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 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$.
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:
| Original Function | T-Charts | Inverse Relation | ||||||||||||||||||||
 Domain: $ \left[ {-4,5} \right]$ Range: $ \left[ {-7,5} \right]$ | Function:
Inverse (Not a function):
|
Domain: $ \left[ {-7,5} \right]$ Range: $ \left[ {-4,5} \right]$ | ||||||||||||||||||||
|
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left[ {0,\infty } \right)$ | Function:
Inverse (Not a function):
|  Domain: $ \left[ {0,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ | ||||||||||||||||||||
|
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ | Function:
Inverse (Function):
|
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ | ||||||||||||||||||||
|
Domain: $ \left( {-\infty ,0} \right)\cup \left( {0,\infty } \right)$ Range: $ \left( {-\infty ,0} \right)\cup \left( {0,\infty } \right)$ | Function:
Inverse (Function):
|
The same function! Domain: $ \left( {-\infty ,0} \right)\cup \left( {0,\infty } \right)$ Range: $ \left( {-\infty ,0} \right)\cup \left( {0,\infty } \right)$ |
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:
| Original Function with Domain and Range | Switch $ x$ and $ y$. Solve for “new” $ y$. | Inverse Relation with Domain and Range | Notes |
| $ \begin{array}{l}f\left( x \right)=3x+8\\\text{or }y=3x+8\end{array}$
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ | $ \displaystyle \begin{align}y&=3x+8\\\,\,\,x&=3y+8\\\,\,\,3y&=x-8\\\,\,\,y&=\frac{{x-8}}{3}\end{align}$ | $ \displaystyle {{f}^{{-1}}}\left( x \right)=\frac{{x-8}}{3}$
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ | Switch the $ x$ and the $ y$ and solve for the “new $ y$”.
Both the original and inverse are functions (lines); the functions are one-to-one. |
| $ \displaystyle f\left( x \right)=-\frac{5}{4}x-2$ $ \displaystyle \text{or }y=-\frac{5}{4}x-2$
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ | $ \displaystyle \begin{align}y&=-\frac{5}{4}x-2\\x&=-\frac{5}{4}y-2\\\frac{5}{4}y&=-x-2\\y&=\left( {-x-2} \right)\left( {\frac{4}{5}} \right)\\y&=\frac{{-4x-8}}{5}\end{align}$ | $ \displaystyle {{f}^{{-1}}}\left( x \right)=\frac{{-4x-8}}{5}=-\frac{{4x+8}}{5}$
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ | Switch the $ x$ and the $ y$ and solve for the “new $ y$”. Watch out for the fractions and the negative signs!
Both the original and inverse are functions (lines); the functions are one-to-one. |
| $ \begin{array}{l}f\left( x \right)={{\left( {x-1} \right)}^{2}}\\\text{or }y={{\left( {x-1} \right)}^{2}}\end{array}$
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left[ {0,\infty } \right)$
Note: Range is $ \left[ {0,\infty } \right)$ because the square function will be positive. | $ \begin{align}y&={{\left( {x-1} \right)}^{2}}\\x&={{\left( {y-1} \right)}^{2}}\\\pm \sqrt{x}&=\sqrt{{{{{\left( {y-1} \right)}}^{2}}}}\\y-1&=\pm \sqrt{x}\\y&=\pm \sqrt{x}+1\end{align}$ | $ {{f}^{{-1}}}\left( x \right)=\pm \sqrt{x}+1$
Domain: $ \left[ {0,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$
Note that this inverse is not a function, since the original doesn’t pass the horizontal line test! To make it a function, we could restrict it to either the plus or minus, but not both. Thus, the original function is not one-to-one! | Work from the outside to the inside to solve for $ y$. Remember that when we take an even root, we have to include both the positive and negative solutions, since the root sign only gives the positive value. We can simply switch the domain and range of the original function to get the domain and range of the inverse. Note that since we took the plus and minus with the inverse, we will have the complete “sideways parabola”.
