Derivatives > How to find critical numbers

A critical number is a number “c” that either makes the derivative equal to zero or it makes the derivative undefined. Critical numbers indicate where a change in the graph is taking place. For example:

- An increasing to decreasing point, or
- A decreasing to increasing point.

The number “c” also has to be in the domain of the original function (the one you took the derivative of). Finding critical numbers is an easy task if your algebra skills are strong; unfortunately, **if you have weak algebra skills you might have trouble finding critical numbers**. Why? Because each function is different, and algebra skills will help you to spot undefined domain possibilities such as division by zero. If your algebra isn’t up to par—now is the time to restudy the old rules (check out Chegg Study if you need a tutor).

## How to find critical numbers

**Sample question:** Find the critical numbers for the following function: ^{x2}⁄_{x2-9}

Step 1: **Take the derivative of the function**. Using the quotient rule, we get:

^{-18x}⁄_{(x2 – 9)2}.

Step 2: **Figure out where the derivative equals zero.** This is where a little algebra knowledge comes in handy, as each function is going to be different. For this particular function, the derivative equals zero when -18x = 0 (making the numerator zero), so one critical number for x is 0 (because -18(0) = 0). Another set of critical numbers can be found by setting the denominator equal to zero, you’ll find out where the derivative is undefined:

(x^{2} – 9) = 0

(x – 3)(x + 3) = 0

x = ±3

Step 3: **Plug any critical numbers you found in Step 2 into your original function** to check that they are in the domain of the original function. For this particular function, the critical numbers were 0, -3 and 3. Let’s plug in 0 first and see what happens:

f(x) = ^{02}⁄_{02-9} = 0. Therefore, 0 is a critical number.

For +3 or -3, if you try to put these into the denominator of the original function, you’ll get **division by zero**, which is **undefined**.

f(x) = ^{32}⁄_{32-9} = 9/0. Therefore, 3 is not a critical number.

That means these numbers are not in the domain of the original function and are not critical numbers.

*That’s it!*

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