This section covers:

**Introduction to Inverses****Finding Inverses Graphically****Finding Inverses Algebraically****Finding Inverses with Restricted Domains****Using Compositions of Functions to Determine if Functions are Inverses****More Practice**

Note that there is an example of **Transformation of Inverse functions** here, **Inverses of Exponential and Log functions **here, and **Inverses of the Trigonometric Functions** here.

# What is the Inverse of a Function?

Getting the **inverse of a function** is simply switching the \(x\) and the \(y\), plotting the new graph (or doing the algebra to get the “new” \(y\)), and seeing what you get! Frequently, the inverse of a function won’t even be a function, since a function can’t have 2 “answers” \((y)\) for the same “question” \((x)\), but it can have 2 “questions” \((x)\) with the same “answer” \((y)\). Switching the \(x\)** **and \(y\) totally reverses these relationships and thus we may not always end up with another function; we may just have a “relation”.

We use the **inverse notation **\({{f}^{{-1}}}\left( x \right)\) to say we want the “normal” \((x)\) value back when \((y)\) is a certain number or expression. (Note that the notation has nothing to do with a negative exponent, which looks similar.)

So for example, if the original function is \(f\left( x \right)=3x-4\), and we wanted \({{f}^{{-1}}}\left( 5 \right)\), we could get this by solving the equation \(5=3x-4\) to get \(3\). In this case then, \({{f}^{{-1}}}\left( 5 \right)=3\). But later we’ll just switch the \((x)\) and \((y)\) variables and just solve for the new \((y)\) to get answers like this.

Inverses can be a little confusing, so let’s start out with an example. Let’s say you and your foreign exchange student Justine (from France) are discussing the weather. Justine says it’s **26 degrees Celsius** outside and you want to convert that to **degrees Fahrenheit**. Then you say it’s supposed to be **90 degrees Fahrenheit** for the weekend and Justine wants to know what that is in **degrees Celsius**.

We actually saw these functions **here in the ****Solving Algebraic Equations** Section where we were solving one variable in terms of another.

Here are those functions and both their graphs on the same set of axes:

Let’s use this graph to answer the questions in the problem above:

“Justine says it’s **26 degrees Celsius** outside and you want to convert that to degrees Fahrenheit.” Since we want to know the temperature in degrees Fahrenheit, we look at the “** F **=” graph above. We can see that when

**is 26 degrees,**

*x***is about 79 degrees.**

*y*“It’s supposed to be **90 degrees Fahrenheit** for the weekend and Justine wants to know what that is in degrees Celsius.” Since we want to know the temperature in degrees Celsius, we look at the “** C** =” graph above. We can see that when

**is 90 degrees,**

*x***is about 32 degrees. I’ve included these points on the graph.**

*y*Notice that the two functions (and, yes, they are both functions!) are **symmetrical** around the line “** y = x**”. Symmetrical means that if you were to fold the piece of paper around that line, the two graphs would sit on top of each other; they are actually equidistant from the line.

This makes sense, since to get the inverse of a function, we are just

**switching the**.

*x*and the*y*# Finding Inverses Graphically

Since we can just switch the ** x** and

**to get the inverse of a function, we can easily do this with**

*y***. Here are some examples of functions and their inverses:**

*t*-chartsLet’s note a few things about the graphs above:

- The
**domains**and**ranges**in the original functions are**reversed**for the Inverses. So the domain for the original function is the range for the inverse, and the range for the original function is the domain for the inverse. - For the \(y={{x}^{2}}\) function, in order to match up the inverse (square root) function, we had to also take the “minus” of the square root function; otherwise, we’d just have “half” the graph. We’ll take about these special cases later.
- In the first two examples above, the original function fails a
**Horizontal Line Test**, meaning that you can draw a horizontal line somewhere across the function, and you’ll pass through more than one point. Notice that the inverse relations then fail the**Vertical Line Test**, since we are just switching points. Thus, the inverses in these cases are**not functions**.

Note that if a function has an inverse that is also a function (thus the original function passes the **Horizontal Line Test, **and the inverse passes the **Vertical Line Test**), the functions are called **one-to-one**. This is because there is only one “answer” for each “question” for both the original function and the inverse function.

# Finding Inverses Algebraically

If we are given the original function, finding the inverse of the function isn’t too bad. The steps involved are:

- Switch the \(x\)’s and \(y\)’s in the equation.
- Solve for the “new” \(y\). When you get the “new” \(y\), replace it with \({{f}^{{-1}}}\left( x \right)\); this is inverse notation. (Note that this has nothing to do with an exponent).
- For original functions with
**even roots**, we will have to add a**restriction**(where we have too much of the function when we get the inverse, so we have to restrict the domain.) - For original functions with
**even powers**, we will have to include the other half of the inverse by using a**plus/minus sign when we take a root**. Note that the inverse will not be a function in this case, since the original function doesn’t pass the**horizontal line test**(thus the inverse will not pass the**vertical line test**).

Note that a lot of times, **to get the range of the original function, it’s easier to solve for the inverse, and see what that domain is **(since the domains and ranges are switched for the inverse). Remember again that we have to restrict the domain in a relation or function if (but not only if):

- There is an
*x*in a denominator, and that denominator could somehow be 0 - There is an
and that*x*inside an even radical sign**radicand**(inside the radical sign)**could be negative** - There is an
**indication anywhere in the problem that the domain is restricted**

Here are some examples:

And here’s an example that is a one-to-one** rational function**, which is a function that has variables in the denominator of a fraction. Don’t worry if you haven’t seen this; we’ll learn about these types of functions and asymptotes in the **Graphing Rational Functions, including Asymptotes **section. Note that the **vertical asymptote **of the original function is the **horizontal asymptote** of the inverses function, and vice versa.

# Finding Inverses with Restricted Domains

Here are more advanced examples. In the first example, we have a **restricted domain** when given the original; thus we have to **restrict the range **when we take the inverse.

In the second example, we are given a function, but asked to **restrict the domain of the function so it is one-to-one**, and then **graph the inverse**. Remember that **one-to-one **means that both the original and the inverse are functions.

One way to think about these is that the original function must pass the **Horizontal Line Test**, so the inverse function can pass the **Vertical Line Test**.

As you can see, sometimes it is easiest to get the domain and range by **drawing the functions**, and in both cases above, the original function passes the horizontal line test, and thus the inverse function passes the vertical line test. For now, we can use **t-charts** to draw the graphs.

Don’t worry; it will be easier to draw “transformed” or “moved” functions when we learn about **Parent Functions and Transformations.**

Here are more examples where we want to find the inverse function, and domain and range of the original and inverse.

The second example is another **rational function**, and we’ll use a t-chart (or graphing calculator) to graph the original, restrict the domain, and then graph the inverse with the domain restriction:

# Using Compositions of Functions to Determine if Functions are Inverses

You can actually determine algebraically if two functions \(f\left( x \right)\) and \(g\left( x \right)\) are inverses of each other by using composition of functions: if \(f\left( {g\left( x \right)} \right)=g\left( {f\left( x \right)} \right)=x\), then \(f\left( x \right)\) and \(g\left( x \right)\) **are inverses**.

Why? Because you’re first plugging in an ** x** to get out

**, and then you plug in that**

*y***in the inverse, and out pops the original**

*y***again! If you don’t get this, don’t worry; just remember that it works!**

*x*Let’s do some problems:

**Learn these rules, and practice, practice, practice!**

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the **Mathway** site, where you can register for the **full version** (steps included) of the software. You can even get math worksheets.

You can also go to the **Mathway** site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!

On to **Parent Functions and Transformations** – you are ready!