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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. | | <span class="exam"> Show that the equation <math style="vertical-align: -2px">x^3+2x-2=0</math> has exactly one real root. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
| + | [[009A Sample Final 2, Problem 7 Solution|'''<u>Solution</u>''']] |
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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:''' | + | [[009A Sample Final 2, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !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.
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | See solution above.
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| − | |}
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| | [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
