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 7: / Line 7: @@
 <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>''']]
-|-
-|'''1.''' '''Intermediate Value Theorem'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;If &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous on a closed interval &nbsp;<math style="vertical-align: -5px">[a,b]</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; is any number
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;between &nbsp;<math style="vertical-align: -5px">f(a)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(b),</math>&nbsp; then there is at least one number &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; in the closed interval such that &nbsp;<math style="vertical-align: -5px">f(x)=c.</math>
-|-
-|'''2.'''  '''Mean Value Theorem'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Suppose &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is a function that satisfies the following:
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous on the closed interval &nbsp;<math style="vertical-align: -5px">[a,b].</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is differentiable on the open interval &nbsp;<math style="vertical-align: -5px">(a,b).</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;Then, there is a number &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; such that &nbsp;<math style="vertical-align: 0px">a<c<b</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
-|}
-'''Solution:'''
+[[009A Sample Final 1, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First note that &nbsp; <math style="vertical-align: -5px">f(0)=7.</math>
-|-
-|Also, &nbsp;<math style="vertical-align: -5px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
-|-
-|Since &nbsp;<math style="vertical-align: -5px">-1\leq \sin(x) \leq 1,</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>-2\leq -2\sin(x) \leq 2.</math>
-|-
-|Thus, &nbsp;<math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math>&nbsp; and hence &nbsp;<math style="vertical-align: -5px">f(-5)<0.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Since &nbsp;<math style="vertical-align: -5px">f(-5)<0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(0)>0,</math>&nbsp; there exists &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">-5<x<0</math>&nbsp; such that
-|-
-|<math style="vertical-align: -5px">f(x)=0</math>&nbsp; by the Intermediate Value Theorem. Hence, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has at least one zero.
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|Suppose that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has more than one zero. So, there exist &nbsp;<math style="vertical-align: -4px">a,b</math>&nbsp; such that &nbsp;<math style="vertical-align: -5px">f(a)=f(b)=0.</math>
-|-
-|Then, by the Mean Value Theorem, there exists &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">a<c<b</math>&nbsp; such that &nbsp;<math style="vertical-align: -5px">f'(c)=0.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|We have &nbsp;<math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math>&nbsp; Since &nbsp;<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>&nbsp; So, &nbsp;<math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
-|-
-|which contradicts &nbsp;<math style="vertical-align: -5px">f'(c)=0.</math>&nbsp; Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has at most one zero.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; Since &nbsp;<math style="vertical-align: -5px">f(-5)<0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(0)>0,</math>&nbsp; there exists &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">-5<x<0</math>&nbsp; such that
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">f(x)=0</math>&nbsp; by the Intermediate Value Theorem. Hence, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has at least one zero.
-|-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; 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

Navigation menu

Search