Invertible functions and inverse of a function

What "invertible" means. A function $f : A \to B$ is called invertible if there exists a function $g : B \to A$ such that composing them either way gives you back where you started:

where: $A$ is the domain of $f$, $B$ is its codomain, $x$ ranges over $A$, and $y$ ranges over $B$. When such a $g$ exists it is unique, and it is called the inverse of $f$, written $f^{-1}$. It is worth pausing on why uniqueness matters: if two different functions $g_1$ and $g_2$ both satisfied these conditions, they would have to agree at every point of $B$, so really there is only ever one candidate worth looking for.

Why bijection is the exact condition. One-one-ness is what lets you reverse $f$ without ambiguity: if two different inputs could give the same output $y$, then "the" inverse would not know which one to send $y$ back to. Onto-ness is what lets the inverse be defined everywhere on $B$: if some $y \in B$ were never hit by $f$, then $g(y)$ would have nothing sensible to equal. A function missing either property cannot be invertible in the strict two-sided sense above; a function with both properties always can be, by simply reading the graph of $f$ backwards.

How to actually find $f^{-1}$. Once you know $f$ is invertible, the mechanical recipe is short: write $y = f(x)$, solve this equation algebraically for $x$ in terms of $y$, and then rename the letters so the answer reads as a function of its own input. That renamed expression is $f^{-1}$. The figure below shows the picture behind the algebra — the arrows of $f$ simply run in reverse for $f^{-1}$.

A common instinct is to try writing an inverse for any formula that looks solvable, but the definition demands the two-sided composition condition above hold for the whole domain and codomain, not just for a few convenient points. If $f$ fails one-one-ness or onto-ness on the sets you were given, "$f^{-1}$" simply does not exist for that $f$ as stated — the fix, when one is possible, is to shrink the domain and codomain until both properties hold, which is exactly the move in the worked example below.

Let $f : [0,\infty) \to [0,\infty)$ be defined by $f(x) = x^2$. Show $f$ is invertible on this restricted domain and codomain, and find $f^{-1}$.

Step 1. Recall the trap. On the full sets $\mathbb{R} \to \mathbb{R}$, $f(x) = x^2$ is neither one-one (since $f(-2) = f(2) = 4$) nor onto (negative numbers are never reached). Restricting both the domain and the codomain to $[0,\infty)$ is what rescues it.

Step 2. Check one-one on $[0,\infty)$. Suppose $x_1, x_2 \ge 0$ and $f(x_1) = f(x_2)$, i.e. $x_1^2 = x_2^2$. Factor: $x_1^2 - x_2^2 = (x_1-x_2)(x_1+x_2) = 0$. So either $x_1 = x_2$, or $x_1 + x_2 = 0$. Because $x_1, x_2 \ge 0$, the second case is only possible when $x_1 = x_2 = 0$ — which again gives $x_1 = x_2$. Either way, $x_1 = x_2$, so $f$ is one-one on $[0,\infty)$.

Step 3. Check onto $[0,\infty)$. Take any $y \ge 0$. Choosing $x = \sqrt{y}$ (a well-defined non-negative real number since $y \ge 0$) gives $f(\sqrt{y}) = (\sqrt{y})^2 = y$. So every $y \ge 0$ is hit, and $f$ is onto.

Step 4. Write and check the inverse. With both properties established, $f$ is invertible on $[0,\infty) \to [0,\infty)$, and

Check: $f^{-1}(f(x)) = \sqrt{x^2} = |x| = x$ for $x \ge 0$, and $f(f^{-1}(x)) = (\sqrt{x})^2 = x$ for $x \ge 0$. The inverse of $f(x) = x^2$ on $[0,\infty)$ is $f^{-1}(x) = \sqrt{x}$.

Competitive stretch

The same restriction idea reappears, with a twist, for $f(x) = \sin x$: on the full real line it is neither one-one nor onto $[-1,1]$, but restricted to the domain $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ and codomain $[-1,1]$ it becomes a bijection, and its inverse there is exactly the principal value branch $\sin^{-1}x$ you will meet in the next chapter — the restricted interval is chosen precisely so one-one-ness and onto-ness both hold.

• Writing down "$f^{-1}$" before checking $f$ is one-one and onto — invertibility must be established, not assumed from the formula alone.
• Confusing the inverse function $f^{-1}(x)$ with the reciprocal $\dfrac{1}{f(x)}$; the superscript $-1$ here means "undo the function," not "raise to the power minus one."
• Ignoring the stated domain and codomain and testing only the "natural" formula — as Worked Example shows, $x^2$ is invertible on $[0,\infty)$ even though it is not invertible on all of $\mathbb{R}$.
• Skipping the final check $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$ after solving for the inverse — this is the step that catches sign errors and algebraic slips before they cost marks.

Ask the tutor: Want to see how this same domain-restriction trick is used to define $\sin^{-1}x$, $\cos^{-1}x$ and the other inverse trigonometric functions, or want more practice finding inverses of algebraic functions step by step?