Difference between revisions of "009A Sample Final 1, Problem 6"

Latest revision as of 15:53, 2 December 2017

Consider the following function:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=3x-2\sin x+7}

(a) Use the Intermediate Value Theorem to show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} has at least one zero.

(b) Use the Mean Value Theorem to show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} has at most one zero.

Solution

Detailed Solution

Return to Sample Exam

@@ Line 1: / Line 1: @@
 <span class="exam"> Consider the following function:
-::::::<math>f(x)=3x-2\sin x+7</math>
+::<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: -3px">f(x)</math> has at least one zero.
+<span class="exam">(a) Use the Intermediate Value Theorem to show that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has at least one zero.
-<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 &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has at most one zero.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 1, Problem 6 Solution|'''<u>Solution</u>''']]
-|-
-|Recall:
-|-
-|'''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
-|-
-|
-::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>.
-|-
-|'''2. Mean Value Theorem''' Suppose <math style="vertical-align: -5px">f(x)</math> is a function that satisfies the following:
-|-
-|
-::<math style="vertical-align: -5px">f(x)</math> is continuous on the closed interval <math style="vertical-align: -5px">[a,b]</math>.
-|-
-|
-::<math style="vertical-align: -5px">f(x)</math> is differentiable on the open interval <math style="vertical-align: -5px">(a,b)</math>.
-|-
-|
-::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>.
-|}
-'''Solution:'''
-'''(a)'''
+[[009A Sample Final 1, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First note that <math style="vertical-align: -3px">f(0)=7</math>.
-|-
-|Also, <math style="vertical-align: -3px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5)</math>.
-|-
-|Since <math style="vertical-align: -3px">-1\leq \sin(x) \leq 1</math>,
-|-
-|
-::<math>-2\leq -2\sin(x) \leq 2</math>.
-|-
-|Thus, <math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math> and hence <math style="vertical-align: -5px">f(-5)<0</math>.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|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
-|-
-|<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.
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|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>.
-|-
-|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>.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|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>,
-|-
-|<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>,
-|-
-|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.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|'''(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
-|-
-|<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.
-|-
-|'''(b)''' See '''Step 1''' and '''Step 2''' above.
-|}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009A Sample Final 1, Problem 6"

Latest revision as of 15:53, 2 December 2017

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