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.
