \, \, 0&lt;x&lt;1[/mathjaxinline]</text>
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Comparing log with a cubic polynomial
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Let us compare the function [mathjaxinline]\ln (1-x)[/mathjaxinline] with a cubic polynomial which is the best cubic approximation at [mathjaxinline]x=0[/mathjaxinline]. </p>
<p>
Which of the following inequalities are true?<br/>(Check all that apply.) <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_antider_1-tab14-problem3_2_1">
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<input type="checkbox" name="input_antider_1-tab14-problem3_2_1[]" id="input_antider_1-tab14-problem3_2_1_choice_0" class="field-input input-checkbox" value="choice_0"/><label id="antider_1-tab14-problem3_2_1-choice_0-label" for="input_antider_1-tab14-problem3_2_1_choice_0" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab14-problem3_2_1"> <text>[mathjaxinline]\ln (1-x)&lt;-x-\frac{x^2}{2}-\frac{x^3}{3} \text {for all } 0&lt;x&lt;1[/mathjaxinline]</text>
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<p><b class="bfseries">Bounding the average rate of change</b></p><p>
The MVT says that the average rate of change over an interval is equal to the derivative at some point in the interval. This implies that the average rate of change must be within the range of possible values of the derivative.<br/></p><p>
More precisely, if </p><table id="a0000000180" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000181"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle m[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle x'(c)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle M[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {for all}\, \, c\, \, \text {with}\, \, a<c<b,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.97)</td></tr></table><p>
that is, [mathjaxinline]m[/mathjaxinline] is a lower bound and [mathjaxinline]M[/mathjaxinline] is an upper bound on [mathjaxinline]x'(c)[/mathjaxinline] over the interval [mathjaxinline](a,b)[/mathjaxinline], then the MVT implies that </p><table id="a0000000182" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000183"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle m[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{x(b)-x(a)}{b-a}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle M[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {(Bounds on the average rate of change).}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.98)</td></tr></table><p>
In other words, a lower bound on the derivative is also a lower bound on the average rate of change, and an upper bound on the derivative is also an upper bound on the average rate of change.<br/></p><p>
Since [mathjaxinline]\, (b-a)>0[/mathjaxinline], multiplying the inequality above of the average rate of change by [mathjaxinline]\, (b-a)[/mathjaxinline] will not change the inequality signs, and we get the following inequality of the total change of the function. </p><table id="a0000000184" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000185"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle m\, \cdot \, (b-a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle x(b)-x(a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle M \, \cdot \, (b-a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {(Bounds on the total change)}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.99)</td></tr></table><p>
In other words, the total change of the function from [mathjaxinline]a[/mathjaxinline] to [mathjaxinline]b[/mathjaxinline] is greater than or equal to any lower bound [mathjaxinline]m[/mathjaxinline] on the derivative over the interval [mathjaxinline](a,b)[/mathjaxinline], times the length of the interval, [mathjaxinline](b-a)[/mathjaxinline]. And it is also less than or equal to any upper bound [mathjaxinline]M[/mathjaxinline] on the derivative over the interval [mathjaxinline](a,b)[/mathjaxinline], times [mathjaxinline](b-a)[/mathjaxinline].<br/></p><p>
Sometimes we know the maximum and minimum values of [mathjaxinline]x'(t)[/mathjaxinline]. In this case, we can use the maximum as an upper bound and the minimum as a lower bound on [mathjaxinline]x'(t)[/mathjaxinline], and obtain the following inequality for the average rate of change. </p><table id="a0000000186" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000187"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \min _{a \leq t \leq b} x'(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{x(b)-x(a)}{b-a}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \max _{a\leq t \leq b} x'(t)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.100)</td></tr></table><p>
