Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 10

a) You need to determine the monotony of the function, hence, you need to verify where f'(x)>0 or f'(x)<0.
Hence, you need to differentiate the function with respect to x, such that:
f'(x) = (4x^3 + 3x^2 - 6x + 1)'
f'(x) = 12x^2 + 6x - 6
You need to set the equation f'(x) = 0, such that:
12x^2 + 6x - 6 = 0
Dividing by 6 yields:
2x^2 + x - 1 = 0
Re-write the equation 2x^2 + x - 1 = 0 such as:
x^2 + x^2 + x - 1 = 0
Group the terms, such that:
(x^2 - 1) + (x^2 + x) = 0
You need to write the difference of squares x^2 - 1 such that:
(x - 1)(x + 1) + (x^2 + x) = 0
You need to factor out x in the group x^2 + x , such that:
(x - 1)(x + 1) + x(x + 1) = 0
Factor out x + 1:
(x + 1)(x - 1 + x) = 0
Put x + 1 = 0 => x = -1
Put 2x - 1 = 0 => 2x = 1 => x = 1/2
You need to notice that f'(x)<0 for x in (-1,1/2) and f'(x)>0 for
x in (-oo,-1)U(1/2,+oo).
Hence, the function decreases for x in (-1,1/2) and it increases for x in (-oo,-1)U(1/2,+oo).
b) The local maximum and minimum of the function are reached at values of x for f'(x) = 0.
From previous point a) yields that f'(x) = 0 for x = -1 and x = 1/2, hence, the function reaches it's maximum at x = -1 and it's minimum at x = 1/2.
Hence, the maximum of the function is the point (-1,f(-1)) and the minimum is the point (1/2,f(1/2)).
c) The function is concave up if f''(x)>0 and it is concave down if f''(x)<0. You need to evaluate the second derivative:
f''(x) = (12x^2 + 6x - 6)'
f''(x) = 24x + 6
You need to put f''(x) = 0:
24x + 6 = 0
Dividing by 6 yields:
4x + 1 = 0
x = -1/4
Hence, the function has an inflection point at x = -1/4 and it is concave down for x in (-oo,1/4) and it is concave up if x in (1/4,oo).