Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 33

Given f(x)=x^3-12x+2:
f'(x)=3x^2-12 and f''(x)=6x
(a) The function increases when f'(x)>0 and decreases when f'(x)<0, so
3x^2-12>0 ==> x^2>4 ==> x<-2 or x>2
3x^2-12<0 ==> x^2<4 ==< -2So the function increases on (-infty,-2) and (2,infty) and decreases on (-2,2).
(b) f'(x)=0 ==> x=-2 or 2. Since the function increases to the left of -2 and decreases to the right, x=-2 is a local maximum. (Also, f''(-2)<0.) Thus x=2 is a local minimum.(f''(2)<0)
(c) f''(x)=0 ==> x=0.
For x<0 f''(x)<0 so the function is concave down.
For x>0 f''(x)>0 so the function is concave up.
x=0 is an inflection point.
(d) the graph: