Question 676223
Take the derivative to get 15x^4 - 60x^3 - 75x^2

Factor out a 15x^2 to get  x^2-4x-5 This factors into 15x^2(x-5)(x+1).

Then  you would get 0,-1,5 are your critical points. But 0 and -1 are the only things on your interval.

Then you plug in the end points:

f(-2) = -136
f(3) = -1161

And then plug in your critical values:

f(-1) = 7
f(0) = 0

Then 7 is your absolute maximum and  -1161 is your minimum.