1.4 Inverse Functions

Inverse functions reverse the input-output relationship of a function, letting you "undo" what the original function does. Understanding them is essential for solving equations and sets the stage for key Calculus I topics like derivatives of inverse functions. This guide covers how to test for invertibility, find and graph inverses, and work with inverse trigonometric functions.

Inverse Functions

Horizontal line test for invertibility

A function has an inverse only if it's one-to-one, meaning every output comes from exactly one input. The horizontal line test gives you a quick visual check for this.

Draw horizontal lines across the graph of the function:

• If any horizontal line hits the graph more than once, the function is not one-to-one and doesn't have an inverse. For example, $f(x) = x^2$ (on all of $\mathbb{R}$) fails because the line $y = 4$ hits the parabola at both $x = 2$ and $x = -2$.
• If no horizontal line hits the graph more than once, the function is one-to-one and has an inverse. Exponential functions like $f(x) = 2^x$ pass this test.

A function that passes the horizontal line test is called injective. A function that's strictly increasing or strictly decreasing over its entire domain (a monotonic function) will always pass the horizontal line test and therefore always has an inverse.

Computation and graphing of inverses

To find the inverse of a function $f(x)$:

1. Replace $f(x)$ with $y$
2. Swap $x$ and $y$
3. Solve the new equation for $y$
4. Write the result as $f^{-1}(x)$

Example: Find the inverse of $f(x) = 3x - 5$.

1. $y = 3x - 5$

2. $x = 3y - 5$

3. $x + 5 = 3y$, so $y = \frac{x + 5}{3}$

4. $f^{-1}(x) = \frac{x + 5}{3}$

The domain of $f^{-1}$ equals the range of $f$, and the range of $f^{-1}$ equals the domain of $f$. For instance, $f(x) = x^2$ with domain $[0, \infty)$ has range $[0, \infty)$, and its inverse $f^{-1}(x) = \sqrt{x}$ has domain $[0, \infty)$ and range $[0, \infty)$.

Graphing: The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$. Every point $(a, b)$ on $f$ becomes $(b, a)$ on $f^{-1}$. So if $(2, 4)$ is on $f$, then $(4, 2)$ is on $f^{-1}$.

Applications of inverse trigonometric functions

Trig functions take angles and produce ratios. Inverse trig functions do the opposite: they take a ratio and return an angle.

• $\arcsin(x)$ (or $\sin^{-1}(x)$) returns the angle whose sine is $x$
• $\arccos(x)$ (or $\cos^{-1}(x)$) returns the angle whose cosine is $x$
• $\arctan(x)$ (or $\tan^{-1}(x)$) returns the angle whose tangent is $x$

Because sine, cosine, and tangent aren't one-to-one on their full domains, we restrict their domains so that each inverse function gives a single output. You need to know these restricted domains and ranges:

Functions vs. their inverses

Composing a function with its inverse gives you back the original input. This is the defining property of inverse functions:

• $f(f^{-1}(x)) = x$ for all $x$ in the domain of $f^{-1}$
• $f^{-1}(f(x)) = x$ for all $x$ in the domain of $f$

This is a useful way to verify that two functions are actually inverses of each other. Just compose them both ways and check that you get $x$.

The derivative of an inverse function is related to the derivative of the original:

$\left(f^{-1}\right)'(x) = \frac{1}{f'\!\left(f^{-1}(x)\right)}$

This formula says: evaluate the derivative of $f$ at the point $f^{-1}(x)$, then take the reciprocal. You'll use this frequently later in the course. Note that $f'(f^{-1}(x))$ can't equal zero for this to work.

Equations with inverse functions

Inverse functions are the main tool for isolating a variable trapped inside a function. The process is straightforward:

1. Start with an equation like $f(x) = a$
2. Apply $f^{-1}$ to both sides: $f^{-1}(f(x)) = f^{-1}(a)$
3. Simplify the left side: $x = f^{-1}(a)$

For trigonometric equations, you need to account for periodicity. Inverse trig functions give you only the principal value (the one angle in the restricted range), but trig equations typically have infinitely many solutions.

Example: Solve $\cos(x) = -\frac{\sqrt{2}}{2}$.

1. The principal value is $\arccos\!\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}$

2. Cosine is also negative and equals $-\frac{\sqrt{2}}{2}$ at $\frac{5\pi}{4}$, which is $2\pi - \frac{3\pi}{4}$

3. The general solutions are $x = \frac{3\pi}{4} + 2\pi n$ or $x = \frac{5\pi}{4} + 2\pi n$, where $n$ is any integer

Function properties and invertibility

A few properties tie together the ideas in this section:

• A surjective (onto) function maps to every element in its codomain. A function that's both injective (one-to-one) and surjective is called bijective, and bijective functions always have inverses.
• A monotonic function is strictly increasing or strictly decreasing over its entire domain. Monotonic functions are always one-to-one, so they're always invertible.
• The composition of two invertible functions is also invertible: if $h(x) = f(g(x))$, then $h^{-1}(x) = g^{-1}(f^{-1}(x))$. Notice the order reverses.