022 Exam 2 Sample A, Problem 9

Find all relative extrema and points of inflection for the function $g(x)={\frac {2}{3}}x^{3}+x^{2}-12x$. 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 $x_{0}$ is $0$, and the second derivative is negative (indicating it is concave-down, like an upside-down parabola), then the point $\left(x_{0},f(x_{0})\right)$ is a local maximum.
On the other hand, if the second derivative is positive, the point $\left(x_{0},f(x_{0})\right)$ 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
$g\,'(x)\,=\,{\frac {2}{3}}\cdot 3x^{2}+2x-12\,=\,2x^{2}+2x-12.$
Similarly, from the first derivative we find
$g\,''(x)\,=\,4x+2.$

Step 2:
Find the roots of the derivatives: We can rewrite the first derivative as
$g\,'(x)\,=\,2x^{2}+2x-12\,=\,2(x^{2}+x-6)\,=\,2(x+3)(x-2),$
from which it should be clear we have roots $2$ and $-3$.
On the other hand, for the second derivative, we have
$g\,''(x)\,=\,4x+2\,=\,4\left(x+{\frac {1}{2}}\right).$
This has a single root: $x=-{\frac {1}{2}}$.

Step 3:
Test the potential extrema: We know that $x=2,-3$ are the candidates. We check the second derivative, finding
$g\,''(2)\,=\,4\cdot 2+2\,>\,0,$
while
$g\,''(-3)\,=\,2(-3)+2\,<\,0.$
Note that
$g(2)\,=\,{\frac {2}{3}}(8)+4-24\,=\,-{\frac {44}{3}},$
while
$g(-3)\,=\,{\frac {2}{3}}(-27)+9-12(-3)\,=\,27.$
By the second derivative test, the point $(2,g(2))=\left(2,-{\frac {44}{3}}\right)$ is a relative minimum, while the point $(-3,g(-3))=(-3,27)$ is a relative maximum.

Step 4:
Test the potential inflection point: We know that $g\,''(-1/2)=0$. On the other hand, it should be clear that if $x<-1/2$, then $g\,''(x)<0$. Similarly, if $x>-1/2$, then $g\,''(x)>0$. Thus, the second derivative "splits" around $x=-1/2$  (i.e., changes sign), so the point $\left(-1/2,g(-1/2)\right)$  is an inflection point.
Since
$g\left(-{\frac {1}{2}}\right)\,=\,{\frac {2}{3}}\left(-{\frac {1}{8}}\right)+{\frac {1}{4}}-12\left(-{\frac {1}{2}}\right)\,=\,{\frac {37}{6}},$
our inflection point is
$\left(-{\frac {1}{2}},{\frac {37}{6}}\right).$

Final Answer:
There is a local minimum at $\left(2,-{\frac {44}{3}}\right)$, a local maximum at $(-3,27)$ and an inflection point at $\left(-{\frac {1}{2}},{\frac {37}{6}}\right).$

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