I know that the answer is -8 but I keep getting 0 when I do $g(0), g(2)$, and $g(-8)$. I know that the critical points are 0, -2, and -8. I'm about to pull my hair out with this problem. Can someone please guide/explain to me how to do this problem? Thank yoU!
But due to the even power in the above expression, when passing from $\;x<2\to x>2\;,\;\;g'(x)\;$ keeps the same sign and, thus, this point is not a local extremum point.
But on the other hand, for $\;x<-8\;$ and very close to $\;-8\;$ , we get that $\;g'(x)>0\;$ , whereas for very close to $\;-8\;$ but $\;x>-8 \;$ we have that $\;g'(x)<0\implies\;$ at $\;x=-8\;$ we have a local maximum.
The same as above is true for $\;x=0\;$ so also here we have a local extremum point.