Inverse Functions

What an inverse does

A function's inverse undoes what the original function does. If $f$ takes an input $x$ to an output $y$, then $f^{-1}$ takes that same $y$ back to $x$. The two functions run the same input-output relationship in opposite directions, and that single idea explains almost everything on this topic.

Because the roles of input and output trade places, so do the domain and range. The domain of $f^{-1}$ equals the range of $f$, and the range of $f^{-1}$ equals the domain of $f$. If a question gives you the domain and range of $f$ and asks for the domain of $f^{-1}$, you do not need a formula at all. Just read off the range of $f$. The most common wrong answer is the domain of $f$ itself.

Finding an inverse algebraically

To find an inverse from a formula, swap $x$ and $y$, then solve for the new $y$.

Worked Example. Find the inverse of $f(x) = \dfrac{2x+3}{5}$.

Write $y = \dfrac{2x+3}{5}$ and swap the letters:

$x = \dfrac{2y+3}{5}$

$5x = 2y + 3$

$y = \dfrac{5x-3}{2}$

So $f^{-1}(x) = \dfrac{5x-3}{2}$. Notice the pattern: $f$ multiplied by 2, added 3, then divided by 5. The inverse multiplies by 5, subtracts 3, then divides by 2. An inverse performs the opposite operations in the reverse order. That is a fast way to find or check an inverse in your head, and it also catches the usual sign slip, where a "$+7$" in $f$ needs to become a "$-7$" in $f^{-1}$.

The big shortcut: $f(f^{-1}(x)) = x$

By definition, $f(f^{-1}(x)) = x$ for every $x$ in the domain of $f^{-1}$, and $f^{-1}(f(x)) = x$ for every $x$ in the domain of $f$. A question that asks for $f(f^{-1}(5))$ or $f^{-1}(f(5))$ is testing whether you recognize this property directly. The answer is just $5$, and you never need either formula.

The same idea hides inside questions that hand you two functions that are inverses of each other. If $f(x) = 5x - 2$ and $g(x) = \dfrac{x+2}{5}$, then $g(f(7))$ must equal $7$. You can still compute it step by step as a check: $f(7) = 33$ and $g(33) = \dfrac{35}{5} = 7$.

Evaluating an inverse at a point without finding its formula

To find $f^{-1}(21)$ for $f(x) = 4x + 9$, you can build the inverse $f^{-1}(x) = \dfrac{x-9}{4}$ and plug in $21$ to get $3$. But you can also ask: which input does $f$ send to $21$? Solve $4x + 9 = 21$ to get $x = 3$. Both routes give the same answer. The second route is usually quicker, and it is the only sensible route for functions like $2x^3 - 1$, where writing the inverse takes extra work.

Cube example. Let $f(x) = 2x^3 - 1$. Find $f^{-1}(15)$. Solve $2x^3 - 1 = 15$, so $x^3 = 8$ and $x = 2$. Stop there. Reporting $8$ (the value of $x^3$) is a classic slip.

Restricted domains

A function like $x^2$ is not one-to-one over all real numbers, because $2$ and $-2$ both give $4$. Restricting the domain to $x \ge 0$ (or $x \le 0$) makes an inverse possible. The restriction carries through to the inverse: the sign in front of the square root in the inverse formula is chosen to match the piece the original function came from. If the original domain is $x \ge 0$, the inverse takes the positive root. If it is $x \le 0$, the inverse takes the negative root.

Worked Example. Let $f(x) = x^2 - 5$, restricted to $x \ge 0$. Find $f^{-1}(11)$.

The restriction to $x \ge 0$ means the inverse takes the positive square root:

$f^{-1}(x) = \sqrt{x+5}$ on this restricted domain

$f^{-1}(11) = \sqrt{16} = 4$

If the same function had been restricted to $x \le 0$, the answer would be $-4$. Always check the restriction before you commit to a sign.

Where this appears as a model

Inverses show up whenever a word problem asks you to convert a relationship back the other way. A Celsius-to-Fahrenheit conversion $F(C) = \tfrac{9}{5}C + 32$ gives $F(20) = 68$. Its inverse, $C(F) = \tfrac{5}{9}(F - 32)$, recovers $C(68) = 20$, confirming the two functions undo each other. When a problem gives you an output and asks for the input, you are using the inverse, whether or not the word appears.

Strategy checklist

• Domain and range of an inverse: swap them. Domain of $f^{-1}$ is the range of $f$.
• Composition of a function with its own inverse: the answer is the input. No computation needed.
• "Find $f^{-1}(k)$": solve $f(x) = k$. It is often the fastest route.
• Finding a formula: swap $x$ and $y$, solve for $y$, then check that the operations were reversed in the opposite order.
• Squares: read the domain restriction to pick the sign of the root.
• Common errors: reporting $f(k)$ instead of $f^{-1}(k)$; dropping a sign when moving a term; stopping at $x^3$ or $x^2$ before taking the root; giving the domain of $f$ instead of the domain of $f^{-1}$.