1.4 Inverse Functions

Learning Objectives

Think of a function as a machine: you drop in an input, and it produces exactly one output. An inverse function reverses that process — it takes the output and hands back the original input. But not every function can be reversed cleanly. In this section we figure out exactly when a function has an inverse, how to find it, how to draw it, and how to apply the idea to the six trigonometric functions — giving us a toolkit of six new inverse trig functions that are indispensable in calculus.

1.4.1 Existence of an Inverse Function

Note that \(f^{-1}\) is read as "f inverse." The \(-1\) is NOT an exponent: \(f^{-1}(x) \ne \frac{1}{f(x)}\). Figure 1.37 shows the domain-and-range relationship between \(f\) and \(f^{-1}\).

Figure 1.37 — Given a function \(f\) and its inverse \(f^{-1}\), \(f^{-1}(y) = x\) if and only if \(f(x) = y\). The range of \(f\) becomes the domain of \(f^{-1}\) and the domain of \(f\) becomes the range of \(f^{-1}\).

A quick graphical check exists: the horizontal line test. If every horizontal line crosses the graph at most once, the function is one-to-one. (Compare this to the vertical line test, which checks whether a curve is a function at all. The horizontal line test checks whether that function is one-to-one.)

Rule: Horizontal Line Test

A function \(f\) is one-to-one if and only if every horizontal line intersects the graph of \(f\) no more than once.

Figure 1.38 — (a) The function \(f(x) = x^2\) is not one-to-one because it fails the horizontal line test. (b) The function \(f(x) = x^3\) is one-to-one because it passes the horizontal line test.

We start with a concrete example. Consider \(f(x) = x^3 + 4\). If we get an output \(y\), we can solve \(y = x^3 + 4\) for \(x\): subtract 4, then take the cube root to get \(x = \sqrt[3]{y - 4}\). That formula defines \(x\) as a function of \(y\), and it perfectly "undoes" whatever \(f\) did. We call this the inverse function and write \(f^{-1}(y) = \sqrt[3]{y - 4}\). Notice \(f^{-1}(f(x)) = \sqrt[3]{(x^3 + 4) - 4} = x\) — the composition brings us right back to where we started.

Not every function can be undone this cleanly. Try \(f(x) = x^2\): solving \(y = x^2\) gives \(x = \pm\sqrt{y}\). Two answers! For any positive \(y\) there are two inputs (\(\sqrt{y}\) and \(-\sqrt{y}\)) that produce it, so we cannot point to a single "original" input. The culprit is that \(f(x) = x^2\) sends two different inputs to the same output. Functions that avoid this problem — where each output came from exactly one input — are called one-to-one functions.

Try It Now 1.4.1

Is the function \(f\) graphed below one-to-one?

Try It Now 1.4.1 — f(x) = x^3 - x for the horizontal-line-test exercise.

Solution

Apply the horizontal line test. The function \(f(x) = x^3 - x\) is a cubic with a local max and a local min, which means some horizontal lines cross it three times. Therefore, \(f\) is not one-to-one on its natural domain.

Answer: Not one-to-one.

Example 1.4.1: Determining Whether a Function Is One-to-One

For each of the following functions, use the horizontal line test to determine whether it is one-to-one.

1. 
2. 

Solution

1. Since the horizontal line \(y = n\) for any integer \(n \ge 0\) intersects the graph more than once, this function is not one-to-one.

Example 1.4.1 solution graph #1 — the step function with two horizontal lines, each crossing an entire tread, demonstrating HLT failure.

1. Since every horizontal line intersects the graph at most once, this function is one-to-one.

Example 1.4.1 solution graph #2 — the hyperbola 1/x with three horizontal lines each crossing exactly once, demonstrating HLT passage.

1.4.2 Finding a Function's Inverse

Because a one-to-one function sends each input to a unique output, we can always work backwards: given any output \(y\) in the range, there is exactly one input \(x\) in the domain with \(f(x) = y\). To find that input algebraically:

Problem-Solving Strategy: Finding an Inverse Function

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

Step 2. Solve the equation for \(x\) in terms of \(y\). The result is \(x = f^{-1}(y)\).

Step 3. Interchange \(x\) and \(y\) so the inverse is written as \(y = f^{-1}(x)\).

Step 4. State the domain and range of \(f^{-1}\): the domain of \(f^{-1}\) is the range of \(f\), and the range of \(f^{-1}\) is the domain of \(f\). Verify by checking \(f^{-1}(f(x)) = x\).

Graphing Inverse Functions

The graphs of \(f\) and \(f^{-1}\) are mirror images of each other across the line \(y = x\). Here is why: if the point \((a, b)\) is on the graph of \(f\) (meaning \(b = f(a)\)), then \(a = f^{-1}(b)\), so the point \((b, a)\) is on the graph of \(f^{-1}\). Swapping \(x\)- and \(y\)-coordinates is exactly the reflection over \(y = x\).

Figure 1.39 — (a) The graph of this function \(f\) shows point \((a, b)\) on the graph of \(f\). (b) Since \((a, b)\) is on the graph of \(f\), the point \((b, a)\) is on the graph of \(f^{-1}\). The graph of \(f^{-1}\) is a reflection of the graph of \(f\) about the line \(y = x\).

Example 1.4.3: Sketching Graphs of Inverse Functions

For the graph of \(f\) shown below, sketch a graph of \(f^{-1}\) by using symmetry about the line \(y = x\). Identify the domain and range of \(f^{-1}\).

Example 1.4.3 prompt graph — f(x) = sqrt(x + 2) on the visible window, given for the inverse-sketching exercise.

Answer: Domain of \(f^{-1}\) is \([0, \infty)\); range of \(f^{-1}\) is \([-2, \infty)\).