Again multiplying both sides by [mathjaxinline]\, (b-a)[/mathjaxinline], we get an inequality on the total change of the function. </p><table id="a0000000188" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000189"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \min _{a \leq t \leq b} x'(t) \, \cdot \, (b-a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle x(b)-x(a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \max _{a\leq t \leq b} x'(t) \, \cdot \, (b-a)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.101)</td></tr></table><p>
In other words, the average rate of change must be in between the maximum and the minimum of the derivative, and the total change must be in between the maximum and minimum of the derivative multiplied by the length of the interval.<br/></p><p>
Moreover, the maximum and minimum on [mathjaxinline]\, x'(t)[/mathjaxinline] are the best bounds on [mathjaxinline]x'(t)[/mathjaxinline]. That is, for any upper bound [mathjaxinline]M[/mathjaxinline] and lower bound [mathjaxinline]m[/mathjaxinline] on [mathjaxinline]\, x'(t)[/mathjaxinline] from [mathjaxinline]a[/mathjaxinline] to [mathjaxinline]b[/mathjaxinline], </p><table id="a0000000190" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000191"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle m[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \min _{a \leq t \leq b} x'(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle x'(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \max _{a\leq t \leq b} x'(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle M.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.102)</td></tr></table><p>
In other words, the maximum is the least upper bound, the smallest number which is an upper bound, and the minimum is the greatest lower bound, the biggest number which is a lower bound.<br/></p><p>
The MVT then gives us the following inequalities for the average rate of change and the total change. For any upper bound [mathjaxinline]M[/mathjaxinline] and lower bound [mathjaxinline]m[/mathjaxinline] on the derivative [mathjaxinline]\, x'(t)[/mathjaxinline] from [mathjaxinline]a[/mathjaxinline] to [mathjaxinline]b[/mathjaxinline], </p><table id="a0000000192" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000193"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle m[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \min _{a \leq t \leq b} x'(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{x(b)-x(a)}{b-a}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \max _{a\leq t \leq b} x'(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle M[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.103)</td></tr><tr id="a0000000194"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle m\, \cdot \, (b-a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \min _{a \leq t \leq b} x'(t) \, \cdot \, (b-a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle x(b)-x(a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \max _{a\leq t \leq b} x'(t)\, \cdot \, (b-a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle M\, \cdot \, (b-a)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.104)</td></tr></table>
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<h2 class="hd hd-2 unit-title">16. An inequality of sine</h2>
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An inequality of sine
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Our goal is to determine the smallest constant [mathjaxinline]C[/mathjaxinline] such that the following inequality holds. </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000000195" style="table-layout:auto" width="100%">
<tr id="a0000000196">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \left|\sin (b)-\sin (a)\right|\leq C\, \left|b-a\right| \qquad \text {for any }\, a,b[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.105)</td>
</tr>
</table>
<p>
Let us first assume [mathjaxinline]a&lt;b.\, \,[/mathjaxinline]<br/>What are the greatest lower bound and the least upper bound on [mathjaxinline]\displaystyle \frac{\sin (b)-\sin (a)}{b-a}[/mathjaxinline] for any [mathjaxinline]a&lt;b[/mathjaxinline]?<br/>In other words, what is the biggest number [mathjaxinline]m[/mathjaxinline] such that [mathjaxinline]\displaystyle m\leq \frac{\sin (b)-\sin (a)}{b-a}[/mathjaxinline] for all choices of [mathjaxinline]a&lt;b[/mathjaxinline], and what is the smallest number [mathjaxinline]M[/mathjaxinline] such that [mathjaxinline]\displaystyle \frac{\sin (b)-\sin (a)}{b-a}\leq M[/mathjaxinline] for all choices [mathjaxinline]a&lt;b[/mathjaxinline]?<br/></p>
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[mathjaxinline]\displaystyle \leq \, \frac{\sin (b)-\sin (a)}{b-a} \, \leq[/mathjaxinline]</td>
<td style="text-align:right; border:none">
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<p>
What does this imply about the least upper bound on [mathjaxinline]\displaystyle \left|\frac{\sin (b)-\sin (a)}{b-a}\right|[/mathjaxinline] over any [mathjaxinline]a&lt;b[/mathjaxinline]? </p>
