# Relative extrema question

• Mar 3rd 2011, 06:28 PM
rowdy3
Relative extrema question
Find the x-values of all points where the functions defined as follows have any relative extrema. Find the value(s) of any relative extrema.

f(x)= x^3 + 3x^2 - 24x + 2
I did f'(x)= 3X^2 + 6X - 24
0= 3X^2 + 6X - 24 / 3
0=x^2 +2x-8
(x+4)(x-2) x=-4 and x= 2
I did a number line and picked -5, 0, and 3. I plug those numbers to the original problem and -5 came out to be a positive, 0 came out to be a negative, and 3 came out to be a positive.
f(-4)= (-4)^3 + 3(-4)^2 -24(-4) + 2= . Would -4 be my relative max?
f(2)= (2)^3 + 3(2)^2 -24(2) +2=-26. Would 2 be my relative min?
• Mar 3rd 2011, 06:50 PM
FernandoRevilla
Quote:

Originally Posted by rowdy3
Would -4 be my relative max?

We get relative maximum at \$\displaystyle x=-4\$ and the relative maximum is \$\displaystyle f(-4)\$.

Same comments for the relative minimum.
• Mar 4th 2011, 04:10 AM
HallsofIvy
rowdy3, I call your attention to the very precise language FernandoRevilla is using. You said (or implied) "the maximum is -4". He is saying the maximum is at x= -4 and the maximum itself is the value of f(-4).