How Do You Find the Inverse of a Function With a Restricted Domain?

First ensure the restricted function is one-to-one. Then interchange $x$ and $y$, solve for $y$, and state the inverse's domain and range; the restriction determines which solution branch is valid.

The Method

A function has an inverse only if each output corresponds to exactly one input. Restricting its domain removes repeated outputs, allowing the function to pass the horizontal line test.

In IB Mathematics: Applications and Interpretation HL, this is the key purpose of a domain restriction: it turns a many-to-one relation into a one-to-one function. The inverse then exchanges the original domain and range.

Consider

$$f(x)=(x-2)^2+1, \qquad x\geq 2.$$

Without the restriction, inputs on both sides of $x=2$ produce the same outputs. The restriction $x\geq 2$ selects the increasing branch, making the function one-to-one.

1. Write the function using $y$:

$$y=(x-2)^2+1.$$

2. Interchange the variables:

$$x=(y-2)^2+1.$$

3. Solve for $y$:

$$(y-2)^2=x-1$$ $$y=2\pm\sqrt{x-1}.$$

The original domain is $x\geq 2$, so the inverse must have range $y\geq 2$. Therefore, select the positive branch:

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

The original range is $y\geq 1$, so this becomes the inverse's domain.

If the restriction had instead been $x\leq 2$, the decreasing branch would be selected, giving $f^{-1}(x)=2-\sqrt{x-1}$, with domain and range . Thus, the algebra produces two candidates, but the restriction identifies the actual inverse.

You can verify the result by checking that $f(f^{-1}(x))=x$ for $x\geq 1$, and $f^{-1}(f(x))=$ for . The graphs are reflections in .

A common misconception is that $f^{-1}(x)=\frac{1}{f(x)}$. This is incorrect: $$ reverses the function, while is its reciprocal.

Exam Technique

This is HL-only content from AHL 2.7: Composite functions and inverse functions, assuming a Paper 1 short-response style. Show the variable interchange, justify the selected branch using the restriction, and state the inverse's domain. An unsupported $\pm$ answer is incomplete. A GDC can confirm the reflection or compositions, but it does not replace the branch justification.