Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 9

a) You need to determine the monotony of the function, hence, you need to remember that f(x) increases if f'(x)>0 and f(x) decreases if f'(x)<0.
You need to evaluate f'(x), such that:
f'(x) = 6x^2+6x-36
Putting f'(x) = 0, yields:
6x^2+6x-36 = 0 => x^2 + x - 6 = 0
You need to use quadratic formula to evaluate the zeroes of the equation:
x_(1,2) = (-1+-sqrt(1+24))/2 => x_(1,2) = (-1+-5)/2
x_1 = 2 ; x_2 = -3
You need to notice that the function increases for x in (-oo,-3)U(2,oo) and it decreases if x in (-3,2) .
b) The function has maximum and minimum points at x values such as f'(x) = 0.
From point a) yields that f'(x) = 0 for x = -3 and x = 2.
Hence, the function has a maximum point at (-3,f(-3)) and it has a minimum point at (2,f(2)).
c) The function is concave up for f''(x)>0 and it is concave down for f''(x)<0.
f''(x) = 12x + 6
Putting f''(x) = 0, yields:
12x + 6 = 0 => x = -6/12 => x = -1/2
Hence, the function has an inflection point at (-1/2,f(-1/2)) and the function is concave up for x in (-1/2,oo) and it is concave down for x in (-oo,-1/2).