Inverse Function Rules

Why This Matters

Inverse functions show up constantly in Pre-Calculus, and they connect to everything from solving equations to trigonometry. The core idea is straightforward: an inverse function "undoes" what the original function does. But knowing when an inverse exists, how to find it, and why certain functions need restrictions before they can have one is where the real understanding lives.

This guide covers the rules you need to know, the methods you need to practice, and the common pitfalls you need to avoid.

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Foundational Concepts: When Does an Inverse Exist?

Before you can find an inverse, you need to determine whether one exists at all. The key requirement is that the function must be one-to-one.

One-to-One Functions and the Horizontal Line Test

A function is one-to-one if every output comes from exactly one input. No two different $x$-values can produce the same $y$-value.

The horizontal line test is the graphical version of this check. Draw horizontal lines across the graph. If any horizontal line crosses the graph more than once, the function is not one-to-one, and it doesn't have an inverse (at least not without a domain restriction).

Think about $f(x) = x^2$. The horizontal line $y = 4$ hits the graph at both $x = 2$ and $x = -2$. Two inputs, same output. That fails the test.

Domain and Range Swap

When a function does have an inverse, the domain and range trade places:

• The domain of $f$ becomes the range of $f^{-1}$
• The range of $f$ becomes the domain of $f^{-1}$

For example, if $f(x)$ has domain $[-3, 5]$ and range $[0, 10]$, then $f^{-1}(x)$ has domain $[0, 10]$ and range $[-3, 5]$.

This swap is automatic. You don't choose it or calculate it; it's built into what "inverse" means.

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Graphical and Algebraic Methods

These are the two main ways you'll work with inverse functions: seeing them on a graph and finding them with algebra.

Reflection Over $y = x$

The graph of $f^{-1}$ is the mirror image of the graph of $f$ across the line $y = x$.

Every point $(a, b)$ on the graph of $f$ corresponds to the point $(b, a)$ on the graph of $f^{-1}$. You're flipping the coordinates.

This is useful for quick visual checks. If you've graphed a function and its supposed inverse, they should be symmetric across $y = x$. If they aren't, something went wrong.

For linear functions, this reflection has a specific consequence: if $f$ has slope $m$, then $f^{-1}$ has slope $\frac{1}{m}$. A line with slope 3 reflects to a line with slope $\frac{1}{3}$.

Finding an Inverse Algebraically

The algebraic process has three steps:

1. Start with $y = f(x)$. Solve the equation for $x$ in terms of $y$.
2. Swap the variables: replace $x$ with $y$ and $y$ with $x$.
3. Verify using composition: confirm that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.

Here's a concrete example. Find the inverse of $f(x) = 2x + 5$:

• Write $y = 2x + 5$
• Solve for $x$: subtract 5, then divide by 2. You get $x = \frac{y - 5}{2}$
• Swap variables: $f^{-1}(x) = \frac{x - 5}{2}$

Verify: $f(f^{-1}(x)) = 2\left(\frac{x-5}{2}\right) + 5 = (x - 5) + 5 = x$. It checks out.

Composition Property

The defining algebraic property of inverse functions is:

$f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$

Both compositions must equal $x$. If you only check one direction, you could be fooled by functions that partially "undo" each other but aren't true inverses. Always check both.

The result of these compositions is the identity function, which simply returns its input unchanged.

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Domain Restrictions: Making Non-Invertible Functions Invertible

Some of the most important functions in Pre-Calculus aren't naturally one-to-one. To give them inverses, you restrict their domain to a portion where they are one-to-one.

The Standard Example: $f(x) = x^2$

Restrict the domain to $x \geq 0$, and the function becomes one-to-one. Its inverse is then $f^{-1}(x) = \sqrt{x}$.

You could also restrict to $x \leq 0$, but the convention is $x \geq 0$.

Inverse Trigonometric Functions

Trig functions are periodic, so they fail the horizontal line test badly. Each one needs a carefully chosen restricted domain.

$\sin(x)$ is restricted to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ On this interval, sine goes from $-1$ to $1$ and is one-to-one. The inverse, $\sin^{-1}(x)$, has range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

$\cos(x)$ is restricted to $[0, \pi]$ On this interval, cosine goes from $1$ to $-1$ and is one-to-one. The inverse, $\cos^{-1}(x)$, has range $[0, \pi]$.

$\tan(x)$ is restricted to $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ On this interval, tangent covers all real numbers and is one-to-one. The inverse, $\tan^{-1}(x)$, has range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

These restricted domains and the resulting inverse trig ranges need to be memorized. They come up repeatedly.

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Key Inverse Pairs

Exponential and Logarithmic Functions

The functions $f(x) = a^x$ and $g(x) = \log_a(x)$ are inverses of each other, provided $a > 0$ and $a \neq 1$.

The two identities that follow from this relationship:

$a^{\log_a(x)} = x$ (for $x > 0$) and $\log_a(a^x) = x$

Unlike trig functions, exponentials are naturally one-to-one for valid bases. No domain restriction is needed. The function $a^x$ always passes the horizontal line test (it's always increasing or always decreasing depending on the base).

This is a key contrast with trig functions. Exponentials and logs are naturally invertible; trig functions are not.

Derivative of an Inverse Function (Calculus Preview)

For those moving toward Calculus, there's a useful formula:

$(f^{-1})'(y) = \frac{1}{f'(x)}$ where $y = f(x)$

The rate of change of the inverse is the reciprocal of the rate of change of the original function. This connects to implicit differentiation and the chain rule, and it's worth being aware of even if you aren't tested on it yet.

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Quick Reference Table

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Self-Check Questions

1. What do the horizontal line test and the one-to-one requirement have in common, and why are both necessary for a function to have an inverse?

2. If $f(x)$ has domain $[-3, 5]$ and range $[0, 10]$, what are the domain and range of $f^{-1}(x)$?

3. Compare and contrast the process of verifying inverses graphically (reflection) versus algebraically (composition). When would you use each method?

4. Why does $f(x) = \sin(x)$ require a domain restriction to have an inverse, while $f(x) = e^x$ does not? What restriction is used for sine?

5. Given $f(x) = 2x + 5$, find $f^{-1}(x)$ algebraically and verify your answer using composition. What does the reflection property predict about the slopes of $f$ and $f^{-1}$?