Answer:
(a) increasing: (-∞, -1) U (0, 1); decreasing: (-1, 0) U (1, ∞)
(b) local minimum: (0, 6); local maxima (also global maxima): (-1, 7), (1, 7)
(c) points of inflection: (-(√3)/3, 6 5/9), ((√3)/3, 6 5/9)
(d) concave upward: (-(√3)/3, (√3)/3); concave downward: (-∞, -(√3)/3) U ((√3)/3, ∞)
Step-by-step explanation:
Most of the answers can be learned from the first derivative:
f'(x) = 4x -4x^3 = 4x(1-x^2)
This has zeros at x={-1, 0, 1}.
For points of infection, we also need to know the points where the second derivative is zero:
f''(x) = 4(-3x^2 +1)
This has zeros at x = ±1/√3
The attached plot shows the function f(x) (red) and the first derivative f'(x) (dashed blue), along with labels on most of the points of interest.
___
(a) The function is increasing where the first derivative is positive, for x < -1 and 0 < x < 1. The function is decreasing elsewhere. It is neither increasing nor decreasing where the first derivative is zero, at x = {-1, 0, 1}, so these points are left out of the intervals for increase or decrease.
__
(b) The local extremes are where the first derivative is zero, at x = {-1, 0, 1}. The function values at those points are, respectively, {7, 6, 7}. Hence the outer ones are maxima and the one at (0, 6) is a local minimum. Since the function is an even function that opens downward, the maxima are also global maxima.
__
(c) The points of inflection are where the second derivative is zero—where the first derivative has local extremes. The x-values of these points are ±1/√3 = ±(√3)/3, and the y-values are computed to be ...
f(±1/√3) = 6 + 2/3 - 1/9 = 6 5/9
__
(d) The graph is concave upward where the first derivative has positive slope (between the points of inflection), and concave downward elsewhere. Of course, it is neither concave upward nor downward at the points of inflection, so those are left out of the intervals.
