xaktly | Algebra | Functions

Inverse functions

Functions that undo the actions of other functions

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.

Inverse function notation

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.

Pro tip

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 $\frac{1}{f(x)}.$

A simple example to start:   $f(x) = 2x - 4$   &   $f^{-1}(x) = \frac{1}{2}x + 2$

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 x back, "untouched." Likewise, f(f-1(x)) should just equal x.

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) "operates" on f(x), and f(x) "operates" on 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!

Finding inverses

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:

  1. Write the function as y = f(x)
  2. Swap x and y
  3. Solve for y
  4. 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.

A graphical interpretation of inverses

A function and its inverse are mirror images across the line y = x

Take a good look at this graph. f(x) and f-1(x) are both plotted, along with the line y = x. Notice that the function and its inverse cross on that line, and that all points on the function have a mirror image across y = x on the inverse.

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.

A function and its inverse

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

Practice problems

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

Inverse of a quadratic function

Now let's find the inverse of a curved function, the quadratic function f(x) = x2. 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,

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

Important inverse functions

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 arcsin, arccos and arctan.

Function Inverse
$f(x) = sin(x)$ $f^{-1}(x) = sin^{-1}(x)$
$f(x) = cos(x)$ $f^{-1}(x) = cos^{-1}(x)$
$f(x) = tan(x)$ $f^{-1}(x) = tan^{-1}(x)$
$f(x) = 10^x$ $f^{-1}(x) = log_{10}(x)$
$f(x) = e^x$ $f^{-1}(x) = ln(x)$

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