If we think of a function as a thing that does something to an independent variable (like square it, or add 7 to it), then an inverse function simply un-does that action. A function performs some action, and an inverse function just undoes that action.

Now, why would we want to do that? It turns out that it's very useful very often, particularly with functions you usually learn later in your study of functions, like exponential and trigonometric functions.

Here's one example. You don't have to understand the function types right now, just the logic. We are very interested in using exponential growth functions to understand compound interest in the world of finance. In an exponential function, the independent variable is *in* the exponent. So the question is, how can we solve for it if it's in the exponent? The answer lies in exponential functions, and they're just crucial to many fields.

Learning about inverse functions is absolutely crucial if you want to understand inverse trigonometric functions or logarithms. If that's your goal, you're starting in the right place!

Before we start, we'll need to know how to write inverse functions, and there's a convenient way. Notice that this does not mean **1/f(x)**. When the -1 superscript is attached to a function name like **f** in **f(x)**, it has a different meaning.

If a function is $f(x),$ its inverse is written as $f^{-1}(x).$ In this special case, the exponent, $-1,$ does not mean "take the reciprocal." The inverse of $f(x)$ is *not* $\frac{1}{f(x)}.$

If these two functions are really inverses, then one should "undo" the action of the other, and vice-versa. But what does that mean? If **f(x)** does something to x, then placing the result of **f(x)** into **f'(x)**, in other words, calculating **f ^{-1}(f(x))**, should just give

Because we've studied compositions of functions (I hope you have!), we can check.

If we put the function inside the inverse, we get

$$ \begin{align} f^{-1}(f(x)) &= \frac{1}{2}(2x - 4) + 2 \\ &= (x - 2) + 2 \\ &= x \end{align}$$

and if we put the inverse inside of the function, we get

$$ \begin{align} f(f^{-1}(x)) &= 2 \left( \frac{1}{2} x + 2 \right) - 4 \\ &= (x + 4) - 4 \\ &= x \end{align}$$

Sometimes we use a different language for compositions of functions: We say that **f ^{-1}(x)** "

So at least for this simple example, we've proven that **f(x)** and **f ^{-1}(x)** are indeed inverses, at least as we've defined them. Onward!

There is a pretty easy method for calculating inverses. It works for many, but not all functions. We'll try it first on linear functions. First, here are the simple steps:

- Write the function as
**y = f(x)** - Swap
**x**and**y** - Solve for
**y** - The new y is
**f**^{-1}(x)

To start, we'll use the function from the last example, $f(x) = 2x - 4$, and use this method to find the inverse we already know.

1. Write the function as **y = f(x)**

$$y = 2x - 4$$

2. Swap **x** and **y**

$$x = 2y - 4$$

3. Solve for y

$$2y = x + 4$$

4. The inverse is:

$$f^{-1}(x) = \frac{1}{2}x + 2$$

That's the function we expected, so the method worked. Now let's use the same function and inverse to explore what they mean graphically.

Take a good look at this graph. **f(x)** and **f ^{-1}(x)** are both plotted, along with the line

Take a look at the point (2, 0) on **f(x)**. The point (0, 2) is on **f ^{-1}(x)**. It's the same for (0, 4) on the function and (-4, 0) on the inverse, and for all points on both functions. It's always this way for functions and inverses.

For any point (x, y) on a function, there will be a point (y, x) on its inverse, and the other way around.

The graphs of a function and its inverse are always mirror images across the line **y = x**.

Find the inverses of these linear functions. Prove that $f(f^{-1}((x)) = f^{-1}(f(x)) = x$, and sketch a graph of each showing that the two are symmetric across the line **y = x**.

1. | $f(x) = -7x + 3$ | |

2. | $f(x) = \frac{1}{3}x - \frac{2}{3}$ | |

3. | $f(x) = 4x + 4$ |

4. | $f(x) = 7x - 3$ | |

5. | $f(x) = \frac{1}{4}x - \frac{2}{5}$ | |

6. | $f(x) = 9x - 3$ |

Now let's find the inverse of a curved function, the quadratic function **f(x) = x ^{2}**. Remember that the graph of a quadratic function is a parabola.

We follow the same steps as for linear functions (they're always the same).

1. Write the function as **y = f(x)**

$$f(x) = x^2 \; \longrightarrow \; y = x^2$$

2. Swap **x** and **y**

$$x = y^2$$

3. Solve for y

$$y = ±\sqrt{x}$$

Now here's the trick. This really can't be the inverse, because if we graph this function (see below), it's double-valued – for every **x** ≠ 0, there are two values of **y**, and we can't have that for a function.

In these cases, it's common to "throw out" one half of the function. In this one, we'll throw out the lower half to make the inverse

$$f^{-1}(x) = \sqrt{x}$$

Notice the mirror symmetry between these two functions. If (2, 4) is on **f(x)**, then (4, 2) is on **f ^{-1}(x)**.

Finally, it's obvious in this case that these two functions, a squaring function and a square root function are inverses,

$$\sqrt{x^2} = (\sqrt{x})^2 = x$$

Ahead of you, as you study **trigonometry** (or *more* trigonometry) and **exponential functions**, you'll find that you need some very important inverse functions in order to solve for variables "trapped" inside trig. or exponential functions. The most important ones are shown in the table.

**Note**: The inverse trig. functions like **sin ^{-1}(x)**, &c., are sometimes referred to as

Function | Inverse |
---|---|

$f(x) = sin(x)$ | $f^{-1}(x) = sin^{-1}(x)$ |

$f(x) = cos(x)$ | $f^{-1}(x) = cos^{-1}(x)$ |

$f(x) = tan(x)$ | $f^{-1}(x) = tan^{-1}(x)$ |

$f(x) = 10^x$ | $f^{-1}(x) = log_{10}(x)$ |

$f(x) = e^x$ | $f^{-1}(x) = ln(x)$ |

**xaktly.com** by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.