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

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
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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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::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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|'''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'''  
 +
|-
 +
|&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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::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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|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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|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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::<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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|-
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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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|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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|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 16:48, 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   has at most one zero.

Foundations:  
1. Intermediate Value Theorem
       If   is continuous on a closed interval and is any number

       between   and , then there is at least one number in the closed interval such that

2. Mean Value Theorem
        Suppose   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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