We will talk about restricting the domain below, but If we wanted this inverse to be a function, we could use “half” of the parabola with an original domain and range of $ \left[ {0,\infty } \right)$ and $ \left[ {0,\infty } \right)$. Then we’d use the positive value of the inverse, $ {{f}^{{-1}}}\left( x \right)=\,\sqrt{x}+1$, making the inverse domain $ \left[ {0,\infty } \right)$ and range $ \left[ {0,\infty } \right)$. |
| $ \begin{array}{l}f\left( x \right)=\sqrt{{x+3}}\\\text{or }y=\sqrt{{x+3}}\end{array}$
Domain: $ \left[ {-3,\infty } \right)$ Range: $ \left[ {0,\infty } \right]$
Note: Domain is $ \left[ {-3,\infty } \right)$ since under an even radical has to be $ \ge 0$. Range is $ \left[ {0,\infty } \right)$ because the square root function is only positive by definition. | $ \begin{align}\,y&=\sqrt{{x+3}}\\\,x&=\sqrt{{y+3}}\\{{\left( x \right)}^{2}}&={{\left( {\sqrt{{y+3}}} \right)}^{2}}\\\,{{x}^{2}}&=y+3\\\,y&={{x}^{2}}-3\end{align}$ | $ \,{{f}^{{-1}}}\left( x \right)={{x}^{2}}-3,\,\,\,x\ge 0$
Domain: $ \left[ {0,\infty } \right]$ Range: $ \left[ {-3,\infty } \right)$ | Note that when we state the inverse function, we need to include the domain’s restriction ($ x\ge 0$) to “erase” the extra part of the graph.
Since the square root function is only “half” a parabola, our inverse must be “half” a parabola too, and our restricted domain (from the restricted range of the original) will take care of that. Both the original and inverse are functions; they are one-to-one. |
| $ \begin{array}{l}f\left( x \right)={{x}^{2}}-12x+20\\\text{or }y={{x}^{2}}-12x+20\end{array}$
For this quadratic, it’s easier to complete the square first to get vertex form: $ \begin{array}{c}y={{x}^{2}}-12x+20\\y={{\left( {x-6} \right)}^{2}}+20-36\\y={{\left( {x-6} \right)}^{2}}-16\end{array}$
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left[ {-16,\infty } \right)$ | Now, switch the $ x$ and the $ y$ and solve for new $ y$:
$ \begin{array}{c}x={{\left( {y-6} \right)}^{2}}-16\\x+16={{\left( {y-6} \right)}^{2}}\\y-6=\pm \sqrt{{x+16}}\\y=\pm \sqrt{{x+16}}+6\end{array}$ | $ {{f}^{{-1}}}\left( x \right)=\pm \sqrt{{x+16}}+6$
Domain: $ \left[ {-16,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$
Note that this inverse is not a function, since the original doesn’t pass the horizontal line test! To make it a function, we could restrict it to either the plus or minus, but not both. Thus, the original function is not one-to-one! | Note again that the inverse is not a function; we have a “sideways parabola”.
We will talk about restricting the domain below, but If we wanted this inverse to be a function, we could use “half” of the parabola with an original domain and range of $ \left[ {6,\infty } \right)$ and $ \left[ {-16,\infty } \right)$. Then we’d use the positive value of the inverse, $ {{f}^{{-1}}}\left( x \right)=\sqrt{{x+16}}+6$, making the inverse domain $ \left[ {-16,\infty } \right)$ and range $ \left[ {6,\infty } \right)$. |
| Let’s try this piece-wise function:
$ \displaystyle f\left( x \right)=\,\left\{ \begin{array}{l}x-2,\text{ }\,\,\text{ }\,\text{ }x\le -2\\{{\left( {x+1} \right)}^{3}}\text{, }\,x>-2\end{array} \right.$
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left( {-\infty ,-4} \right]\cup \left( {-1,\infty } \right)$ | Find the inverse for each of the pieces of the function (switch $ x$ and $ y$):
$ \begin{array}{c}y\le -2\,\,(x\le -4):\\x=y-2\,\\y=x+2\,\end{array}$
$ \begin{array}{c}y>-2\,\left( {x>-1} \right):\\x={{\left( {y+1} \right)}^{3}}\,\\\sqrt[3]{x}=y+1\\y=\sqrt[3]{x}-1\end{array}$ | Note that it really helps to graph the original function to see the boundaries (switch the $ x$’s and $ y$’s to get inverse graph).
