022 Exam 2 Sample A, Problem 9

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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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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},}
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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