What Is a Self-Inverse Function?

A self-inverse function is a function that is equal to its own inverse. Applying the function twice returns the original input, so $f^{-1}=f$ and $f(f(x))=x$.

The Reasoning

For a function $f$ and its inverse $f^{-1}$, the defining relationship is

$$f^{-1}(f(x))=x.$$

If $f$ is self-inverse, then $f^{-1}$ can be replaced by $f$:

$$f(f(x))=x.$$

Equivalently, using the identity function $I(x)=x$,

$$(f\circ f)(x)=I(x).$$

Consider $f(x)=3-x$, with domain and range $\mathbb{R}$. Composing the function with itself gives

$$f(f(x))=f(3-x)=3-(3-x)=x.$$

Therefore, $f^{-1}(x)=f(x)=3-x$, so the function is self-inverse.

Another example is $g(x)=\frac{1}{x}$, defined on $\mathbb{R}\setminus\{0\}$. Then . The exclusion of matters because the formula is undefined there; self-inverse claims must use a domain on which both compositions are valid.

Graphically, the graphs of a function and its inverse are reflections in the line $y=x$. A self-inverse function is unchanged by this reflection, meaning its graph is symmetric about $y=x$.

A function must be one-to-one to have an inverse function, and its domain and range must be handled carefully. For example, $f(x)=x^2$ on $\mathbb{R}$ is not one-to-one, so it cannot be self-inverse.

The composition test is usually strongest because it proves both directions at once. However, checking only a few numerical inputs is not a proof; the simplification must hold algebraically for every permitted input. Also confirm that every output remains in the stated domain.

A common misconception is that $f^{-1}(x)=\frac{1}{f(x)}$. This is false: $$ denotes the inverse function, not the reciprocal.

Exam Technique

This is AA HL content in AHL 2.14 and may be tested analytically on Paper 1 or with graphical support on Paper 2. For the command term verify, show the composition $f(f(x))=x$; simply stating that the function is self-inverse is insufficient.