Difference between revisions of "022 Exam 2 Sample A, Problem 9"
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'''Second Derivative Test:''' If the first derivative at a point <math style="verticalalign: 12%">x_0</math> is <math style="verticalalign: 0%">0</math>, and the second derivative is negative (indicating it is concavedown, like an upsidedown parabola), then the point <math style="verticalalign: 20%">\left(x_0,f(x_0)\right)</math> is a local maximum.  '''Second Derivative Test:''' If the first derivative at a point <math style="verticalalign: 12%">x_0</math> is <math style="verticalalign: 0%">0</math>, and the second derivative is negative (indicating it is concavedown, like an upsidedown parabola), then the point <math style="verticalalign: 20%">\left(x_0,f(x_0)\right)</math> is a local maximum.  
    
−  On the other hand, if the second derivative is positive, the point <math style="verticalalign: 20%">\left(x_0,f(x_0)\right)</math> 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.  +  On the other hand, if the second derivative is positive, the point <math style="verticalalign: 20%">\left(x_0,f(x_0)\right)</math> 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. 
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−  ::<math>g\,''(x)\,=\,4x+2\,=\,  +  ::<math>g\,''(x)\,=\,4x+2\,=\,4\left(x+\frac{1}{2}\right).</math> 
    
This has a single root: <math style="verticalalign: 60%">x=\frac{1}{2}</math>.  This has a single root: <math style="verticalalign: 60%">x=\frac{1}{2}</math>.  
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−  ::<math>g\left(\frac{1}{2}\right)\,=\,\frac{2}{3}\  +  ::<math>g\left(\frac{1}{2}\right)\,=\,\frac{2}{3}\left(\frac{1}{8}\right)+\frac{1}{4}12\left(\frac{1}{2}\right)\,=\,\frac{37}{6},</math> 
    
our inflection point is  our inflection point is  
    
    
−  ::<math>\left(\frac{1}{2},\frac{  +  ::<math>\left(\frac{1}{2},\frac{37}{6}\right).</math> 
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!Final Answer:  !Final Answer:  
    
−  There is a local minimum at <math style="verticalalign: 70%">\left(2,\frac{44}{3}\right)</math>, a local maximum at <math style="verticalalign: 22%">(3,27)</math> and an inflection point at <math style="verticalalign: 70%">\left(\frac{1}{2},\frac{  +  There is a local minimum at <math style="verticalalign: 70%">\left(2,\frac{44}{3}\right)</math>, a local maximum at <math style="verticalalign: 22%">(3,27)</math> and an inflection point at <math style="verticalalign: 70%">\left(\frac{1}{2},\frac{37}{6}\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>''']] 
Latest revision as of 08:12, 16 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 concavedown, like an upsidedown 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 