Inverse Functions and Domain Restrictions

Example For $f(x) = 3x + 5$ on all real numbers, $3a + 5 = 3b + 5$ forces $a = b$, so $f$ is one-to-one. The square function is not one-to-one on all real numbers, since $(-2)^2 = 2^2 = 4$ with $-2 \ne 2$. On $[0, \infty)$ it is one-to-one, because two nonnegative numbers with equal squares are equal.

Example The line $y = 4$ meets the parabola $y = x^2$ at $(-2, 4)$ and $(2, 4)$, so the square function fails on all real numbers. The graph of $y = x^3$ is increasing everywhere, so no horizontal line meets it twice and the cube function passes. In fact each horizontal line meets it exactly once, since every real number is a cube. A sketch makes the test easy to read, but the algebra of the previous section is what proves it.

Example Take $f(x) = 3x + 5$ on all real numbers, with range all real numbers. Solving $y = 3x + 5$ for $x$ gives $x = \frac{y - 5}{3}$, so

Since $f(2) = 11$, the inverse must send $11$ back to $2$, and indeed $\frac{11 - 5}{3} = 2$.

Example The function $f(x) = 3x + 5$ multiplies by $3$ and then adds $5$. Undoing in reverse order means subtracting $5$ and then dividing by $3$, which is the formula found above. Both checks work for every real $x$.

Example The function $f(x) = \sqrt{x + 1}$ has domain $[-1, \infty)$ and range $[0, \infty)$. Solving $y = \sqrt{x + 1}$ gives $x = y^2 - 1$, and because the inverse accepts only outputs of $f$,

The points $(-1, 0)$, $(0, 1)$ and $(3, 2)$ on the graph of $f$ become $(0, -1)$, $(1, 0)$ and $(2, 3)$ on the graph of the inverse. The condition $x \ge 0$ is part of the answer. Without it the formula would send $-2$ to $3$, while $f(3) = 2$, not $-2$.

Example For $f(x) = 2x - 4$ the inverse is $f^{-1}(x) = \frac{x + 4}{2}$. The point $(3, 2)$ is on the graph of $f$ and $(2, 3)$ is on the graph of the inverse. The segment between them has midpoint $(2.5, 2.5)$, which lies on $y = x$, and slope

so it crosses the line $y = x$, of slope $1$, at a right angle.

Example The shifted square function $f(x) = (x - 2)^2 + 1$ sends both $0$ and $4$ to $5$, so it is not one-to-one. Its vertex is $(2, 1)$, and it is increasing on $[2, \infty)$, where its outputs fill $[1, \infty)$. On that branch $x - 2 \ge 0$, so solving $(x - 2)^2 = y - 1$ takes the nonnegative root, $x - 2 = \sqrt{y - 1}$. The chosen inverse is

with range $[2, \infty)$. As a check, $f(4) = 5$ and $f^{-1}(5) = 2 + 2 = 4$. On the other branch $(-\infty, 2]$ the inverse would be $2 - \sqrt{x - 1}$, which sends $5$ to $0$. Applying the chosen inverse to the unrestricted function gives $f^{-1}(f(x)) = 2 + \lvert x - 2 \rvert$, which equals $x$ only when $x \ge 2$, exactly the chosen domain.

Example Let $f(x) = x^3 + 1$. Its inverse is $f^{-1}(x) = \sqrt[3]{x - 1}$, while its reciprocal is $\frac{1}{x^3 + 1}$. At $x = 9$ the inverse gives $\sqrt[3]{8} = 2$, while the reciprocal gives $\frac{1}{730}$.