Inverse of Absolute Value Function

An absolute value function (without domain restriction) has an inverse that is NOT a function. That’s why by “default”, an absolute value function does not have an inverse function (as you will see in the first example below).

In order to guarantee that the inverse must also be a function, we need to restrict the domain of the absolute value function so that it passes the horizontal line test which implies that it is a one-to-one function.

Let’s take a look at some examples!

## Examples of How to Find the Inverse of Absolute Value Functions

**Example 1:** Find the inverse of [latex]f\left( x \right) = \left| x \right|[/latex].

I am sure that you are familiar with the graph of an absolute value function. It resembles a “V” shape. Without any restriction to its domain, the graph of [latex]f\left( x \right) = \left| x \right|[/latex] would fail the horizontal line test because a horizontal line will intersect at it more than once. Since this not a one-to-one function, its inverse is **not a function**. There’s no reason for moving forward to find its inverse algebraically because we know already that the inverse is not a function.

**Example 2:** Find the inverse of [latex]f\left( x \right) = \left| {x + 2} \right|[/latex] for [latex]x \le – 2[/latex].

If we are going to graph this absolute value function without any restriction to its domain, it will look like this. This is the graph of [latex]f\left( x \right) = \left| x \right|[/latex] shifted two units to the left.

However, if we apply the restriction of [latex]x \le – 2[/latex], the graph of [latex]f\left( x \right) = \left| {x + 2} \right|[/latex] has been modified to be just the **left half** of the original function. The left half of [latex]f\left( x \right) = \left| {x + 2} \right|[/latex] can be expressed as the line [latex]f\left( x \right) = – \left( {x + 2} \right)[/latex] for [latex]x \le – 2[/latex].

Therefore, to find the inverse of [latex]f\left( x \right) = \left| {x + 2} \right|[/latex] for [latex]x \le – 2[/latex] is the same as finding the inverse of the line [latex]f\left( x \right) = – \left( {x + 2} \right)[/latex] for [latex]x \le – 2[/latex]. Obviously, this “new” function will have an inverse because it passes the horizontal line test.

Let’s now apply the basic procedures on how to find the inverse of a function algebraically.

- Replace [latex]f\left( x \right)[/latex] by [latex]y[/latex].

- Swap the roles of [latex]\color{blue}x[/latex] and [latex]\color{red}y[/latex]. Then solve for [latex]\color{red}y[/latex] to get the inverse.

Okay, so we have found the inverse function. However, don’t forget to include the domain of the inverse function as part of the final answer. The domain of the inverse function is the range of the original function. If you refer to the graph again, you’ll see that the range of the given function is [latex]y \ge 0[/latex].

That means our final answer is

**Example 3:** Find the inverse of [latex]f\left( x \right) = \left| {x – 3} \right| + 2[/latex] for [latex]x \ge 3[/latex].

The first step is to graph the function. Notice that the restriction in the domain divides the absolute value function into two halves. For [latex]x \ge 3[/latex], we are interested in the right half of the absolute value function.

Therefore, to find the inverse of [latex]f\left( x \right) = \left| {x – 3} \right| + 2[/latex] for [latex]x \ge 3[/latex] is the same as finding the inverse of the line [latex]f\left( x \right) = \left( {x – 3} \right) + 2[/latex] for [latex]x \ge 3[/latex]. You also need to observe the range of the given function which is [latex]y \ge 2[/latex] because this will be the domain of the inverse function.

Let’s solve the inverse of this function algebraically.

- Replace [latex]f\left( x \right)[/latex] by [latex]y[/latex].

- Switch [latex]x[/latex] and [latex]y[/latex], then solve for [latex]y[/latex] to get the inverse.

Since the range of the original function is [latex]y \ge 2[/latex], the domain of the inverse function must be [latex]x \ge 2[/latex].

That means our final answer is

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