Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 36

Given: f(x)=-x^4+4x^3+8x^2
Find the critical x values by setting the first derivative equal to zero and solving for the x value(s).
f'(x)=-4x^3+12x^2+16x=0
-4x(x^2-3x-4)=0
-4x(x-4)(x+1)=0
x=0, x=4,x=-1
If f'(x)>0 the function is increasing over an interval.
If f'(x)>0 the function is decreasing over an interval.
Choose a value for x that is less than -1.
f'(-2)=48 Since f'(2)>0 the function increases in the interval (-oo,-1).
Choose a value for x that is between -1 and 0.
f'(-.5)=-4.5 Since f'(-.5)<0 the function decreases in the interval (-1, 0).
Choose a value for x that is between 0 and 4.
f'(1)=24 Since f'(1)>0 the function increases in the interval (0, 4).
Choose a value for x that is greater than 4.
f'(5)=-120 Since f'(5)<0 the function decreases in the interval (4, oo).
Because the function changes direction from increasing to decreasing a relative maximum will exist at x=-1 and at x=4. The relative maximum points are
(-1, 3) and (4, 128).
Because the function changes direction from decreasing to increasing a relative minimum value will exist at x=0. The relative minimum point is (0, 0).