# Thread: First derivative test

1. ## First derivative test

Okay so here is the deal.

I have the function.

G(x)=(x^3-1)^3

G'(x)= 3 (x^3-1)^2 * 3x

And i need to find out where the function is increasing and decreasing. I know how to do that, I have to set it to zero to find the critical numbers. However, I have two questions I guess. One being is my first derv right, and then if it is....how to i solve for zero. Should I divide by 2x?? then 3?? I am confused.

2. $\displaystyle G'(x)= 3 (x^3-1)^2 \times 3x^2$

Finding where $\displaystyle G'(x)=0$ is a good start. This where the stationary points are. In between these are where the function is either increasing or decreasing, take some test points to find this out.

3. You forgot to take the derivative of x^3 properly when using the chain rule - you have 3x instead of 3x^2:

$\displaystyle g'(x) = 9x^2(x^3-1)^2 = 0$

You can use the same methods you do with equations like $\displaystyle (x-a)(x-b) = 0$ so either $\displaystyle 9x^2 = 0$ or $\displaystyle (x^3-1)^2 = 0$

Then take some points either side to test