009A Sample Final 1, Problem 10 - Revision history

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MathAdmin: Created page with "Consider the following continuous function: :::::: $f(x)=x^{1/3}(x-8)$ defined on the closed, bounded interval
2016-03-05T07:11:43Z

Created page with "Consider the following continuous function: ::::::<math>f(x)=x^{1/3}(x-8)</math> defined on the closed, bounded interval <math style="ve..."

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Consider the following continuous function:
::::::<math>f(x)=x^{1/3}(x-8)</math>

defined on the closed, bounded interval <math style="vertical-align: -5px">[-8,8]</math>.

a) Find all the critical points for <math style="vertical-align: -5px">f(x)</math>.

b) Determine the absolute maximum and absolute minimum values for <math style="vertical-align: -5px">f(x)</math> on the interval <math style="vertical-align: -5px">[-8,8]</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' To find the critical points for <math style="vertical-align: -5px">f(x),</math> we set <math style="vertical-align: -5px">f'(x)=0</math> and solve for <math style="vertical-align: -1px">x.</math>
|-
|
::Also, we include the values of <math style="vertical-align: -1px">x</math> where <math style="vertical-align: -5px">f'(x)</math> is undefined.
|-
|'''2.''' To find the absolute maximum and minimum of <math style="vertical-align: -5px">f(x)</math> on an interval <math>[a,b],</math>
|-
|
::we need to compare the <math style="vertical-align: -5px">y</math> values of our critical points with <math style="vertical-align: -5px">f(a)</math> and <math style="vertical-align: -5px">f(b).</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the critical points, first we need to find <math style="vertical-align: -5px">f'(x).</math>
|-
|Using the Product Rule, we have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{3}x^{-\frac{2}{3}}(x-8)+x^{\frac{1}{3}}}\\
&&\\
& = & \displaystyle{\frac{x-8}{3x^{\frac{2}{3}}}+x^{\frac{1}{3}}.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Notice <math style="vertical-align: -5px">f'(x)</math> is undefined when <math style="vertical-align: -1px">x=0.</math>
|-
|Now, we need to set <math style="vertical-align: -5px">f'(x)=0.</math>
|-
|So, we get
|-
|
::<math>-x^{\frac{1}{3}}\,=\,\frac{x-8}{3x^{\frac{2}{3}}}.</math>
|-
|We cross multiply to get <math style="vertical-align: 1px">-3x=x-8.</math>
|-
|Solving, we get <math style="vertical-align: -1px">x=2.</math>
|-
|Thus, the critical points for <math style="vertical-align: -5px">f(x)</math> are <math style="vertical-align: -5px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6)).</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We need to compare the values of <math style="vertical-align: -5px">f(x)</math>  at the critical points and at the endpoints of the interval.
|-
|Using the equation given, we have <math style="vertical-align: -5px">f(-8)=32</math>  and <math style="vertical-align: -5px">f(8)=0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Comparing the values in Step 1 with the critical points in '''(a)''', the absolute maximum value for <math style="vertical-align: -5px">f(x)</math>  is <math style="vertical-align: -1px">32</math>
|-
|and the absolute minimum value for <math style="vertical-align: -5px">f(x)</math>  is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math style="vertical-align: -4px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6))</math>
|-
|'''(b)'''  The absolute minimum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 1: / Line 1: @@
 <span class="exam">Consider the following continuous function:
-::::::<math>f(x)=x^{1/3}(x-8)</math>
+::<math>f(x)=x^{1/3}(x-8)</math>
-<span class="exam">defined on the closed, bounded interval <math style="vertical-align: -5px">[-8,8]</math>.
+<span class="exam">defined on the closed, bounded interval &nbsp;<math style="vertical-align: -5px">[-8,8]</math>.
-::<span class="exam">a) Find all the critical points for <math style="vertical-align: -5px">f(x)</math>.
+<span class="exam">(a) Find all the critical points for &nbsp;<math style="vertical-align: -5px">f(x)</math>.
-::<span class="exam">b) Determine the absolute maximum and absolute minimum values for <math style="vertical-align: -5px">f(x)</math> on the interval <math style="vertical-align: -5px">[-8,8]</math>.
+<span class="exam">(b) Determine the absolute maximum and absolute minimum values for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; on the interval &nbsp;<math style="vertical-align: -5px">[-8,8]</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
-|Recall:
