Difference between revisions of "009A Sample Midterm 2, Problem 2"

Latest revision as of 15:35, 9 November 2017

The function $f(x)=3x^{7}-8x+2$ is a polynomial and therefore continuous everywhere.

(a) State the Intermediate Value Theorem.

(b) Use the Intermediate Value Theorem to show that $f(x)$ has a zero in the interval $[0,1].$

Solution

Detailed Solution

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@@ Line 1: / Line 1: @@
-<span class="exam">The function <math>f(x)=3x^7-8x+2</math> is a polynomial and therefore continuous everywhere.
+<span class="exam">The function &nbsp;<math style="vertical-align: -5px">f(x)=3x^7-8x+2</math>&nbsp; is a polynomial and therefore continuous everywhere.
-::<span class="exam">a) State the Intermediate Value Theorem.
-::<span class="exam">b) Use the Intermediate Value Theorem to show that <math>f(x)</math> has a zero in the interval <math>[0,1].</math>
+<span class="exam">(a) State the Intermediate Value Theorem.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<span class="exam">(b) Use the Intermediate Value Theorem to show that &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has a zero in the interval &nbsp;<math style="vertical-align: -5px">[0,1].</math>
-!Foundations: &nbsp;
+<hr>
-|-
+[[009A Sample Midterm 2, Problem 2 Solution|'''<u>Solution</u>''']]
-|?
-|}
-'''Solution:'''
+[[009A Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|'''Intermediate Value Theorem'''
-|-
-|&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>
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; and <math style="vertical-align: 0px">c</math> is any number between <math style="vertical-align: -5px">f(a)</math>&thinsp; and <math style="vertical-align: -5px">f(b)</math>,
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; 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>
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, <math>f(x)</math> is continuous on the interval <math>[0,1]</math> since <math>f(x)</math> is continuous everywhere.
-|-
-|Also,
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>f(0)=2</math>
-|-
-|and
-&nbsp; &nbsp; &nbsp; &nbsp; <math>f(1)=3-8+2=-3.</math>.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Since <math>0</math> is between <math>f(0)=2</math> and <math>f(1)=-3,</math>
-|-
-|the Intermediate Value Theorem tells us that there is at least one number <math>x</math>
-|-
-|such that <math>f(x)=0.</math>
-|-
-|This means that <math>f(x)</math> has a zero in the interval <math>[0,1].</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; See solution above.
-|-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; See solution above.
-|}
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009A Sample Midterm 2, Problem 2"

Latest revision as of 15:35, 9 November 2017

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