$ \displaystyle {{f}^{{-1}}}\left( x \right)=\left\{ \begin{array}{l}x+2,\text{ }\,\,\text{ }x\le -4\\\sqrt[3]{x}-1,\text{ }x>-1\end{array} \right.$
Domain: $ \left( {-\infty ,-4} \right]\cup \left( {-1,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ | $ f\left( x \right)\,\,\text{and}\,{{f}^{{-1}}}\left( x \right):$
|
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.
| Original Rational Function | Inverse Function |
| $ \displaystyle f\left( x \right)=\frac{{2x-5}}{{x+3}}$
This is a rational function, since the denominator contains a variable. The denominator can’t be 0, so $ x=-3$ is the vertical asymptote (VA) of the function, and –3 is the domain restriction.
$ y=2$ is the horizontal asymptote (End Behavior Asymptote, or EBA) of the function (exponents are the same, so divide coefficients). From this asymptote, we know that the range can’t be 2 (Without knowing the EBA, we can get the range restriction by first solving for the inverse of the function, and then finding the domain of this inverse.) Thus, for the original function, we have: Domain: $ \left( {-\infty ,-3} \right)\cup \left( {-3,\infty } \right)$ Range: $ \left( {-\infty ,2} \right)\cup \left( {2,\infty } \right)$ | Switch the $ x$ and the $ y$ and solve for the “new” $ y$ to get the inverse function: $ \displaystyle y=\frac{{2x-5}}{{x+3}}\,\,\,\,\,\,\,\,x=\frac{{2y-5}}{{y+3}}$ $ \displaystyle \begin{array}{c}x\left( {y+3} \right)=2y-5\\\,\,xy+3x=2y-5\\\,\,2y-xy=3x-5\\y\left( {2-x} \right)=3x-5\end{array}$ $ \displaystyle y=\frac{{3x-5}}{{2-x}};\,\,\,\,\,\,{{f}^{{-1}}}\left( x \right)=\frac{{3x-5}}{{2-x}}$ (We can see that the domain restriction of this inverse function is 2, which matches the range restriction of the original function.)
Switch the domain of the range of the original to get the domain and range of the inverse function. Domain: $ \left( {-\infty ,2} \right)\cup \left( {2,\infty } \right)$ Range: $ \left( {-\infty ,-3} \right)\cup \left( {-3,\infty } \right)$ Note that for the inverse function, asymptotes are switched: the horizontal (EBA) asymptote is $ y=-3$ and the VA is $ x=2$. |
Inverse Function Word Problems
Here are some types of word problems you might see to make sure you understand how inverses work.
