009A Sample Final 1, Problem 6 - Revision history

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MathAdmin at 22:15, 18 April 2016

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MathAdmin at 19:30, 18 April 2016

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MathAdmin at 19:15, 18 April 2016

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MathAdmin: Created page with " Consider the following function: :::::: $f(x)=3x-2\sin x+7$ a) Use the Intermediate Value Theorem to show that
2016-03-05T07:09:46Z

Created page with "<span class="exam"> Consider the following function: ::::::<math>f(x)=3x-2\sin x+7</math> <span class="exam">a) Use the Intermediate Value Theorem to show that <math style="..."

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<span class="exam"> Consider the following function:

::::::<math>f(x)=3x-2\sin x+7</math>

<span class="exam">a) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at least one zero.

<span class="exam">b) Use the Mean Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at most one zero.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
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|Recall:
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|'''1. Intermediate Value Theorem:''' If <math style="vertical-align: -5px">f(x)</math>  is continuous on a closed interval <math style="vertical-align: -5px">[a,b]</math> and <math style="vertical-align: 0px">c</math> is any number
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::between <math style="vertical-align: -5px">f(a)</math>  and <math style="vertical-align: -5px">f(b)</math>, then there is at least one number <math style="vertical-align: 0px">x</math> in the closed interval such that <math style="vertical-align: -5px">f(x)=c.</math>
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|'''2. Mean Value Theorem:''' Suppose <math style="vertical-align: -5px">f(x)</math>  is a function that satisfies the following:
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::<math style="vertical-align: -5px">f(x)</math>  is continuous on the closed interval  <math style="vertical-align: -5px">[a,b].</math>
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::<math style="vertical-align: -5px">f(x)</math>  is differentiable on the open interval <math style="vertical-align: -5px">(a,b).</math>
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::Then, there is a number <math style="vertical-align: 0px">c</math> such that  <math style="vertical-align: 0px">a<c<b</math>  and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
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'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|First note that  <math style="vertical-align: -5px">f(0)=7.</math>
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|Also,  <math style="vertical-align: -5px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
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|Since  <math style="vertical-align: -5px">-1\leq \sin(x) \leq 1,</math>
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::<math>-2\leq -2\sin(x) \leq 2.</math>
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|Thus,  <math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math>  and hence  <math style="vertical-align: -5px">f(-5)<0.</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|Since <math style="vertical-align: -5px">f(-5)<0</math>  and  <math style="vertical-align: -5px">f(0)>0,</math>  there exists <math style="vertical-align: 0px">x</math> with  <math style="vertical-align: 0px">-5<x<0</math>  such that
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|<math style="vertical-align: -5px">f(x)=0</math>  by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>  has at least one zero.
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'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exist <math style="vertical-align: -4px">a,b</math> such that  <math style="vertical-align: -5px">f(a)=f(b)=0.</math>
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|Then, by the Mean Value Theorem, there exists <math style="vertical-align: 0px">c</math> with  <math style="vertical-align: 0px">a<c<b</math> such that  <math style="vertical-align: -5px">f'(c)=0.</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|We have <math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math>  Since  <math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
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|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math>  So, <math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
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|which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math>  has at most one zero.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|'''(a)''' Since <math style="vertical-align: -5px">f(-5)<0</math>  and  <math style="vertical-align: -5px">f(0)>0,</math>  there exists <math style="vertical-align: 0px">x</math> with  <math style="vertical-align: 0px">-5<x<0</math>  such that
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|<math style="vertical-align: -5px">f(x)=0</math>  by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>  has at least one zero.
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|'''(b)''' See '''Step 1''' and '''Step 2''' above.
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[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]