# 022 Exam 2 Sample A, Problem 9

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Find all relative extrema and points of inflection for the function ${\displaystyle g(x)={\frac {2}{3}}x^{3}+x^{2}-12x}$. 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 ${\displaystyle x_{0}}$ is ${\displaystyle 0}$, and the second derivative is negative (indicating it is concave-down, like an upside-down parabola), then the point ${\displaystyle \left(x_{0},f(x_{0})\right)}$ is a local maximum.
On the other hand, if the second derivative is positive, the point ${\displaystyle \left(x_{0},f(x_{0})\right)}$ 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
${\displaystyle g\,'(x)\,=\,{\frac {2}{3}}\cdot 3x^{2}+2x-12\,=\,2x^{2}+2x-12.}$
Similarly, from the first derivative we find
${\displaystyle g\,''(x)\,=\,4x+2.}$
Step 2:
Find the roots of the derivatives: We can rewrite the first derivative as
${\displaystyle g\,'(x)\,=\,2x^{2}+2x-12\,=\,2(x^{2}+x-6)\,=\,2(x+3)(x-2),}$
from which it should be clear we have roots ${\displaystyle 2}$ and ${\displaystyle -3}$.
On the other hand, for the second derivative, we have
${\displaystyle g\,''(x)\,=\,4x+2\,=\,2\left(x+{\frac {1}{2}}\right).}$
This has a single root: ${\displaystyle x=-{\frac {1}{2}}}$.
Step 3:
Test the potential extrema: We know that ${\displaystyle x=2,-3}$ are the candidates. We check the second derivative, finding
${\displaystyle g\,''(2)\,=\,4\cdot 2+2\,>\,0,}$
while
${\displaystyle g\,''(-3)\,=\,2(-3)+2\,<\,0.}$
Note that
${\displaystyle g(2)\,=\,{\frac {2}{3}}(8)+4-24\,=\,-{\frac {44}{3}},}$
while
${\displaystyle g(-3)\,=\,{\frac {2}{3}}(-27)+9-12(-3)\,=\,27.}$
By the second derivative test, the point ${\displaystyle (2,g(2))=\left(2,-{\frac {44}{3}}\right)}$ is a relative maximum, while the point ${\displaystyle (-3,g(-3))=(-3,27)}$ is a relative maximum.
Step 4:
Test the potential inflection point: We know that ${\displaystyle g\,''\left(-{\frac {1}{2}}\right)=0}$. On the other hand, it should be clear that if ${\displaystyle x<-{\frac {1}{2}}}$, then ${\displaystyle g\,''(x)<0}$. Similarly, if ${\displaystyle x>-{\frac {1}{2}}}$, then ${\displaystyle g\,''(x)>0}$. Thus, the second derivative "splits" around ${\displaystyle x=-{\frac {1}{2}}}$  (i.e., changes sign), so the point ${\displaystyle \left(-{\frac {1}{2}},g\left(-{\frac {1}{2}}\right)\right)}$  is an inflection point.
Since
${\displaystyle g\left(-{\frac {1}{2}}\right)\,=\,{\frac {2}{3}}\cdot -{\frac {1}{8}}+{\frac {1}{4}}-12\left(-{\frac {1}{2}}\right)\,=\,{\frac {19}{4}},}$
our inflection point is
${\displaystyle \left(-{\frac {1}{2}},{\frac {19}{4}}\right).}$