Critical Points Examples

Find the critical points of the function f (x) = 5x² + 4x - 2.

First we take the derivative:

f '(x) = 10x + 4.

Since f '(x) is defined everywhere, we can move right on to finding the roots of f '(x). We need to solve the equation

f '(x) = 0.

Here we go:

This means -2/5 is a critical point—in fact, our only critical point. If we look at the graph, this makes sense, so long as we zoom in enough to see what the graph looks like around x = -2/5:

Find the critical points of the function

f (x) = 3x^2/3 + 3x^5/3

First we take the derivative.

By some mildly tricky rewriting, we can factor this formula.

Now that the derivative is nicely factored, we'll do the rest of the job.

f ' is undefined when x = 0, since

is undefined. Since f (0) is defined at 0, x = 0 is a critical point.

Finally we want to find the roots of the derivative. We have

f '(x) = x^-1/3(2 + 5x).

The factor x^-1/3 is never 0. The factor (2 + 5x) is 0 when , so we've found another critical point.

The final answer: there are critical points at x = 0 and at .

If we zoom in enough on the graph, this makes sense. The derivative doesn't exist at 0, and there appears to be a horizontal tangent line at :

Find the critical points of the function f (x) = e^x.

First we take the derivative:

f '(x) = e^x.

Since the derivative f '(x) is never undefined and has no roots, the function f (x) has no critical points. If we graph the function, we see no corners and no places with a horizontal tangent line:

Find the critical points of the function f (x) = 1/x.

First we take the derivative. Since f (x) = x^-1,

f '(x) = -x^-2 = 1/x²

Although f '(0) is undefined, x = 0 isn't a critical point because f (0) is also undefined. A critical point must be a valid point on the function, which means the original function must be defined there.

Since f ' has no roots, we conclude that f (x) has no critical points. On the graph, we see no places where f is defined and simultaneously the derivative is undefined or zero: