Question 143597
f'(x)=-6x^2-6x
Find where it is=0

0=-6x^2-6x
0=6x(x+1)
implies x=0 or x=-1. These are our two critical values. Find out what happens outside and between these two values for the derivative f'.

f'(-2)=-6(-2)^2-6(-2)=-24+12=-12. It is decreasing from ({{{-infinity}}},-1).
f'(-1/2)=-6(-1/2)^2-6(-1/2)=-3/2+3=3/2. It is increasing from (-1,0).
f(1)=-6-6=-12. It is decreasing from (0,{{{infinity}}}).


Plot:
{{{graph( 300, 200, -2, 2, -20, 20, -2x^3-3x^2+12 )}}}