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| | <span class="exam"> Consider the following function: | | <span class="exam"> Consider the following function: |
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| − | ::::::<math>f(x)=3x-2\sin x+7</math>
| + | ::<math>f(x)=3x-2\sin x+7</math> |
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| − | <span class="exam">a) Use the Intermediate Value Theorem to show that <math style="vertical-align: -3px">f(x)</math> has at least one zero. | + | <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. |
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| − | <span class="exam">b) Use the Mean Value Theorem to show that <math style="vertical-align: -3px">f(x)</math> has at most 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. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009A Sample Final 1, Problem 6 Solution|'''<u>Solution</u>''']] |
| − | |- | |
| − | |Recall:
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| − | |-
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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: -1px">c</math> is any number
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| − | |-
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| − | |
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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: -1px">x</math> in the closed interval such that <math style="vertical-align: -5px">f(x)=c.</math>
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| − | |-
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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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| − | |-
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| − | |
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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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| − | |-
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| − | |
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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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| − | |-
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| − | |
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| − | ::Then, there is a number <math style="vertical-align: -1px">c</math> such that <math style="vertical-align: -1px">a<c<b</math> and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
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| − | |}
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| − | '''Solution:'''
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| − | '''(a)''' | + | [[009A Sample Final 1, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First note that <math style="vertical-align: -3px">f(0)=7.</math>
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| − | |-
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| − | |Also, <math style="vertical-align: -3px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
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| − | |-
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| − | |Since <math style="vertical-align: -3px">-1\leq \sin(x) \leq 1,</math>
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| − | |-
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| − | |
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| − | ::<math>-2\leq -2\sin(x) \leq 2.</math>
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| − | |-
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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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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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: -1px">x</math> with <math style="vertical-align: -1px">-5<x<0</math> such that
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| − | |-
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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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| − | |}
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| − |
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| − | '''(b)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exists <math style="vertical-align: -5px">a,b</math> such that <math style="vertical-align: -5px">f(a)=f(b)=0.</math>
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| − | |-
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| − | |Then, by the Mean Value Theorem, there exists <math style="vertical-align: -1px">c</math> with <math style="vertical-align: -1px">a<c<b</math> such that <math style="vertical-align: -5px">f'(c)=0.</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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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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| − | |-
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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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| − | |-
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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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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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: -1px">x</math> with <math style="vertical-align: -1px">-5<x<0</math> such that
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| − | |-
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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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| − | |-
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| − | |'''(b)''' See '''Step 1''' and '''Step 2''' above.
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| − | |}
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| | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