| Inverse Word Problem | Solution | ||||||||
| If $ f\left( x \right)=2x+6$, what is $ {{f}^{{-1}}}\left( 3 \right)$? | The expression $ {{f}^{{-1}}}\left( 3 \right)$ is an inverse, so it’s an $ x$-value, and the 3 is the $ y$-value. Therefore, we have: $ \displaystyle 3=2x+6;\,\,\,2x=-3;\,\,\,x=-\frac{3}{2}$. Thus, $ \displaystyle {{f}^{{-1}}}\left( 3 \right)=-\frac{3}{2}$. We could have also switched the $ x$ and the $ y$ (or $ f\left( x \right)$), to solve for the “new” $ y$ (or $ {{f}^{{-1}}}\left( x \right)$): $ \displaystyle x=2y+6;\,\,\,\,y=\frac{{x-6}}{2};\,\,\,\,\,{{f}^{{-1}}}\left( x \right)=\frac{{x-6}}{2};\,\,\,\,\,{{f}^{{-1}}}\left( 3 \right)=\frac{{3-6}}{2}\,=\,-\frac{3}{2}$ | ||||||||
| If $ f\left( x \right)={{x}^{5}}+{{x}^{3}}$, what is
$ {{\left( {{{f}^{{-1}}}\left( 4 \right)} \right)}^{5}}+{{\left( {{{f}^{{-1}}}\left( 4 \right)} \right)}^{3}}-4$? | Since $ f\left( x \right)={{x}^{5}}+{{x}^{3}}$, switch the $ x$ and the $ y$, to solve for the “new” $ y$ (or $ {{f}^{{-1}}}\left( x \right)$): $ x={{y}^{5}}+{{y}^{3}}={{\left( {{{f}^{{-1}}}\left( x \right)} \right)}^{5}}+{{\left( {{{f}^{{-1}}}\left( x \right)} \right)}^{3}}$ Since $ {{\left( {{{f}^{{-1}}}\left( x \right)} \right)}^{5}}+{{\left( {{{f}^{{-1}}}\left( x \right)} \right)}^{3}}=x,\,\,\,{{\left( {{{f}^{{-1}}}\left( 4 \right)} \right)}^{5}}+{{\left( {{{f}^{{-1}}}\left( 4 \right)} \right)}^{3}}=4$. Therefore, $ {{\left( {{{f}^{{-1}}}\left( 4 \right)} \right)}^{5}}+{{\left( {{{f}^{{-1}}}\left( 4 \right)} \right)}^{3}}-4=4-4=0$. Tricky! | ||||||||
| If $ f\left( x \right)$ is an increasing function with an $ x$-intercept of 5, then what is true about the graph of $ {{f}^{{-1}}}\left( x \right)$?
Fill in the blanks: The graph of $ {{f}^{{-1}}}\left( x \right)$ is a(n) ____________ function with a(n) _____-intercept of _____. | The easiest way to see this is to sketch a graph of an increasing function with an $ x$-intercept of 5, and see how it reflects across the line $ y=x$ :
Can you see how $ {{f}^{{-1}}}\left( x \right)$ would be an increasing graph with a $ y$-intercept of 5? | ||||||||
Use the information in the table above to find: a) $ {{f}^{{-1}}}\left( 5 \right)$ b) $ {{f}^{{-1}}}\left( {f\left( 3 \right)} \right)$ c) $ {{f}^{{-1}}}\left( {{{f}^{{-1}}}\left( 4 \right)} \right)$ | a) The expression $ {{f}^{{-1}}}\left( 5 \right)$ is an inverse, so it’s an $ x$-value. The $ x$-value in the table when $ f\left( x \right)=5$ (the $ y$-value) is $ 1$, so $ {{f}^{{-1}}}\left( 5 \right)=1$.
b) This is a composition of function, like we saw in here in the Advanced Functions section. Work from the inside out to see that $ f\left( 3 \right)=0$, and $ {{f}^{{-1}}}\left( 0 \right)$ is the $ x$-value where $ y=0$. This is just $ 3$; we’re back to where we started! So, $ {{f}^{{-1}}}\left( {f\left( 3 \right)} \right)=3$.
c) Work from the inside out to see that $ {{f}^{{-1}}}\left( 4 \right)=0$, since this is the $ x$-value when the $ y$-value is $ 4$. The outside inverse function is the $ x$-value where $ y=0$; the answer is $ 3$ again. So, $ {{f}^{{-1}}}\left( {{{f}^{{-1}}}\left( 4 \right)} \right)=3$. Note that this is the same as finding $ x$ where $ f\left( {f\left( x \right)} \right)=4$. Tricky! |
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.
| Original Function | Inverse Function | ||
| $ f\left( x \right)={{\left( {x+2} \right)}^{2}}-3,\,x>-2$ Find inverse, and get the domain and range of both functions.
Domain: $ \left( {-2,\infty } \right)$ Range: ?