+|'''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>
 |-
+|
-::'''1.''' To find the critical points for <math style="vertical-align: -5px">f(x),</math> we set <math style="vertical-align: -5px">f'(x)=0</math> and solve for <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>
 |-
+|
-:::we need to compare the <math style="vertical-align: -5px">y</math> values of our critical points with <math style="vertical-align: -5px">f(a)</math> and <math style="vertical-align: -5px">f(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:'''
@@ Line 33: / Line 30: @@
 !Step 1: &nbsp;
 |-
-|To find the critical points, first we need to find <math style="vertical-align: -5px">f'(x).</math>
+|To find the critical points, first we need to find &nbsp;<math style="vertical-align: -5px">f'(x).</math>
 |-
 |Using the Product Rule, we have
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{f'(x)} & = & \displaystyle{\frac{1}{3}x^{-\frac{2}{3}}(x-8)+x^{\frac{1}{3}}}\\
 &&\\
@@ Line 48: / Line 45: @@
 !Step 2: &nbsp;
 |-
-|Notice <math style="vertical-align: -5px">f'(x)</math> is undefined when <math style="vertical-align: -1px">x=0.</math>
+|Notice &nbsp;<math style="vertical-align: -5px">f'(x)</math>&nbsp; is undefined when &nbsp;<math style="vertical-align: -1px">x=0.</math>
 |-
-|Now, we need to set <math style="vertical-align: -5px">f'(x)=0.</math>
+|Now, we need to set &nbsp;<math style="vertical-align: -5px">f'(x)=0.</math>
 |-
 |So, we get
 |-
+|
-::<math>-x^{\frac{1}{3}}\,=\,\frac{x-8}{3x^{\frac{2}{3}}}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>-x^{\frac{1}{3}}\,=\,\frac{x-8}{3x^{\frac{2}{3}}}.</math>
 |-
-|We cross multiply to get
+|We cross multiply to get &nbsp;<math style="vertical-align: 1px">-3x=x-8.</math>
 |-
+|Solving, we get &nbsp;<math style="vertical-align: -1px">x=2.</math>
 |-
-|Solving, we get <math style="vertical-align: -1px">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)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6)).</math>
 |}
@@ Line 72: / Line 66: @@
 !Step 1: &nbsp;
 |-
-|We need to compare the values of <math style="vertical-align: -5px">f(x)</math>&thinsp; at the critical points and at the endpoints of the interval.
+|We need to compare the values of &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at the critical points and at the endpoints of the interval.
 |-
-|Using the equation given, we have <math style="vertical-align: -5px">f(-8)=32</math>&thinsp; and <math style="vertical-align: -5px">f(8)=0.</math>
+|Using the equation given, we have &nbsp;<math style="vertical-align: -5px">f(-8)=32</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(8)=0.</math>
 |}
@@ Line 80: / Line 74: @@
 !Step 2: &nbsp;
 |-
-|Comparing the values in Step 1 with the critical points in '''(a)''', the absolute maximum value for <math style="vertical-align: -5px">f(x)</math>&thinsp; is <math style="vertical-align: -1px">32</math>
+|Comparing the values in Step 1 with the critical points in (a), the absolute maximum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">32</math>
 |-
-|and the absolute minimum value for <math style="vertical-align: -5px">f(x)</math>&thinsp; is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
+|and the absolute minimum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)'''&thinsp; <math style="vertical-align: -4px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6))</math>
+|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math style="vertical-align: -4px">(0,0)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6))</math>
 |-
-|&nbsp;&nbsp; '''(b)'''&thinsp; The absolute minimum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
+|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; The absolute maximum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">32</math>&nbsp; and the absolute minimum value for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 17: / Line 17: @@
 |-
+|
-::Also, we include the values of <math style="vertical-align: -1px">x</math> where <math style="vertical-align: -5px">f'(x)</math> is undefined.
+:::Also, we include the values of <math style="vertical-align: -1px">x</math> where <math style="vertical-align: -5px">f'(x)</math> is undefined.
 |-
+|
@@ Line 23: / Line 23: @@
 |-
+|
-::we need to compare the <math style="vertical-align: -5px">y</math> values of our critical points with <math style="vertical-align: -5px">f(a)</math> and <math style="vertical-align: -5px">f(b).</math>
+:::we need to compare the <math style="vertical-align: -5px">y</math> values of our critical points with <math style="vertical-align: -5px">f(a)</math> and <math style="vertical-align: -5px">f(b).</math>
 |}