To find the absolute extrema of f(x), we need to consider both the endpoints and the critical points. (We have to consider the endpoints because the interval is closed).
First, let's find the critical points:
f(x)=x^3+x^2-4x-4
f'(x)=3x^2+2x-4=0
We solve this equation using the quadratic formula:
x=(-2+-sqrt(4-4(-4)(3)))/(2(3))
x=(-2+-sqrt(52))/(6)=(-2+-2sqrt(13))/(6)
x=(-1+-sqrt13)/3
Notice that x=(-1+sqrt13)/3 is not in the interval [-3,0], so we can disregard it.
We have critical points at x=(-1-sqrt13)/3 and
we have endpoints at x=-3,0.
Now, we just plug all those x-values in and find which ones are the highest and lowest.
f((-1-sqrt13)/3)=((-1-sqrt13)/3)^3+((-1-sqrt13)/3)^2-4((-1-sqrt13)/3)-4~~0.8794
f(-3)=-10
f(0)=-4
We can see that f((-1-sqrt13)/3)~~0.8794 results in the highest value and
f(-3)=-10 results in the lowest value, making them the absolute maximum and minimum respectively.