# Thread: Curve Sketching - Concavity - Problem 2

1. ## Curve Sketching - Concavity - Problem 2

I'm having a really hard time with this. Could anyone help me figuring out the solution for this? Thank you very much in advance!

2. Could anyone here possibly help me? I still don't get it. Thanks.

3. $\displaystyle g(x)=-2x^3-(-24)x^2+(-90)x-2$

$\displaystyle g'(x)=-6x^2-(-48)x+(-90)$

Can you find $\displaystyle x_1,x_2$ now?

Spoiler:

[tex]g'(x)=0[tex] gives $\displaystyle x_1=-5$ and $\displaystyle x_2=-3$.

$\displaystyle g''(x)=-12x-(-24)$

$\displaystyle g''(x_1)>0$

there is relative minimum at $\displaystyle x_2,x_1$

$\displaystyle g''(x_2)>0$

4. Take the your derivative function, set that equal to zero and solve for x. Those values are your critical numbers. You can then find the second derivative and put your critical values into the second derivative. If that results in a value greater than zero the function at that given point is a relative minimum, and if the value is less than zero then you have a relative maximum.
Hope that helps.