Inverse Functions and Domain Restriction

Example Take the function on $\{1, 2, 3\}$ with $f(1) = 4$, $f(2) = 7$ and $f(3) = -1$. Reversing each arrow gives $f^{-1}(4) = 1$, $f^{-1}(7) = 2$ and $f^{-1}(-1) = 3$. If instead $f(3)$ were $4$, the reversed rule would have to send $4$ to both $1$ and $3$, and there would be no inverse function.

Example The function $f(x) = 2x + 3$ doubles and then adds $3$, so its inverse subtracts $3$ and then halves, $f^{-1}(x) = \frac{x - 3}{2}$. Its reciprocal is $\frac{1}{2x + 3}$. At $x = 5$ they give different numbers.

Example The function $g(x) = \sqrt{x - 2}$ has domain $[2, \infty)$ and range $[0, \infty)$. Squaring $y = \sqrt{x - 2}$ and solving gives $x = y^2 + 2$, so its inverse is

with range $[2, \infty)$. The condition $x \ge 0$ is part of the answer. Without it the formula would also send $-3$ to $11$, yet $g(11) = 3$, not $-3$.

Example The function $f(x) = x^3$ is one-to-one with inverse $f^{-1}(x) = \sqrt[3]{x}$. The points $(1, 1)$, $(2, 8)$ and $(-2, -8)$ on the cube curve become $(1, 1)$, $(8, 2)$ and $(-8, -2)$ on the cube root curve. The segment from $(2, 8)$ to $(8, 2)$ has its midpoint $(5, 5)$ on the line $y = x$ and slope $-1$, so it crosses that line at a right angle. Points on the line itself, such as $(1, 1)$, stay where they are.

Example Let $f(x) = \frac{2x + 1}{x - 3}$ for $x \ne 3$. Set $y$ equal to it, clear the denominator and collect the $x$ terms.

The last step divides by $y - 2$, so it needs $y \ne 2$. Indeed $f(x) = 2$ has no solution, because it would force $2x - 6 = 2x + 1$. For every other $y$ the formula gives an allowed input, since $x = 3$ would force $3y + 1 = 3y - 6$. So the range of $f$ is every real number except $2$, and

As a check, the inverse gives $f^{-1}(5) = \frac{16}{3}$, and

Example The function $f(x) = (x - 1)^2 + 2$ sends both $0$ and $2$ to $3$, so it is not one-to-one. Its graph is a parabola with lowest point $(1, 2)$, and it is increasing on $[1, \infty)$. On that branch, solving $y = (x - 1)^2 + 2$ gives $(x - 1)^2 = y - 2$, and since $x - 1 \ge 0$ there we take the nonnegative square root.

On the other branch $(-\infty, 1]$ we have $x - 1 \le 0$, so the inverse there is $1 - \sqrt{x - 2}$, also with domain $[2, \infty)$. The first inverse sends $3$ to $2$ and the second sends $3$ to $0$, the two inputs that $f$ sent to $3$.

Example Take $f(x) = x^3 - 1$ and $g(x) = \sqrt[3]{x + 1}$, both defined for all real numbers. Then

so $g = f^{-1}$. Domains matter for $f(x) = x^2$ restricted to $[0, \infty)$, whose inverse is $\sqrt{x}$. There $\sqrt{x^2} = x$ for every $x \ge 0$, but at $x = -3$ the same expression gives $\sqrt{(-3)^2} = 3$. That is no contradiction, because $-3$ is not in the domain of the restricted function.