Difference between revisions of "022 Exam 2 Sample A, Problem 9"

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(Created page with "<span class="exam"> Find all relative extrema and points of inflection for the function <math style="vertical-align: -45%">g(x) = \frac{2}{3}x^3 + x^2 - 12x</math>. Be sure to...")
 
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::<math>g(-3)\,=\,\frac{2}{3}(-27)+9-12(-3)\,=\,27.</math>
 
::<math>g(-3)\,=\,\frac{2}{3}(-27)+9-12(-3)\,=\,27.</math>
 
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|By the second derivative test, the point <math style="vertical-align: -70%">(2,g(2))=\left(2,-\frac{44}{3}\right)</math> is a relative maximum, while the point <math style="vertical-align: -22%">(-3,g(-3))=(-3,27)</math> is a relative maximum.
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|By the second derivative test, the point <math style="vertical-align: -70%">(2,g(2))=\left(2,-\frac{44}{3}\right)</math> is a relative minimum, while the point <math style="vertical-align: -22%">(-3,g(-3))=(-3,27)</math> is a relative maximum.
 
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!Step 4: &nbsp;
 
!Step 4: &nbsp;
 
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|'''Test the potential inflection point:''' We know that <math style="vertical-align: -70%">g\,''\left(-\frac{1}{2}\right)=0</math>. On the other hand, it should be clear that if <math style="vertical-align: -60%">x<-\frac{1}{2}</math>, then <math style="vertical-align: -23%">g\,''(x)<0</math>.  Similarly, if <math style="vertical-align: -60%">x>-\frac{1}{2}</math>, then <math style="vertical-align: -23%">g\,''(x)>0</math>. Thus, the second derivative "splits" around <math style="vertical-align: -60%">x=-\frac{1}{2}</math>&thinsp; (i.e., changes sign), so the point <math style="vertical-align: -70%">\left(-\frac{1}{2},g\left(-\frac{1}{2}\right) \right)</math>&thinsp; is an inflection point.
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|'''Test the potential inflection point:''' We know that <math style="vertical-align: -25%">g\,''(-1/2)=0</math>. On the other hand, it should be clear that if <math style="vertical-align: -25%">x<-1/2</math>, then <math style="vertical-align: -23%">g\,''(x)<0</math>.  Similarly, if <math style="vertical-align: -25%">x>-1/2</math>, then <math style="vertical-align: -23%">g\,''(x)>0</math>. Thus, the second derivative "splits" around <math style="vertical-align: -25%">x=-1/2</math>&thinsp; (i.e., changes sign), so the point <math style="vertical-align: -25%">\left(-1/2,g(-1/2)\right)</math>&thinsp; is an inflection point.
 
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|Since
 
|Since
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
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|The area is maximized when both the length and width are 12 meters.
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|There is a local minimum at <math style="vertical-align: -70%">\left(2,-\frac{44}{3}\right)</math>, a local maximum at <math style="vertical-align: -22%">(-3,27)</math> and an inflection point at <math style="vertical-align: -70%">\left(-\frac{1}{2},\frac{19}{4}\right).</math>
 
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[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]
 
[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

Revision as of 22:38, 15 May 2015

Find all relative extrema and points of inflection for the function . Be sure to give coordinate pairs for each point. You do not need to draw the graph.

Foundations:  
Since our function is a polynomial, the relative extrema occur when the first derivative is zero. We then have two choices for finding if it is a local maximum or minimum:
Second Derivative Test: If the first derivative at a point is , and the second derivative is negative (indicating it is concave-down, like an upside-down parabola), then the point is a local maximum.
On the other hand, if the second derivative is positive, the point is a local minimum. You can also use the first derivative test, but it is usually a bit more work! For inflection points, we need to find when the second derivative is zero, as well as check that the second derivative "splits" on both sides.

 Solution:

Step 1:  
Find the first and second derivatives: Based on our function, we have
Similarly, from the first derivative we find
Step 2:  
Find the roots of the derivatives: We can rewrite the first derivative as
from which it should be clear we have roots and .
On the other hand, for the second derivative, we have
This has a single root: .
Step 3:  
Test the potential extrema: We know that are the candidates. We check the second derivative, finding
while
Note that
while
By the second derivative test, the point is a relative minimum, while the point is a relative maximum.
Step 4:  
Test the potential inflection point: We know that . On the other hand, it should be clear that if , then . Similarly, if , then . Thus, the second derivative "splits" around   (i.e., changes sign), so the point   is an inflection point.
Since
our inflection point is
Final Answer:  
There is a local minimum at , a local maximum at and an inflection point at

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