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2017-04-10T16:24:58Z

Created page with "<span class="exam"> Show that the equation <math style="vertical-align: -2px">x^3+2x-2=0</math> has exactly one real root. {| class="mw-collapsible mw-collapsed"..."

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<span class="exam"> Show that the equation  <math style="vertical-align: -2px">x^3+2x-2=0</math>  has exactly one real root.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
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|'''1.''' '''Intermediate Value Theorem'''
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|       If  <math style="vertical-align: -5px">f(x)</math>  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>  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'''
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|        Suppose  <math style="vertical-align: -5px">f(x)</math>  is a function that satisfies the following:
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       <math style="vertical-align: -5px">f(x)</math>  is continuous on the closed interval  <math style="vertical-align: -5px">[a,b].</math>
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       <math style="vertical-align: -5px">f(x)</math>  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  <math style="vertical-align: 0px">a<c<b</math>  and  <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
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'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|First, we note that
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|       <math>f(0)=-2.</math>
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|Also,
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|       <math>f(1)=1.</math>
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|Since  <math style="vertical-align: -5px">f(0)<0</math>  and  <math style="vertical-align: -5px">f(1)>0,</math> 
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|there exists  <math style="vertical-align: 0px">x</math>  with  <math style="vertical-align: -1px">0<x<1</math>  such that
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|       <math style="vertical-align: -5px">f(x)=0</math> 
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|by the Intermediate Value Theorem.
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|Hence,  <math style="vertical-align: -5px">f(x)</math>  has at least one zero.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|Suppose that  <math style="vertical-align: -5px">f(x)</math>  has more than one zero.
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|So, there exist  <math style="vertical-align: -4px">a,b</math>  such that
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|       <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  <math style="vertical-align: 0px">a<c<b</math>  such that
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|       <math style="vertical-align: -5px">f'(c)=0.</math>
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|We have  <math style="vertical-align: -5px">f'(x)=3x^2+2.</math> 
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|Since  <math style="vertical-align: -5px">x^2\ge 0,</math>
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|       <math style="vertical-align: -5px"> f'(x) \ge 2.</math>
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|Therefore, it is impossible for  <math style="vertical-align: -5px">f'(c)=0.</math>  Hence,  <math style="vertical-align: -5px">f(x)</math>  has at most one zero.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|        See solution above.
|}
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

009A Sample Final 2, Problem 7 - Revision history

MathAdmin: Created page with " Show that the equation x^3+2x-2=0 has exactly one real root. {| class="mw-collapsible mw-collapsed"..."

MathAdmin: Created page with " Show that the equation $x^3+2x-2=0$ has exactly one real root. {| class="mw-collapsible mw-collapsed"..."