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="vertical-align: -12%">x_0</math> is <math style="vertical-align: 0%">0</math>, and the second derivative is negative (indicating it is concave-down, like an upside-down parabola), then the point <math style="vertical-align: -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="vertical-align: -12%">x_0</math> is <math style="vertical-align: 0%">0</math>, and the second derivative is negative (indicating it is concave-down, like an upside-down parabola), then the point <math style="vertical-align: -20%">\left(x_0,f(x_0)\right)</math> is a local maximum.
 
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|On the other hand, if the second derivative is positive, the point <math style="vertical-align: -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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|On the other hand, if the second derivative is positive, the point <math style="vertical-align: -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\,=\,2\left(x+\frac{1}{2}\right).</math>
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::<math>g\,''(x)\,=\,4x+2\,=\,4\left(x+\frac{1}{2}\right).</math>
 
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|This has a single root: <math style="vertical-align: -60%">x=-\frac{1}{2}</math>.
 
|This has a single root: <math style="vertical-align: -60%">x=-\frac{1}{2}</math>.

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 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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