<table cellspacing="0" class="tabular" style="table-layout:auto">
<tr>
<td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \left|\frac{\sin (b)-\sin (a)}{b-a}\right|\leq[/mathjaxinline]</td>
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In conclusion, what is the smallest constant [mathjaxinline]C[/mathjaxinline] such that the following inequality holds? </p>
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[mathjaxinline]\displaystyle |\sin (b)-\sin (a)|\leq C\, \, |b-a| \qquad \text {for any }\, a,b[/mathjaxinline]
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<p style="display:inline">[mathjaxinline]C=\,[/mathjaxinline]</p>
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An inequality of tangent
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Is the following inequality true for any [mathjaxinline]\phi &lt;\theta[/mathjaxinline] in [mathjaxinline]\left( -\frac{\pi }{2},\frac{\pi }{2}\right)[/mathjaxinline]? </p>
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<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \tan (\theta )-\tan (\phi )\, \geq \, \theta -\phi \qquad \left(\text {for all }\, \, -\frac{\pi }{2}&lt;\phi &lt;\theta &lt;\frac{\pi }{2}\right)[/mathjax]</td>
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Is the same inequality with absolute values true for any [mathjaxinline]\theta[/mathjaxinline] and any [mathjaxinline]\phi[/mathjaxinline] in [mathjaxinline]\left( -\frac{\pi }{2},\frac{\pi }{2}\right)[/mathjaxinline]? </p>
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<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \left|\tan (\theta )-\tan (\phi )\right|\geq \left| \theta -\phi \right|\qquad \text { for all } \theta ,\phi \, \, \text {in}\, \, \left(-\frac{\pi }{2},\frac{\pi }{2}\right)[/mathjax]</td>
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An inequality of the inverse of tangent
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True or False: </p>
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<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \left|\arctan (x)-\arctan (u)\right|\leq |x-u| \qquad \text {for all } \, \, x,u.\, \,[/mathjax]</td>
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<h2 class="hd hd-2 unit-title">18. Compare linear approximation and MVT</h2>
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Both the linear approximation and the MVT say something about the change of a function over an interval. Let us now put them side by side and compare them. </p>
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Review: Linear approximation
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Let </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000000227" style="table-layout:auto" width="100%">
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<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \Delta x[/mathjaxinline]
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[mathjaxinline]\displaystyle = \ x(t)-x(a)[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.107)</td>
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<tr id="a0000000229">
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[mathjaxinline]\displaystyle \Delta t[/mathjaxinline]
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[mathjaxinline]\displaystyle = \ t-a[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.108)</td>
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<p>
Use the linear approximation of [mathjaxinline]x(t) \,[/mathjaxinline] at [mathjaxinline]\, t=a[/mathjaxinline] to approximate [mathjaxinline]\Delta x[/mathjaxinline]. </p>
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<text> [mathjaxinline]\Delta x \approx x'(c) \, \Delta t \, \, \, \,[/mathjaxinline], where [mathjaxinline]a&lt;c&lt;t[/mathjaxinline]</text>
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MVT
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As above, let </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000000231" style="table-layout:auto" width="100%">
<tr id="a0000000232">
<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \displaystyle \Delta x[/mathjaxinline]
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[mathjaxinline]\displaystyle = \ x(t)-x(a)[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.109)</td>
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<tr id="a0000000233">
<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \Delta t[/mathjaxinline]
</td>
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[mathjaxinline]\displaystyle = \ t-a[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.110)</td>
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<p>
The definition of the average rate of change gives the following equation for [mathjaxinline]\Delta x[/mathjaxinline]. </p>
