Inverse function notation

One-to-one functions, the notation f⁻¹(x), the composition test f(f⁻¹(x)) = x, restricting a domain so the inverse is a function, and where f meets its inverse.

One-to-one functions and f⁻¹

A function is one-to-one when each $y$-value comes from exactly one $x$-value, which is the horizontal line test from 1F. Only one-to-one functions have an inverse function. Any function that is always increasing, or always decreasing, is one-to-one.

Example 1●●

Find the inverse function of $f(x) = \dfrac{3}{x - 2} + 1$, and state its domain.

$x = \dfrac{3}{y - 2} + 1$

Write y = f(x) and swap x and y.

$x - 1 = \dfrac{3}{y - 2}$, so $y - 2 = \dfrac{3}{x - 1}$.

$f^{-1}(x) = \dfrac{3}{x - 1} + 2$, with domain $x \ne 1$.

The range of f is y ≠ 1, so that becomes the domain.

Restricting the domain

A function that is not one-to-one can be given an inverse function by restricting its domain to a piece where it is one-to-one. Usually you choose the largest such piece containing a given point, split at a turning point.

f(x) = (x − 1)², x ≥ 1, and its inverse

Orange: f with domain x ≥ 1 (the faint left half is removed). Blue: f⁻¹(x) = 1 + √x. Drag P along f and watch its mirror point.

Try this

Where do $f$ and $f^{-1}$ meet? Is that point on the line $y = x$?

Example 2●●

Restrict $f(x) = (x - 1)^2$ to the largest domain containing $x = 3$ for which it has an inverse function. Find $f^{-1}(x)$.

Domain $x \ge 1$, range $y \ge 0$.

The vertex is at x = 1, and 3 is to its right.

$f^{-1}(x) = 1 + \sqrt{x}$, domain $x \ge 0$, range $y \ge 1$.

Swap: x = (y − 1)² with y ≥ 1, so take the positive root.

$f\big(f^{-1}(x)\big) = (1 + \sqrt{x} - 1)^2 = x$ ✓

Check by composing.

Where f meets its inverse

Example 3●●●

$f(x) = x^2 - 2$ with $x \ge 0$. Find where $y = f(x)$ meets $y = f^{-1}(x)$.

Solve $x^2 - 2 = x$: $(x - 2)(x + 1) = 0$.

f is increasing on x ≥ 0.

They meet at $(2, 2)$.

x = −1 is outside the domain.

Flashcards

One-to-one

Each $y$-value comes from only one $x$-value.

$f^{-1}ig(f(x)ig)$

$x$, for every $x$ in the domain of $f$.

$f^{-1}(x)$ vs $dfrac{1}{f(x)}$

Inverse function vs reciprocal. They are different.

Increasing $f$ meets $f^{-1}$

Only on $y = x$: solve $f(x) = x$.

Practice

$f^{-1}(x)$ means:
$f(x) = (x - 1)^2$ with $x \ge 1$. Find $f^{-1}(9)$.
For $x$ in the domain of $f^{-1}$, $f\big(f^{-1}(x)\big)$ equals:
The largest domain containing $x = 3$ on which $f(x) = (x - 1)^2$ is one-to-one is:
The inverse of $f(x) = \dfrac{3}{x - 2} + 1$ is:
The domain of $f^{-1}(x) = \dfrac{3}{x - 1} + 2$ is all real $x$ except one value. Enter it.
$f(x) = x^2 - 2$ with $x \ge 0$. Find the $x$-coordinate where $y = f(x)$ meets $y = f^{-1}(x)$.
$f(x) = 2x + 1$. Find $f^{-1}\big(f(5)\big)$.
Which function is not one-to-one?
An increasing function and its inverse can only meet: