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

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!Foundations:    
 
!Foundations:    
 
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|-
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|Recall:
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|'''1.''' '''Intermediate Value Theorem'''
 
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|-
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|'''1. Intermediate Value Theorem:''' If <math style="vertical-align: -5px">f(x)</math>&thinsp; is continuous on a closed interval <math style="vertical-align: -5px">[a,b]</math> and <math style="vertical-align: 0px">c</math> is any number
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|&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -5px">f(x)</math>&thinsp; is continuous on a closed interval <math style="vertical-align: -5px">[a,b]</math> and <math style="vertical-align: 0px">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>&thinsp; and <math style="vertical-align: -5px">f(b)</math>, then there is at least one number <math style="vertical-align: 0px">x</math> in the closed interval such that <math style="vertical-align: -5px">f(x)=c.</math>
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&nbsp; &nbsp; &nbsp; &nbsp;between <math style="vertical-align: -5px">f(a)</math>&thinsp; and <math style="vertical-align: -5px">f(b)</math>, then there is at least one number <math style="vertical-align: 0px">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>&thinsp; is a function that satisfies the following:
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|'''2.'''  '''Mean Value Theorem'''  
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|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; Suppose <math style="vertical-align: -5px">f(x)</math>&thinsp; is a function that satisfies the following:
 
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|
 
|
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::<math style="vertical-align: -5px">f(x)</math>&thinsp; is continuous on the closed interval &thinsp;<math style="vertical-align: -5px">[a,b].</math>
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&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&thinsp; is continuous on the closed interval &thinsp;<math style="vertical-align: -5px">[a,b].</math>
 
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|-
 
|
 
|
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::<math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable on the open interval <math style="vertical-align: -5px">(a,b).</math>
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&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(x)</math>&thinsp; is differentiable on the open interval <math style="vertical-align: -5px">(a,b).</math>
 
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|-
 
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::Then, there is a number <math style="vertical-align: 0px">c</math> such that &thinsp;<math style="vertical-align: 0px">a<c<b</math>&thinsp; and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
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&nbsp; &nbsp; &nbsp; &nbsp;Then, there is a number <math style="vertical-align: 0px">c</math> such that &thinsp;<math style="vertical-align: 0px">a<c<b</math>&thinsp; and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|First note that&thinsp; <math style="vertical-align: -5px">f(0)=7.</math>
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|First note that &nbsp; <math style="vertical-align: -5px">f(0)=7.</math>
 
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|-
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|Also,&thinsp; <math style="vertical-align: -5px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
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|Also,&nbsp; <math style="vertical-align: -5px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
 
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|-
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|Since&thinsp; <math style="vertical-align: -5px">-1\leq \sin(x) \leq 1,</math>
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|Since&nbsp; <math style="vertical-align: -5px">-1\leq \sin(x) \leq 1,</math>
 
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|-
 
|
 
|
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::<math>-2\leq -2\sin(x) \leq 2.</math>
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&nbsp; &nbsp; &nbsp; &nbsp;<math>-2\leq -2\sin(x) \leq 2.</math>
 
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|-
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|Thus,&thinsp; <math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math>&thinsp; and hence &thinsp;<math style="vertical-align: -5px">f(-5)<0.</math>
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|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>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Since <math style="vertical-align: -5px">f(-5)<0</math>&thinsp; and &thinsp;<math style="vertical-align: -5px">f(0)>0,</math>&thinsp; there exists <math style="vertical-align: 0px">x</math> with &thinsp;<math style="vertical-align: 0px">-5<x<0</math>&thinsp; such that  
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|Since <math style="vertical-align: -5px">f(-5)<0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(0)>0,</math>&nbsp; there exists <math style="vertical-align: 0px">x</math> with &nbsp;<math style="vertical-align: 0px">-5<x<0</math>&nbsp; such that  
 
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|-
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|<math style="vertical-align: -5px">f(x)=0</math>&thinsp; by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>&thinsp; has at least one zero.
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|<math style="vertical-align: -5px">f(x)=0</math>&nbsp; by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>&nbsp; has at least one zero.
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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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 exist <math style="vertical-align: -4px">a,b</math> such that &thinsp;<math style="vertical-align: -5px">f(a)=f(b)=0.</math>
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|Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exist <math style="vertical-align: -4px">a,b</math> such that &nbsp;<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: 0px">c</math> with &thinsp;<math style="vertical-align: 0px">a<c<b</math> such that &thinsp;<math style="vertical-align: -5px">f'(c)=0.</math>
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|Then, by the Mean Value Theorem, there exists <math style="vertical-align: 0px">c</math> with &nbsp;<math style="vertical-align: 0px">a<c<b</math> such that &nbsp;<math style="vertical-align: -5px">f'(c)=0.</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|-
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|We have <math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math>&thinsp; Since &thinsp;<math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
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|We have <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>
 
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|-
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|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math>&thinsp; So, <math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
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|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math>&nbsp; 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>&thinsp; has at most one zero.
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|which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math>&nbsp; has at most one zero.
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)''' Since <math style="vertical-align: -5px">f(-5)<0</math>&thinsp; and &thinsp;<math style="vertical-align: -5px">f(0)>0,</math>&thinsp; there exists <math style="vertical-align: 0px">x</math> with &thinsp;<math style="vertical-align: 0px">-5<x<0</math>&thinsp; such that  
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; Since <math style="vertical-align: -5px">f(-5)<0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(0)>0,</math>&nbsp; there exists <math style="vertical-align: 0px">x</math> with &nbsp;<math style="vertical-align: 0px">-5<x<0</math>&nbsp; such that  
 
|-
 
|-
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|<math style="vertical-align: -5px">f(x)=0</math>&thinsp; by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>&thinsp; has at least one zero.
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|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">f(x)=0</math>&nbsp; by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>&nbsp; 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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|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; See Step 1 and Step 2 above.
 
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[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 17:48, 25 February 2017

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 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.

Foundations:  
1. Intermediate Value Theorem
       If 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 continuous on a closed interval 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 [a,b]} 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 c} is any number

       between 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(a)}   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(b)} , then there is at least one number 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} in the closed interval 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)=c.}

2. Mean Value Theorem
        Suppose 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 a function that satisfies the following:

       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 continuous on the closed interval  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 [a,b].}

       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 differentiable on the open interval 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 (a,b).}

       Then, there is a number 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 c} 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 a<c<b}   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'(c)=\frac{f(b)-f(a)}{b-a}.}


Solution:

(a)

Step 1:  
First note 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(0)=7.}
Also,  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)=-15-2\sin(-5)+7=-8-2\sin(-5).}
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 -1\leq \sin(x) \leq 1,}

       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 -2\leq -2\sin(x) \leq 2.}

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 -10\leq f(-5) \leq -6}   and 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(-5)<0.}
Step 2:  
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)

Step 1:  
Suppose 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 more than one zero. So, there exist 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 a,b} 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(a)=f(b)=0.}
Then, by the Mean Value Theorem, 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 c} 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 a<c<b} 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'(c)=0.}
Step 2:  
We have 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)=3-2\cos(x).}   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 -1\leq \cos(x)\leq 1,}
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 -2 \leq -2\cos(x)\leq 2.}   So, 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 1\leq f'(x) \leq 5,}
which contradicts 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'(c)=0.} 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.


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)     See Step 1 and Step 2 above.

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