<table cellpadding="7" cellspacing="0" class="equation" id="a0000000234" style="table-layout:auto" width="100%">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \Delta x = m_{sec} \, \Delta t \qquad \text {where }\, m_{sec}\, = \text {slope of secant line through } (a,x(a))\text { , }(t,x(t))[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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<p>
So what does the MVT say about [mathjaxinline]\Delta x[/mathjaxinline]?<br/></p>
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<text> [mathjaxinline]\Delta x = x'(c) \, \Delta t[/mathjaxinline], for some [mathjaxinline]c[/mathjaxinline] where [mathjaxinline]a&lt;c&lt;t[/mathjaxinline]</text>
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<h3 class="hd hd-2">Linear approximation vs MVT</h3>
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Linear approximation VS MVT:Geometrical picture
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Let us compare the geometric pictures of linear approximations and the MVT.<br/></p>
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<p style="display:inline">The LEFT figure below corresponds to</p>
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<p style="display:inline">The RIGHT figure below corresponds to</p>
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<text> linear approximation of [mathjaxinline]x(t)[/mathjaxinline] near [mathjaxinline]a[/mathjaxinline]</text>
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<text> linear approximation of [mathjaxinline]x(t)[/mathjaxinline] near [mathjaxinline]t[/mathjaxinline]</text>
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<text> linear approximation of [mathjaxinline]x(t)[/mathjaxinline] near [mathjaxinline]c[/mathjaxinline], where [mathjaxinline]a&lt;c&lt;t[/mathjaxinline]</text>
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<text> an equality for [mathjaxinline]\Delta x[/mathjaxinline] given by the MVT</text>
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<text> linear approximation of [mathjaxinline]x(t)[/mathjaxinline] near [mathjaxinline]a[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab18-problem3_3_1" id="input_antider_1-tab18-problem3_3_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="antider_1-tab18-problem3_3_1-choice_2-label" for="input_antider_1-tab18-problem3_3_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab18-problem3_3_1">
<text> linear approximation of [mathjaxinline]x(t)[/mathjaxinline] near [mathjaxinline]t[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab18-problem3_3_1" id="input_antider_1-tab18-problem3_3_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="antider_1-tab18-problem3_3_1-choice_3-label" for="input_antider_1-tab18-problem3_3_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab18-problem3_3_1">
<text> linear approximation of [mathjaxinline]x(t)[/mathjaxinline] near [mathjaxinline]c[/mathjaxinline], where [mathjaxinline]a&lt;c&lt;t[/mathjaxinline]</text>
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<input type="radio" name="input_antider_1-tab18-problem3_3_1" id="input_antider_1-tab18-problem3_3_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="antider_1-tab18-problem3_3_1-choice_4-label" for="input_antider_1-tab18-problem3_3_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_antider_1-tab18-problem3_3_1">
<text> an equality for [mathjaxinline]\Delta x[/mathjaxinline] given by the MVT</text>
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<img alt="A graph of a function in the x versus t plane is shown in the first quadrant. Two points on the horizontal axis are indicated by a and t where a is less than t. The function is increasing concave up from a to t then concave down for points greater than t. The tangent line at the value a is shown in pink and extends to the point t. The hoizontal component of this line has length delta t. The vertical height of this line is given by x prime of a times delta t. The secant line between a and t is drawn in green. The vertical height of this line is given by delta x. The quantity delta x is approximately equal to x prime of a times delta t." src="/assets/courseware/v1/360aa5879c73b0ca30738fa02cd31b87/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_LinApprox.svg" style="margin: 10px 25px 25px 25px" width="280px"/>
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<img alt="A graph of a function in the x versus t plane is shown in the first quadrant. Three points on the horizontal axis are indicated by a, c, and t where a is less than c and c is less than t. The function is increasing concave up from a to t then concave down for points greater than t. The tangent line at the value c is shown in pink and extends from the point a to the point t. The hoizontal component of this line has length delta t. The vertical height of this line is given by x prime of c times delta t. The secant line between a and t is drawn in green. The vertical height of this line is given by delta x. The tangent line at c is parallel to the secant line between a and t. The quantity delta x is approximately equal to x prime of c times delta t." src="/assets/courseware/v1/4464e613eb72054eddd3a52963cbac57/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_MvtNotLinApprox.svg" style="margin: 10px 25px 25px 25px" width="280px"/>
