Question

Find the critical points of the followng equation:
(x^3-8)/(x-2)

Answer 1

\[(x^3-8)/(x-2)\]

Answer 2

(Using the quotient rule)

Answer 3

Take the derivative.

Critical points are where the function is equal to 0 or DNE.

If the top of the fraction equals 0, then the whole thing is equal to 0.
If the bottom of the fraction is equal to 0, then it DNE.

Answer 4

Do you know how to use the quotient rule?

Answer 5

Yes, but once I found the derivative, I didn't know where to go. For the derivative I got
2x^3−6x^2+8

Answer 6

Hello?

Answer 7

I believe that you equal that result to zero, then solve for the variables and those should be the critical points. I'm struggling with derivatives myself.

Answer 8

If you do the quotient rule correctly, you will have a fraction

Answer 9

To find critical points, set the top of the fraction equal to 0 and solve, then set the bottom equal to 0 and solve.

Answer 10

I didn't get a fraction though. Here are my steps:\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \rightarrow\]

Answer 11

\[((3x^2)(x-2)-(1)(x^3-8))/((x-2)^2)\]

Answer 12

\[\rightarrow (3x^3-6x^2-x^3+8)/(x^2-4x+4)\]

Answer 13

\[\rightarrow 2x^3-6x^2+8\]

Answer 14

Set top equal to 0, solve. Set bottom equal to 0, solve.

Answer 15

would the denominator = 1?

Answer 16

No. There's no reason to simplify that much. Use ((3x2)(x−2)−(1)(x3−8))/((x−2)2).

Answer 17

The original derivative, that is.

Answer 18

Could you help, I'm having trouble solving

Answer 19

hi?

Answer 20

Start with the bottom. X-2=0

Answer 21

For the top, get all of your x terms on one side, and everything else on the other.