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

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!Final Answer:    
 
!Final Answer:    
 
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|'''(a)'''&thinsp; <math style="vertical-align: -4px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6))</math>
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|&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>
 
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|'''(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>
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|&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>
 
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[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 15:16, 18 April 2016

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:  
Recall:
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 minimum value for is

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