# 022 Exam 2 Sample B, Problem 9

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Find all relative extrema and points of inflection for the function ${\displaystyle g(x)=x^{3}-3x}$. Be sure to give coordinate pairs for each point. You do not need to draw the graph. Explain how you know which point is the local minimum and which is the local maximum (i.e., which test did you use?).

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
${\displaystyle g\,'(x)\,=\,3x^{2}-3.}$
Similarly, from the first derivative we find
${\displaystyle g\,''(x)\,=\,6x.}$
Step 2:
Find the roots of the derivatives: We can rewrite the first derivative as
${\displaystyle g\,'(x)\,=\,3x^{2}-3\,=\,3(x^{2}-1)\,=\,3(x-1)(x+1),}$
from which it should be clear we have roots ${\displaystyle \pm 1}$.
On the other hand, for the second derivative, we have only a single root: ${\displaystyle x=0}$.
Step 3:
Test the potential extrema: We know that ${\displaystyle x=\pm 1}$ are the candidates. We check the second derivative, finding
${\displaystyle g\,''(1)\,=\,6\,>\,0,}$
while
${\displaystyle g\,''(-1)\,=\,-6\,<\,0.}$
Note that
${\displaystyle g(1)\,=\,1-3\,=\,-2,}$
while
${\displaystyle g(-1)\,=\,-1+3\,=\,2.}$
By the second derivative test, the point ${\displaystyle (1,g(1))=\left(1,-2\right)}$ is a relative minimum, while the point ${\displaystyle (-1,g(-1))=(-1,2)}$ is a relative maximum.
Step 4:
Test the potential inflection point: We know that ${\displaystyle g\,''(0)=0}$. On the other hand, it should be clear that if ${\displaystyle x<0}$, then ${\displaystyle g\,''(x)<0}$. Similarly, if ${\displaystyle x>0,}$ then ${\displaystyle g\,''(x)>0}$. Thus, the second derivative "splits" around ${\displaystyle x=0}$  (i.e., changes sign), so the point ${\displaystyle \left(0,g(0)\right)=(0,0)}$  is an inflection point.
There is a local minimum at ${\displaystyle (1,-2)}$, a local maximum at ${\displaystyle (-1,2)}$ and an inflection point at ${\displaystyle (0,0).}$