Difference between revisions of "009A Sample Final 3, Problem 9"

Latest revision as of 16:47, 2 December 2017

Let

g(x)=(2x^{2}-8x)^{\frac {2}{3}}

(a) Find all critical points of $g$ over the $x$ -interval $[0,8].$

(b) Find absolute maximum and absolute minimum of $g$ over $[0,8].$

Solution

Detailed Solution

Return to Sample Exam

@@ Line 7: / Line 7: @@
 <span class="exam">(b) Find absolute maximum and absolute minimum of &nbsp;<math style="vertical-align: -4px">g</math>&nbsp; over &nbsp;<math style="vertical-align: -5px">[0,8].</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 3, Problem 9 Solution|'''<u>Solution</u>''']]
-|-
-|'''1.''' To find the critical points for &nbsp;<math style="vertical-align: -5px">f(x),</math>&nbsp; we set &nbsp;<math style="vertical-align: -5px">f'(x)=0</math>&nbsp; and solve for &nbsp;<math style="vertical-align: -1px">x.</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; Also, we include the values of &nbsp;<math style="vertical-align: -1px">x</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">f'(x)</math>&nbsp; is undefined.
-|-
-|'''2.''' To find the absolute maximum and minimum of &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; on an interval &nbsp;<math>[a,b],</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; we need to compare the &nbsp;<math style="vertical-align: -5px">y</math>&nbsp; values of our critical points with &nbsp;<math style="vertical-align: -5px">f(a)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(b).</math>
-|}
-'''Solution:'''
+[[009A Sample Final 3, Problem 9 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|To find the critical points, first we need to find &nbsp;<math style="vertical-align: -5px">g'(x).</math>
-|-
-|Using the Chain Rule, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{g'(x)} & = & \displaystyle{\frac{2}{3}(2x^2-8x)^{-\frac{1}{3}}(2x^2-8x)'}\\
-&&\\
-& = & \displaystyle{\frac{2}{3}(2x^2-8x)^{-\frac{1}{3}}(4x-8)}\\
-&&\\
-& = & \displaystyle{\frac{8x-16}{3\sqrt[3]{2x^2-8x}}.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|First, we note that &nbsp;<math style="vertical-align: -5px">g'(x)</math>&nbsp; is undefined when
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>3\sqrt[3]{2x^2-8x}=0.</math>
-|-
-|Solving for &nbsp;<math style="vertical-align: -4px">x,</math>&nbsp; we get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{0} & = & \displaystyle{2x^2-8x}\\
-&&\\
-& = & \displaystyle{x(2x-8).}
-\end{array}</math>
-|-
-|Therefore, &nbsp;<math style="vertical-align: -5px">g'(x)</math>&nbsp; is undefined when &nbsp;<math style="vertical-align: -4px">x=0,4.</math>&nbsp;
-|-
-|Now, we need to set &nbsp;<math style="vertical-align: -5px">g'(x)=0.</math>
-|-
-|So, we get
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>8x-16=0.</math>
-|-
-|Solving, we get &nbsp;<math style="vertical-align: 0px">x=2.</math>
-|-
-|Thus, the critical points for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; are &nbsp;<math style="vertical-align: -5px">(0,0),(2,4),(4,0).</math>
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|
-|-
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;&nbsp;<math>(0,0),(2,4),(4,0).</math>
-|-
-|'''(b)'''
-|}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009A Sample Final 3, Problem 9"

Latest revision as of 16:47, 2 December 2017

Navigation menu

Search