Difference between revisions of "009A Sample Final 1, Problem 10"

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<span class="exam">Consider the following continuous function:
 
<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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!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>
:::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>
+
&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:'''
 
'''Solution:'''
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!Step 1: &nbsp;  
 
!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
 
|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}}}\\
 
\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{3}x^{-\frac{2}{3}}(x-8)+x^{\frac{1}{3}}}\\
 
&&\\
 
&&\\
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!Step 2: &nbsp;
 
!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  
 
|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>
::<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 &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>
|-
 
|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>
 
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!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>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!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>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 09:18, 10 April 2017

Consider the following continuous function:

defined on the closed, bounded interval  .

(a) Find all the critical points for  .

(b) Determine the absolute maximum and absolute minimum values for    on the interval  .

Foundations:  
1. To find the critical points for    we set    and solve for  

        Also, we include the values of    where    is undefined.

2. To find the absolute maximum and minimum of    on an interval  

        we need to compare the    values of our critical points with    and  


Solution:

(a)

Step 1:  
To find the critical points, first we need to find  
Using the Product Rule, we have

       

Step 2:  
Notice    is undefined when  
Now, we need to set  
So, we get

       

We cross multiply to get  
Solving, we get  
Thus, the critical points for    are    and  

(b)

Step 1:  
We need to compare the values of    at the critical points and at the endpoints of the interval.
Using the equation given, we have    and  
Step 2:  
Comparing the values in Step 1 with the critical points in (a), the absolute maximum value for    is  
and the absolute minimum value for    is  


Final Answer:  
    (a)      and  
    (b)    The absolute maximum value for    is    and the absolute minimum value for    is  

Return to Sample Exam