Get domain of the inverse function first; this will be the range of original. | $ \begin{align}y&={{\left( {x+2} \right)}^{2}}-3\\x&={{\left( {y+2} \right)}^{2}}-3\\{{\left( {y+2} \right)}^{2}}&=x+3\\\sqrt{{{{{\left( {y+2} \right)}}^{2}}}}&=\pm \sqrt{{x+3}}\\\,y+2&=\pm \sqrt{{x+3}}\\\\y&=+\sqrt{{x+3}}-2\text{ }\end{align}$
Note: We have to take the positive square root only because of domain restriction – see dark red function. | $ {{f}^{{-1}}}\left( x \right)=\sqrt{{x+3}}-2$
Domain: $ \left( {-3,\infty } \right)$ Range: $ \left( {-2,\infty } \right)$
We got the range from the domain of original.
Domain of the inverse has to be $ >-3$ since the range has to be $ >-2$ (see green function). |  Now we can get the Domain and Range of the Original Function by switching: Domain: $ \left( {-2,\infty } \right)$ Range: $ \left( {-3,\infty } \right)$ |
| $ f\left( x \right)={{\left( {x-2} \right)}^{2}}-1$ Find inverse so functions are one-to-one.
To make one-to-one, we can only use “half” of the parabola. It actually doesn’t even matter which half, as long as the inverse matches.
Since the vertex is at $ (2,–1)$, let’s take the half on the right side, such that: Domain: $ \left[ {2,\infty } \right)$ Range: $ \left[ {-1,\infty } \right)$ | $ \begin{align}y&={{\left( {x-2} \right)}^{2}}-1\\x&={{\left( {y-2} \right)}^{2}}-1\\{{\left( {y-2} \right)}^{2}}&=x+1\\\sqrt{{{{{\left( {y-2} \right)}}^{2}}}}&=\pm \sqrt{{x+1}}\\y-2&=\pm \sqrt{{x+1}}\\\\y&=+\sqrt{{x+1}}+2\text{ }\end{align}$
Note: We have to take the positive square root only because the functions need to be one-to-one. | $ {{f}^{{-1}}}\left( x \right)=\sqrt{{x+1}}+2$
It can be much easier to get the domain and range after drawing the graph.
So, for the inverse function:
Domain: $ \left[ {-1,\infty } \right)$ Range: $ \left[ {2,\infty } \right)$ |  We can now verify that we got the correct Domain and Range of the Original Function. |
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:
| Original Function | Inverse Function | |||||||||||||||
| $ f\left( x \right)={{x}^{2}}+3;\,\,\,\,x\le -2$ Find inverse, and get the domain and range of both functions.
Domain: $ \left( {-\infty ,-2} \right]$ Range: ? Since the domain has to be $ \le -2$, plug in $ -2$ for $ x$ (and test some points) to see that the range has to be $ \ge 7$.
Thus, for the original function, we get: Domain: $ \left( {-\infty ,-2} \right]$ Range: $ \left[ {7,\infty } \right)$ This will help us draw the graph of the inverse function, since we will switch these for that graph. | $ \begin{align}y&={{x}^{2}}+3\\x&={{y}^{2}}+3\\x-3&={{y}^{2}}\\y&=-\sqrt{{x-3}}\\{{f}^{{-1}}}\left( x \right)&=\,\,-\sqrt{{x-3}}\end{align}$
Note: We have to use the negative square root, since $ x\le -2$ in the original function, means $ y\le -2$ in the inverse. We can also see this from the graph.
Since we switch the domain and range for the inverse, we can see where the graph “starts”. |  We can see that for the inverse function: Domain: $ \left[ {7,\infty } \right)$ Range: $ \left( {-\infty ,-2} \right]$ | ||||||||||||||
| $ \displaystyle f\left( x \right)=\frac{{x+2}}{{{{{\left( {x-1} \right)}}^{2}}}}$ Find inverse so functions are one-to-one.
Note there is a horizontal (end behavior) asymptote at $ y=0$ and a vertical asymptote at $ x=1$. Thus, for the original function without a restriction: Domain: $ \left( {-\infty ,1} \right)\cup \left( {1,\infty } \right)$ Range: $ \left( {-\infty ,0} \right)\cup \left( {0,\infty } \right)$ We can use a T-chart, and see that we can “erase” the left part of the graph to make it one-to-one.
