1. ## Critical Numbers

Hello,
I am trying to find the critical numbers for
f'(x) = 36x^3 - 72x^2 - 72

I got it to:
36(x^3 - 2x^2 - 2)

However, I am confused to as if I am supposed to factor it more?

Or how would I solve for
36(x^3 - 2x^2 - 2) = 0

2. What you are doing will find the zeros if any.

For the critical numbers take the derivative equal to zero and solve for x, these are the critical numbers I think you are after.

3. Try using synthetic division to see if you can break the polynomial down.

Hello,
I am trying to find the critical numbers for
f'(x) = 36x^3 - 72x^2 - 72

I got it to:
36(x^3 - 2x^2 - 2)

However, I am confused to as if I am supposed to factor it more?

Or how would I solve for
36(x^3 - 2x^2 - 2) = 0

If you get f'(x) = 36x^3 - 72x^2 - 72 as part of finding the critical values of f(x) then you're expected to solve x^3 - 2x^2 - 2 = 0. Since there is no simple solution I assume you're expected to use technology such as a CAS: solve x&#94;3 - 2x&#94;2 - 2 &#61; 0 - Wolfram|Alpha

On the other hand, if you have to find the critical values of f'(x) = 36x^3 - 72x^2 - 72, then you need to solve f''(x) = 0.

5. The orignal question is
determine the local maximum and minimum values of f(x) = (3x^2 -6x)(3x^2+-6x)

I then got f'(x) = 36x^3 - 72x^2 - 72

but I am not supposed to use technology to solve.. very confused

6. 1:$\displaystyle \displaystyle f'(x)=2(3x^2-6x)(6x-6)$

2:$\displaystyle \displaystyle f'(x)=(6x-6)(3x^2+6x)+(3x^2-6x)(6x+6)=(3x^2+6x)(6x+6x+12)=(3x^2+6x)(12x+12)$

You have a +- 6 so I did both derivatives since I wasn't sure what was a typo.

Critical points of 1: 1, 0, 2

2: -1, 0, -2

7. sorry, it was supposed to be "+6"

8. Number 2 then.

9. thanks so much!

The orignal question is
determine the local maximum and minimum values of f(x) = (3x^2 -6x)(3x^2+-6x)

I then got f'(x) = 36x^3 - 72x^2 - 72

but I am not supposed to use technology to solve.. very confused
You have a difference times a sum, which gives a difference of squares!

$\displaystyle (3x^2 -6x)(3x^2+-6x)=\left(3x^2\right)^2-\left(6x\right)^2=9x^4-72x^2$.

The derivative will have only two terms.

The following has been edited.
Should have been:

$\displaystyle (3x^2 -6x)(3x^2+6x)=\left(3x^2\right)^2-\left(6x\right)^2=9x^4-72x^2$.

11. Originally Posted by SammyS
You have a difference times a sum, which gives a difference of squares!

$\displaystyle (3x^2 -6x)(3x^2+-6x)=\left(3x^2\right)^2-\left(6x\right)^2=9x^4-72x^2$.

The derivative will have only two terms.
$\displaystyle (3x^2+6x)(12x+12)=\frac{d}{dx}[9x^4-72x^2]$