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

Latest revision as of 15:53, 2 December 2017

Consider the following function:

f(x)=3x-2\sin x+7

(a) Use the Intermediate Value Theorem to show that $f(x)$ has at least one zero.

(b) Use the Mean Value Theorem to show that $f(x)$ has at most one zero.

Solution

Detailed Solution

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@@ 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>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>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>''']]
-|-
-|
-|}
-'''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>f(0)=7</math>.
-|-
-|Also, <math>f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5)</math>.
-|-
-|Since <math>-1\leq \sin(x) \leq 1</math>,
-|-
-|<math>-2\leq -2\sin(x) \leq 2</math>.
-|-
-|Thus, <math>-10\leq f(-5) \leq -6</math> and hence <math>f(-5)<0</math>.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Since <math>f(-5)<0</math> and <math>f(0)>0</math>, there exists <math>x</math> with <math>-5<x<0</math> such that
-|-
-|<math>f(x)=0</math> by the Intermediate Value Theorem. Hence, <math>f(x)</math> has at least one zero.
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We have <math>f'(x)=3-2\cos(x)</math>. Since <math>-1\leq \cos(x)\leq 1</math>,
-|-
-|<math>-2 \leq -2\cos(x)\leq 2</math>. So, <math>1\leq f'(x) \leq 5</math>.
-|-
-|Therefore, <math>f'(x)</math> is always positive.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Since <math>f'(x)</math> is always positive, <math>f(x)</math> is an increasing function.
-|-
-|Thus, <math>f(x)</math> has at most one zero.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|'''(a)''' Since <math>f(-5)<0</math> and <math>f(0)>0</math>, there exists <math>x</math> with <math>-5<x<0</math> such that
-|-
-|<math>f(x)=0</math> by the Intermediate Value Theorem. Hence, <math>f(x)</math> has at least one zero.
-|-
-|'''(b)''' Since <math>f'(x)</math> is always positive, <math>f(x)</math> is an increasing function.
-|-
-|Thus, <math>f(x)</math> has at most one zero.
-|}
 [[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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