Note that there are examples of Transformation of Inverse functions here, Inverses of the Trigonometric Functions here, and Derivatives of Inverse Functions (Calculus) here.
What is the Inverse of a Function?
Getting the inverse of a function is simply switching the domain and range, which is just switching the the independent and dependent variables, to get a new relation or function. We can do this graphically, using t-charts, or algebraically.
Frequently, the inverse of a function isn’t a function but just a relation, since a function can’t have two “answers” $ (y)$ for the same “question” $ (x)$, but it can have two “questions” $ (x)$ with the same “answer” $ (x)$. (We can restrict the domain of the new inverse to make it a function). Functions that have inverses that are also functions are called one-to-one functions or invertible functions.
We use the inverse notation $ {{f}^{{-1}}}\left( x \right)$ to represent a function. For example, when we’ve switched the $ x$- and the $ y$-values and solved for the new $ y$-value, this would be $ {{f}^{{-1}}}\left( x \right)$. (Note that the notation has nothing to do with a negative exponent, which looks similar.) 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:
$ \displaystyle \begin{align}{c}x&=3y-4;\,\,\,\,\,y=\frac{{x+4}}{3}\\{{f}^{{-1}}}\left( x \right)&=\frac{{x+4}}{3};\,\,\,\,\,\,\,{{f}^{{-1}}}\left( 5 \right)=\frac{{5+4}}{3}=3\end{align}$
The interpretation of inverses can be a little confusing, so let’s start out with an example. 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. 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 (literal equations).
Here are those functions and both their graphs on the same set of axes:
| Fahrenheit based on Celsius | Celsius based on Fahrenheit |
| $ \displaystyle F= \frac{9}{5}C+32$ | $ \displaystyle C=\frac{5}{9}\left( {F-32} \right)$ |
 | |
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, look at the “$ F=$” graph above. When $ x$ is 26 degrees, $ y$ is about 79 degrees.
“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, look at the “$ C=$” graph above. When $ x$ is 90 degrees, $ y$ is about 32 degrees. I’ve included these points on the graph.
Notice that the two functions (and, yes, they are both functions!) are symmetrical around the line “$ y=x$”. Symmetrical means that if you were to fold the piece of paper around that line, the two graphs would sit on top of each other; they are actually equidistant from the line. This makes sense, since to get the inverse of a function, we are just switching the independent ($ x$-) and dependent ($ y$) variables.
Finding Inverses Graphically
Since we can just switch the $ x$- and $ y$-values to get the inverse of a function, we can easily do this with $ t$-charts. Here are some examples of functions and their inverses, showing some key points:
Notice from the graphs above:
- The inverse graphs are the original graphs reflected around the line $ y=x$ (meaning straight across the line, perpendicular to it, and the same distance from it).
- The domains and ranges in the original functions are reversed for the inverses. Thus, 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 (although if we include the “minus” part, the inverse is not a function). 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.
- Typically, inverse functions increase/decrease the same as the original functions, if both exist in that interval. For inverse functions as compared to original functions, $ x$- and $ y$-intercepts are switched, and horizontal and vertical asymptotes are switched.
- Note that the last function, the “inverse” Parent Function is an inverse of itself! Pretty cool!
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, or invertible. This is because there is only one “answer” for each “question” for both the original function and the inverse function. (Otherwise, the function is non-invertible). Examples of functions that are not one-to-one are vertical quadratic functions and absolute value functions.
This is interesting: if the graph of a one-to-one function lies entirely in one quadrant, say Quadrant I, the graph of the inverse function lies entirely in Quadrant III. This will be the same case for Quadrants II and IV. Do you see how that happens?
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$ and $ y$ variables in the equation and solve for the “new” $ y$. Notice that when we drill down to the “new” $ y$, we’ll typically work from the outside to the inside by “undoing the algebra”. (Note that some teachers will have you solve for $ x$ and then switch the $ x$ and the $ y$ variables at the end to get the inverse function. This is doing the same thing!)
- 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 powers, include the other half of the inverse by using a plus/minus sign when we take a root (although the inverse will not be a function when we do this, since the original function fails the horizontal line test). To keep the inverse a function, don’t take the plus/minus; just include the positive root.
- For original functions with even roots, add a restriction, since the inverse will give us a function with both “sides”. For example, for the function $ y=\sqrt{x}$, the inverse function is $ y={{x}^{2}};\,\,x\ge 0$.
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 $ x$ inside an even radical sign and that 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:
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.
Inverse Function Word Problems
Here are some types of word problems you might see to make sure you understand how inverses work.
Inverse Functions 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 below, the original function passes the horizontal line test, and thus the inverse function passes the vertical line test. You can always use $ t$-charts to draw the graphs.
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
Note: We learned about Compositions of Functions here in the Advanced Functions: Compositions, Even and Odd, and Extrema section.
You can 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 $ y$, and then you plug in that $ y$ in the inverse, and out pops the original $ x$ again! If you don’t get this, don’t worry; just remember that it works! Here are some problems:
Inverses of Exponential and Logarithmic Functions
As it turns out, Exponential Functions are inverses of Logarithmic Functions and of course vice versa! Let’s show algebraically that the parent exponential and log functions $ (y={{b}^{x}}\,\,and\,\,y={{\log }_{b}}x)$ are inverses – three different ways.
Here are the graphs of the two functions again, so you can see that they are inverses; note symmetry around the line $ y=x$. Also note that their domains and ranges are reversed:
Find the inverses of the following transformed exponential and log functions by switching the $ x$ and the $ y$ and solving for the “new” $ y$:
Learn these rules, and practice, practice, practice!
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On to Parent Functions and Transformations – you are ready!