Find the relative extrema of $g(x)=\frac{1}{9}x^9-x$
$Solution:$
Step 1: Find the critical points, that is, those values of $x$ such that $g'(x)=0$ or $g'(x)$ DNE
$g'(x)=\frac{1}{9}(9)x^8-1=x^8-1=0$
$\rightarrow x^8=1$ and so $x= \pm 1$
Step 2: Figure out what the sign of $g'(x)$ is around the critical poitns
$g'(-2)=(-2)^8-1>0$
$g'(0)=-1<0$
$g'(2)=2^8-1>0$
Step 3: Use this information to figure out which (if any) of the critical points are relative maximum and minimum.
Since $f(x)$ is increasing as $x$ increases towards $-1$, then $f(x)$ begins decreasing as $x$ increases towards $0$, we must have that $-1$ is a relative maximum.
Since $f(x)$ is decreasing as $x$ increases towards $1$, then $f(x)$ begins increasing as $x$ increases towards $\infty$, we must have that $1$ is a relative minimum.
Step 4: Find the correspoding $y$ values for the critical points
$g(-1)=\frac{1}{9}(-1)^9-(-1)=\frac{-1}{9}+1=\frac{8}{9}$
$g(1)=\frac{1}{9}(1)^9-(1)=\frac{1}{9}-1=\frac{-8}{9}$
So we have a relative maximum at $(-1,\frac{8}{9})$
And we have a relative minimum at $(1,\frac{-8}{9})$
This is more or less correct, depending what is expected of you.
A less hand-wavy way of determining whether the points are maxima/minima is the second derivative test.