Question 1112034
f(x)=4x^(3)−6x^(2)−672x−8
f'(x)=12x^2-12x-672; f''(x)=24x-12
Set equal to 0 and factor out a 12
12(x^2-x-56)=0=12(x-8)(x+7)
x=8, -7, which are critical points.
when x=8, f''(x)=180, a minimum
when x=-7, f''(x)=-180, a maximum.  The local maximum is at (-7, 3030)
it goes from -oo to -7, increasing, decreases on the interval (-7, 8) and increases on the interval (8, oo)

{{{graph(300,300,-20,30,-4000,4000,4x^3-6x^2-672x-8)}}}