Domain: $ \left( {1,\infty } \right)$ Range: $ \left( {0,\infty } \right)$ | This function is tough to solve algebraically for an inverse, so we’ll rely on the graphs.
For the inverse function, we will “switch” the asymptotes, so there is a horizontal (end behavior) asymptote at $ y=1$ and a vertical asymptote at $ x=0$.
We can also switch points in the T-chart to help graph. |
We can see that for the inverse function: Domain: $ \left( {0,\infty } \right)$ Range: $ \left( {1,\infty } \right)$ | ||||||||||||||
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:
| Functions | Inverses? |
| $ \displaystyle \begin{align}f\left( x \right)&=-\frac{5}{4}x-2\\g\left( x \right)&=\frac{{-4x-8}}{5}\end{align}$ | $ \displaystyle \require {cancel} \begin{align}f\left( {g\left( x \right)} \right)&=f\left( {\frac{{-4x-8}}{5}} \right)=-\frac{{\cancel{5}}}{4}\left( {\frac{{-4x-8}}{{\cancel{5}}}} \right)-2=\frac{{4x+8}}{4}-2=x+2-2=x\,\,\,\,\,\surd \\g\left( {f\left( x \right)} \right)&=g\left( {-\frac{5}{4}x-2} \right)=\frac{{-4\left( {-\frac{5}{4}x-2} \right)-8}}{5}=\frac{{5x+8-8}}{5}=\frac{{5x}}{5}=x\,\,\,\,\,\surd \end{align}$ Yes; they are inverses! |
| $ \begin{align}f\left( x \right)=3x+8\\g\left( x \right)=\frac{{x+8}}{3}\end{align}$ | $ \displaystyle f\left( {g\left( x \right)} \right)=f\left( {\frac{{x+8}}{3}} \right)=\cancel{3}\left( {\frac{{x+8}}{{\cancel{3}}}} \right)+8=x+8+8=x+16$ No; they are not inverses! You can stop here. |
| $ \begin{align}f\left( x \right)&=\sqrt{{x+3}}\\g\left( x \right)&={{x}^{2}}-3,\,\,\,x\ge 0\end{align}$ | $ \begin{array}{l}f\left( {g\left( x \right)} \right)=f\left( {{{x}^{2}}-3} \right)=\sqrt{{\left( {{{x}^{2}}-3} \right)+3}}=\sqrt{{{{x}^{2}}}}\,\,(\text{where }x\ge 0)=x\,\,\,\,\,\,\surd \\\,\,\,\,\,\,\,\,\,g\left( {f\left( x \right)} \right)=g\left( {\sqrt{{x+3}}} \right)={{\left( {\sqrt{{x+3}}} \right)}^{2}}-3=\left( {x+3} \right)-3=x\,\,\,\,\,\,\,\surd \end{array}$ Yes; they are inverses! |
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.