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[mathjaxinline]\Delta x \approx x'(a) \Delta t[/mathjaxinline]</td>
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[mathjaxinline]\Delta x = x'(c) \Delta t[/mathjaxinline] </td>
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<h2 class="hd hd-2 unit-title">19. Summary</h2>
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<p><b class="bfseries">The Mean Value Theorem (MVT)</b></p><p>
If [mathjaxinline]x(t)[/mathjaxinline] is continuous on [mathjaxinline]a\leq t \leq b[/mathjaxinline], and differentiable on [mathjaxinline]a<t<b[/mathjaxinline], that is, [mathjaxinline]x'(t)[/mathjaxinline] is defined for all [mathjaxinline]t[/mathjaxinline], [mathjaxinline]\, \, a<t<b[/mathjaxinline], then <br/></p><table id="a0000000237" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \frac{x(b)-x(a)}{b-a} \, = x'(c) \qquad \text {for some }c,\, \, \text {with } a<c<b.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Equivalently, in geometric terms, there is at least one point [mathjaxinline]c[/mathjaxinline], with [mathjaxinline]a<c<b[/mathjaxinline], at which the tangent line is parallel to the secant line through [mathjaxinline](a, x(a))[/mathjaxinline] and [mathjaxinline](b, x(b))[/mathjaxinline]: </p><center><img src="/assets/courseware/v1/ff1709fc4f125c8fffb302d051212c2d/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_xtgraph_MVT_statement.svg" width="300px" alt="An increasing, concave down curve is plotted in the x versus t plane. The points on the horizontal axis are indicated by a, c, and b where a is less c and c is less than b. The tangent line to the graph is drawn at c comma x of c." style="margin: 10px 25px 25px 25px"/></center><p><b class="bfseries">Upper and Lower Bounds</b></p><p>
We have introduced the notion of upper and lower bounds.<br/></p><p>
A number [mathjaxinline]M[/mathjaxinline] is an <span style="color:#27408C"><b class="bf">upper bound</b></span> of a function [mathjaxinline]\ f(x)[/mathjaxinline] if </p><table id="a0000000238" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000239"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle f(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle M[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {for all } \, \, x\, \,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
and a number [mathjaxinline]m[/mathjaxinline] is a <span style="color:#27408C"><b class="bf">lower bound</b></span> of a function [mathjaxinline]\ f(x)[/mathjaxinline] if </p><table id="a0000000240" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000241"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle m[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle f(x)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {for all } \, \, x\, \,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
We can consider upper and lower bounds on the entire real number line, or on an interval. </p><center><img src="/assets/courseware/v1/a7e90678f7204ee64f42438d34e042bb/asset-v1:MITx+18.01.2x+3T2019+type@asset+block/images_antider1_upperbounddef2.svg" width="330px" alt="A function y equals f of x is plotted in the x y plane. The function is bounded above by capital M and bounded below by lowercase m. The function's behavior fluctuates, but it never reaches heights above capital M or below lowercase m." style="margin: 10px 25px 25px 25px"/><table id="a0000000242" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle m\leq f(x)\leq M[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></center><p>
In other words, an upper bound of a function is a number that is larger than or equal to all values of the function. A lower bound of a function is a number which is smaller than or equal to all values of the function. </p><p><b class="bfseries">Old news</b></p><p>
We have been relying on the following fundamental facts whenever we try to understand a function using its derivative. But in fact, these facts are consequences of the mean value theorem.<br/></p><ul class="itemize"><li><p>
If [mathjaxinline]x'(t) \geq 0[/mathjaxinline] for all [mathjaxinline]t[/mathjaxinline] in [mathjaxinline](A,B)[/mathjaxinline], then [mathjaxinline]x(t)[/mathjaxinline] is <b class="bfseries">increasing or staying the same</b> over [mathjaxinline][A,B][/mathjaxinline]. </p></li><li><p>
If [mathjaxinline]x'(t) > 0[/mathjaxinline] for all [mathjaxinline]t[/mathjaxinline] in [mathjaxinline](A,B)[/mathjaxinline], then [mathjaxinline]x(t)[/mathjaxinline] is <b class="bfseries">strictly increasing</b> over [mathjaxinline][A,B][/mathjaxinline]. </p></li><li><p>
