Let f (x) = xe^{x}. Use the First Derivative Test to determine if each critical point is a minimum, a maximum, or neither.

First we need to find the critical point(s).

The derivative is

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

This is zero only when x = -1. Since e^{x} is always positive, the sign of f '(x) is the same as the sign of (x + 1). When x < -1, the quantity (x + 1) is negative, so f '(x) is also. When

x > -1 the quantity (x + 1) is positive, so f '(x) is also. We therefore find this numberline:

Therefore f (x) is decreasing until x approaches -1, then increasing again, so at x = -1 there is a minimum.

Example 2

Let f (x) = x^{3}. Use the First Derivative Test to determine if each critical point is a minimum, a maximum, or neither.

The derivative is

f '(x) = 3x^{2},

which is never undefined and is zero only when x = 0. Since x^{2} is positive when x ≠ 0, f '(x) is also positive when x ≠ 0. We find this numberline:

Since f (x) is increasing both to the left and to the right of x = 0, x = 0 is neither a minimum nor a maximum.

Example 3

Let f (x) = sin x on the interval 0 ≤ x ≤ 2π. Use the First Derivative Test to determine if each critical point is a minimum, a maximum, or neither

We take the derivative:

f '(x) = cos x.

The derivative is never undefined and is zero when and when (remember, we're only looking at the interval [0,2π] right now). So far, we have a numberline that looks like this:

In order to use the first derivative test, we need to know what the sign of f ' is everywhere on the numberline. This is possible to see from a graph of f '(x):

The numberline now looks like this:

Since f ' is positive to the left and negative to the right of , the function f has a maximum at . Since f ' is negative to the left and positive to the right of , the function f has a minimum at .