| Show Exp/Log Inverses Method 1 | Show Exp/Log Inverses Method 2 | Show Exp/Log Inverses Method 3 |
| To find the inverse, switch the $ x$ and $ y$, and solve for the “new” $ y$ using the loop method: $ \begin{align}y&={{b}^{x}}\\x&={{b}^{y}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{switch }x\text{ and }y\,\,\\\,\,\,\,\,\,y&={{\log }_{b}}x\,\,\,\,\,\,\,\text{log ”loop”}\\{{f}^{{-1}}}\left( x \right)&={{\log }_{b}}x\end{align}$ | To find the inverse, switch the $ x$ and $ y$, and solve for the “new” $ y$, taking the log of both sides and then using the change of base method: $ \displaystyle \begin{align}\,\,\,\,\,\,\,\,y&={{b}^{x}}\\\,\,\,\,\,\,\,\,x&={{b}^{y}}\\\log x&=\log {{b}^{y}}\,\,\,\,\,\,\,\,\,\text{take logs of both sides}\\\log x&=y\log b\,\,\,\,\,\,\,\,\text{power rule}\\y&=\frac{{\log x}}{{\log b}}\\{{f}^{{-1}}}\left( x \right)&={{\log }_{b}}x\,\,\,\,\,\,\,\,\text{change of base}\end{align}$ | Composition Method: Show $ \displaystyle f\left( {g\left( x \right)} \right)=g\left( {f\left( x \right)} \right)=x$; this means the two functions are inverses: $ \displaystyle \begin{align}f\left( x \right)={{b}^{x}}\,\,&\,\,\,\,\,g\left( x \right)={{\log }_{b}}x\\f\left( {g\left( x \right)} \right)&=f\left( {{{{\log }}_{b}}x} \right)\\&={{b}^{{{{{\log }}_{b}}x}}}=x\,\,\,\,\,\,\surd \\g\left( {f\left( x \right)} \right)&=g\left( {{{b}^{x}}} \right)\\&={{\log }_{b}}{{b}^{x}}=x\,\,\,\,\surd \end{align}$ |
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:
| $ \displaystyle y={{2}^{x}}\,\,\,\,\,\,\text{and}\,\,\,\,\,y={{\log }_{2}}x\,$ | $ \displaystyle y={{e}^{x}}\,\,\,\,\,\,\text{and}\,\,\,\,\,\,y=\ln x\,$ |
 $ \displaystyle y={{2}^{x}}$: Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left( {0,\infty } \right)$ $ \displaystyle y={{\log }_{2}}x$: Domain: $ \left( {0,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ |  $ \displaystyle y={{e}^{x}}$: Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left( {0,\infty } \right)$ $ \displaystyle y=\ln x$: Domain: $ \left( {0,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ |
Find the inverses of the following transformed exponential and log functions by switching the $ x$ and the $ y$ and solving for the “new” $ y$:
| Original Function | Inverse Function | Graph |
| $ y={{3}^{{x-3}}}+2$
Shift right 3, up 2
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left( {2,\infty } \right)$ | $ \begin{align}x&={{3}^{{y-3}}}+2\\x-2&={{3}^{{y-3}}}\\{{\log }_{3}}\left( {x-2} \right)&=y-3\\y&={{\log }_{3}}\left( {x-2} \right)+3\end{align}$
Shift right 2, up 3
Domain: $ \displaystyle \left( {2,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ |  |
| $ \begin{array}{l}y=2\ln \left( {3x+6} \right)\,\,\,\,\,\,\text{or}\\y=2\ln \left( {3\left( {x+2} \right)} \right)\end{array}$
Vertical Stretch of 2, Horizontal Shrink of $ \displaystyle \frac{1}{3}$, Shift left 2
Domain: $ \left( {-2,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ | $ \begin{align}x&=2\ln \left( {3y+6} \right)\\\frac{x}{2}&=\ln \left( {3y+6} \right)\\{{e}^{{\frac{x}{2}}}}&=3y+6\\y&=\frac{1}{3}{{e}^{{\frac{x}{2}}}}-2\end{align}$
Vertical Shrink of $ \displaystyle \frac{1}{3}$ , Horizontal Stretch of 2, Shift down 2
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left( {-2,\infty } \right)$ |  |
| $ \begin{array}{l}y=2{{\left( 3 \right)}^{{1-x}}}\,\,\,\,\,\,\text{or}\\y=2{{\left( 3 \right)}^{{-\left( {x-1} \right)}}}\end{array}$
Vertical Stretch of 2, Reflect $ y$-axis, Shift right 1
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left( {0,\infty } \right)$ | $ \begin{align}x&=2{{\left( 3 \right)}^{{1-y}}}\\\frac{x}{2}&={{3}^{{1-y}}}\\{{\log }_{3}}\left( {\frac{x}{2}} \right)&=1-y\\y&=1-{{\log }_{3}}\left( {\frac{x}{2}} \right)\end{align}$
Horizontal Stretch of 2, Reflect $ x$-axis, Shift up 1
Domain: $ \left( {0,\infty } \right)$ Range: $ \left( {-\infty ,\infty } \right)$ |  |
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