If [mathjaxinline]x'(t) \leq 0[/mathjaxinline] for all [mathjaxinline]t[/mathjaxinline] in [mathjaxinline](A,B)[/mathjaxinline], then [mathjaxinline]x(t)[/mathjaxinline] is <b class="bfseries">decreasing or staying the same</b> over [mathjaxinline][A,B][/mathjaxinline]. </p></li><li><p>
If [mathjaxinline]x'(t) < 0[/mathjaxinline] for all [mathjaxinline]t[/mathjaxinline] in [mathjaxinline](A,B)[/mathjaxinline], then [mathjaxinline]x(t)[/mathjaxinline] is <b class="bfseries">strictly decreasing</b> over [mathjaxinline][A,B][/mathjaxinline]. </p></li><li><p>
If [mathjaxinline]x'(t)=0[/mathjaxinline] for all [mathjaxinline]t[/mathjaxinline] in [mathjaxinline](A,B)[/mathjaxinline], then [mathjaxinline]x(t)[/mathjaxinline] is <b class="bfseries">constant</b> over [mathjaxinline][A,B][/mathjaxinline]. </p></li></ul><p>
These facts need proofs and their proofs are based on the MVT. The subtlety is that the MVT relates the infinitesimal behavior of the function, the derivative, which is defined at a point, to the macroscopic behavior of the function, the total change over an interval. </p><p><b class="bfseries">Bounding the average rate of change</b></p><p>
The equality in the MVT can be used to restrict the range of possible values of the average rate of change and the total change.<br/></p><p>
More precisely, if there are numbers [mathjaxinline]m[/mathjaxinline] and [mathjaxinline]M[/mathjaxinline] such that </p><table id="a0000000243" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000244"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle m[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle x'(c)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle M[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {for all}\, \, c\, \, \text {with}\, \, a<c<b,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.111)</td></tr></table><p>
that is, [mathjaxinline]m[/mathjaxinline] is a lower bound and [mathjaxinline]M[/mathjaxinline] is an upper bounds on [mathjaxinline]x'(c)[/mathjaxinline] over [mathjaxinline](a,b)[/mathjaxinline], then the MVT implies the following. </p><table id="a0000000245" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000246"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle m[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{x(b)-x(a)}{b-a}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle M[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {(Bounds on the average rate of change)}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.112)</td></tr><tr id="a0000000247"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle m\, \cdot \, (b-a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle x(b)-x(a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle M \, \cdot \, (b-a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {(Bounds on the total change)}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.113)</td></tr></table><p>
In other words, a lower bound on the derivative is also a lower bound on the average rate of change, and an upper bound on the derivative is also an upper bound on the average rate of change. Also, the product of a lower bound on the derivative with the length of an interval, is a lower bound on the total change of the function over that interval. Similarly, the product of an upper bound on the derivative with the length of an interval, is an upper bound on the total change of the function over that interval. <br/></p><p>
When we know the maximum and minimum values of [mathjaxinline]x'(t)[/mathjaxinline], we can use them as bounds on [mathjaxinline]x'(t)[/mathjaxinline] and obtain the following.<br/></p><table id="a0000000248" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000249"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \min _{a \leq t \leq b} x'(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{x(b)-x(a)}{b-a}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \max _{a\leq t \leq b} x'(t).[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {(Bounds on the average rate of change)}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.114)</td></tr><tr id="a0000000250"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \min _{a \leq t \leq b} x'(t) \, \cdot \, (b-a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle x(b)-x(a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \max _{a\leq t \leq b} x'(t) \, \cdot \, (b-a)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {(Bounds on the total change)}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(1.115)</td></tr></table><p>
In other words, the average rate of change must be in between the maximum and the minimum of the derivative, and the total change must be in between the maximum and minimum of the derivative multiplied by the length of the interval.<br/></p>
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