Let f (x) = xe^{x}. If possible, use the second derivative test to determine if each critical point is a maximum, a minimum, or neither.

he first derivative is

f '(x) = xe^{x} + e^{x} = e^{x}(x + 1),

so there is one critical point at x = -1. For the second derivative test we need the second derivative, which we can find using the product rule.

At the critical point, we have

Therefore f (x) is concave up at x = -1, so it looks like this:

We must have a minimum at x = -1.

Example 2

Let f (x) = -x^{3} + 3x^{2} – 3x. If possible, use the second derivative test to determine if each critical point is a minimum, maximum, or neither.

The derivative is

This is only zero when x = 1, and never undefined, so x = 1 is the only critical point. The second derivative, using the chain rule, is

f "(x) = -6(x – 1)(1) = -6(x – 1).

Since

f "(1) = 0,

we can't use the second derivative test to determine what type of critical point there is at x = 1. We could use the first derivative test though.

Example 3

Let f (x) = 5 – x^{2}. If possible, use the second derivative test to determine if each critical point is a minimum, maximum, or neither.

The first derivative is

f '(x) = -2x,

which is 0 when x = 0 so we have a critical point at x = 0. The second derivative is

f "(x) = -2.

Since f "(0) = -2 < 0,

the function f is concave down and we have a maximum at x = 0. The graph confirms this:

When doing these problems, remember that we don't need to know the value of the second derivative at each critical point: we only need to know the sign of the second derivative. This might save us some calculation time.