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

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!Step 1:    
 
!Step 1:    
 
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|Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exists <math style="vertical-align: -5px">a,b</math> such that <math style="vertical-align: -5px">f(a)=f(b)=0.</math>
+
|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>
 
|-
 
|-
|Then, by the Mean Value Theorem, there exists <math style="vertical-align: -1px">c</math> with <math style="vertical-align: -1px">a<c<b</math> such that <math style="vertical-align: -5px">f'(c)=0.</math>
+
|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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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|We have <math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math> Since <math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
+
|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>
 
|-
 
|-
|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math> So, <math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
+
|<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>
 
|-
 
|-
|which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math> has at most one zero.
+
|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.
 
|}
 
|}
 +
 
== 3 ==
 
== 3 ==
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

Revision as of 12:15, 4 March 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:  
Recall:
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 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:
  is continuous on the closed interval  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle [a,b].}
  is differentiable on the open interval
Then, there is a number such that    and

Solution:

1

(a)

Step 1:  
First note that 
Also, 
Since 
Thus,    and hence  
Step 2:  
Since   and    there exists with    such that
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=0}   by the Intermediate Value Theorem. Hence,   has at least one zero.

2

(b)

Step 1:  
Suppose that has more than one zero. So, there exist such that  
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.

3

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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