0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ What do the critical points tell you? How do you determine if a critical point is a local maximum or local minimum (if any of these)? $\endgroup$ – Git Gud Apr 25 '14 at 16:25
  • $\begingroup$ You cannot get $0$ when you compute $g(0)$. Actuzally, you cannot compute any value of $g$ itself, only of $g'$ $\endgroup$ – Hagen von Eitzen Apr 25 '14 at 16:48
1
$\begingroup$

$$g'(x)=x^3(x-2)^2(x+8)^9=0\iff x=\begin{cases}x=0\\x=2\\x=-8\end{cases}$$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy