# A critical numbers problem

• afcwestwarrior
In summary, the function f(X) has a critical number at x=3 and the derivative of the function f'(X) is -17.f

#### afcwestwarrior

find the critical number of the function

f(X)= x^3 + 3x^2-24x

f'(X)= 3x^2 + 6x -24

what do i do, next i found the derivative of the function

Well, perhaps you should find out either where the derivative equals 23.67, -17 or perhaps some more important value, say 0.

ok so its supposed to equal zero as the definition states for critical numbers,

so i just plug in any numbers in the x's

so i just plug in any numbers in the x's

Eeh??
Do you know what an equation is, and how to solve it??

i think ido, can u show me an example

yes i know what an equation is, but how do i solve it

Can you name what type of equation that is?

well I am on the section maximum and minum values, it doesn't say what kind of equation it is, id say its a critical number equation

well I am on the section maximum and minum values, it doesn't say what kind of equation it is, id say its a critical number equation  Just look at its form. It's a QUADRATIC equation! Have you not learned about those before? There's a formula for finding the roots of a quadratic equation. If you don't know it, look it up.

ok, so when i take the derivative, i just factor it out

ok, so when i take the derivative, i just factor it out

What?? I don't understand what you mean by that.

well I've learned about quadratic equations but i forgot what it was called,
well when i factored it out, i got 3(x^2+2x-8) then i factored it out

3(x+2)(x-4) ok that's easy, i forgot about that algebra stuff and trig and geo, like my calculus teacher said, it's not the calculus its all that old stuff that will get ya

i mean 3 (x-2) (x+4)

i mean 3 (x-2) (x+4)

OK that's better I was just going to point that mistake out.

yea hey thanks for reminding me about that,

No problem. But I would recommend looking up the quadratic formula and remembering it. It works for those times when factoring isn't obvious.