Difference between revisions of "009A Sample Final 1, Problem 6"
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::::::<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 <math style="vertical-align: -3px">f(x)</math> 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 <math style="vertical-align: -3px">f(x)</math> has at most one zero. |
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
Revision as of 14:13, 22 February 2016
Consider the following function:
a) Use the Intermediate Value Theorem to show that has at least one zero.
b) Use the Mean Value Theorem to show that has at most one zero.
| Foundations: |
|---|
Solution:
(a)
| Step 1: |
|---|
| First note that . |
| Also, . |
| Since , |
| . |
| Thus, and hence . |
| Step 2: |
|---|
| Since and , there exists with such that |
| by the Intermediate Value Theorem. Hence, has at least one zero. |
(b)
| Step 1: |
|---|
| We have . Since , |
| . So, . |
| Therefore, is always positive. |
| Step 2: |
|---|
| Since is always positive, is an increasing function. |
| Thus, has at most one zero. |
| Final Answer: |
|---|
| (a) Since 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(-5)<0} and 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(0)>0} , there exists 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 x} with 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 -5<x<0} such 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)=0} by the Intermediate Value Theorem. Hence, 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) Since 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)} is always positive, 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)} is an increasing function. |
| Thus, 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. |