# Thread: finding critical number for extrema

1. ## finding critical number for extrema

im a little confused on this....here is my problem

f(x)=(x+2)²(x-1)

f'(x) = 3x²+6x critical numbers --- x=0 & x=-2 correct?

2. Yes, x = -2 and x = 0 are the abs. of the two critical points.
At x = -2, local max and at x = 0, local min.

3. Originally Posted by maybnxtseasn
im a little confused on this....here is my problem

f(x)=(x+2)²(x-1)

f'(x) = 3x²+6x critical numbers --- x=0 & x=-2 correct?
Correct.
Why are you confused ?
When you differentiate it, you need to expand and collect like terms.
And you did it correctly.

4. ## dfs

i thought i was but i got that one lol but this problem is getting me....

f(x) = [x^(5)-5x]/5

f'(x) = x-1

i get x-1 was derivative which means critical point is x=1...back of book says critical point should be x= 1 & x= -1 ??? any help what i did wrong

5. Originally Posted by maybnxtseasn
i thought i was but i got that one lol but this problem is getting me....

f(x) = [x^(5)-5x]/5

f'(x) = x-1

i get x-1 was derivative which means critical point is x=1...back of book says critical point should be x= 1 & x= -1 ??? any help what i did wrong
You did not differentiate it correctly.
You have $f(x)=\frac{x^5-5x}{5}=\frac{1}{5}x^5-x$
then, $f'(x)=x^4-1$.
Now, solve $f'=0 \,$.