Question

find critical numbers and absolute extrema of the cubed root of x on the closed interval [-8,8]

Answer 1

Do you know how to find critical numbers?

Answer 2

I do it's just when I take the derivative of the function it comes to be 1/3x^-2/3, which would be the same as 1/3x^2/3

Answer 3

Holy parenthesis, batman!

Answer 4

You got this?\[
f'(x)=\frac 13 x^{-2/3}
\]

Answer 5

yes

Answer 6

What are the roots?

Answer 7

roots?

Answer 8

When does \(f'(x)=0\)?

Answer 9

when x=0, right?

Answer 10

Well, that isn't a root, but it is a critical point nonetheless because the derivative is undefined at \(x=0\)

Answer 11

okay, so i plug that in to the original function right? along with the numbers from the interval

Answer 12

Yes

Answer 13

okay so that's what i did originally but for some reason i didn't think it was right, my professor didn't go over any examples where the derivative function DNE so i wasn't sure

Answer 14

so at the left endpoint it would be (-8,-2), which is the min, at the critial point it'd be (0,0) and the right endpoint would be (8,2) which is the max

Answer 15

critical*