What Does It Mean for a Function to Have an Inverse?

A function has an inverse function if its effect can be reversed: the inverse takes each output of the original function back to its unique input. An inverse exists only when the original function is one-to-one on its stated domain.

The Reasoning

If $f$ maps $x$ to $y$, then its inverse, written $f^{-1}$, maps $y$ back to $x$:

$$f(x)=y \quad \Longleftrightarrow \quad f^{-1}(y)=x.$$

For this reversal to define a function, every output of $f$ must come from exactly one input. Graphically, this is checked using the horizontal line test: every horizontal line must intersect the graph at most once.

This condition is also called injectivity. If two different inputs produce the same output, reversing that output would produce two possible inputs, so the reverse relation would fail the definition of a function. A strictly increasing or strictly decreasing function is one-to-one on its domain.

The domain and range exchange roles:

Therefore, the graphs of $f$ and $f^{-1}$ are reflections in the line $y=x$.

For example, let $f(x)=2x+3$. Write $y=2x+3$, interchange $x$ and , and solve:

$$x=2y+3 \quad \Rightarrow \quad y=\frac{x-3}{2}.$$

Hence, $f^{-1}(x)=\frac{x-3}{2}$. Verification uses composition:

$$f^{-1}(f(x))=\frac{(2x+3)-3}{2}=x.$$

A common misconception is that $f^{-1}(x)=\frac{1}{f(x)}$. This is false: $$ denotes the inverse, not the reciprocal. Also, has no inverse on all real numbers because ; restricting its domain to makes it one-to-one. On this restricted domain, solving gives , so . Its domain is , exactly the range of the restricted original function.

IB Exam Technique

This is Mathematics AA SL 2.2 and may be tested on Paper 1 or Paper 2. State any domain restriction, swap domain and range correctly, and verify using composition or reflection in $y=x$. For a Find question, show the algebraic stages rather than only the final formula. If asked to verify, check both $f^{-1}(f(x))=x$ and $$ on the appropriate domains.