Step 1

Step 2

f'(x) = 0, when x = -1 and x = 1.

There are two interior critical points. We divide the interval (-∞, ∞) on which f is defined into the two sub-intervals (-∞, 0] and [0, ∞). On each of these subintervals, f has just one interior critical point.

Step 3

We shall use the Direct Test for the subinterval (-∞, 0]. At the critical point -1, we have

f( -1) = -½.

By direct computation, we see that

f(-2) = -2/5 and f(0) = 0.

Both of these values are greater than -½. This shows that the restriction of f to the subinterval ( -∞, 0] has a minimum at x = -1. Moreover, f(x) is always ≥0 for x in the other subinterval [0, ∞). Therefore f has a minimum at -1 for the whole interval (-∞, ∞). In a similar way, we can show that f has a maximum at x = 1.