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| | <span class="exam">(b) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math> has a zero in the interval <math style="vertical-align: -5px">[0,1].</math> | | <span class="exam">(b) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math> has a zero in the interval <math style="vertical-align: -5px">[0,1].</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009A Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']] |
| − | !Foundations:
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| − | |What is a zero of the function <math style="vertical-align: -5px">f(x)?</math>
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| − | | A zero is a value <math style="vertical-align: -1px">c</math> such that <math style="vertical-align: -5px">f(c)=0.</math>
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| − | '''Solution:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !(a)
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| − | |'''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>
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| − | | and <math style="vertical-align: 0px">c</math> is any number between <math style="vertical-align: -5px">f(a)</math> and <math style="vertical-align: -5px">f(b),</math>
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| − | 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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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |First, <math style="vertical-align: -5px">f(x)</math> is continuous on the interval <math style="vertical-align: -5px">[0,1]</math> since <math style="vertical-align: -5px">f(x)</math> is continuous everywhere.
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| − | |Also,
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| − | <math>f(0)=2</math>
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| − | |and
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| − | <math>f(1)=3-8+2=-3.</math>.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Since <math style="vertical-align: -1px">0</math> is between <math style="vertical-align: -5px">f(0)=2</math> and <math style="vertical-align: -5px">f(1)=-3,</math>
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| − | |the Intermediate Value Theorem tells us that there is at least one number <math style="vertical-align: -1px">x</math>
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| − | |such that <math style="vertical-align: -5px">f(x)=0.</math>
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| − | |This means that <math style="vertical-align: -5px">f(x)</math> has a zero in the interval <math style="vertical-align: -5px">[0,1].</math>
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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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| − | | '''(a)''' See solution above.
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| − | | '''(b)''' See solution above.
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| − | |}
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| | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
The function 
is a polynomial and therefore continuous everywhere.
(a) State the Intermediate Value Theorem.
(b) Use the Intermediate 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 a zero in the 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 [0,1].}
(insert picture of handwritten solution)
Detailed Solution for this Problem
Return to Sample Exam