Classify the extreme points of the function, using either the first or second derivative test. Explain why you chose to use the test you did.
f (x) = x4 - 32x
The first derivative is
f'(x) = 4x3 - 32.
This is zero when x = 2, so x = 2 is the single critical point of f. If we use the first derivative test, we need to evaluate f' at two different values of x.
Since it's possible to take another derivative, we'll use the second derivative test. The second derivative is
f"(x) = 12x2
which is positive for any x ≠ 0. We didn't even need to evaluate f" at one value of x, which is nice. Since f" is positive at x = 2, the function f is concave up, and therefore has a minimum at x = 2.