Showing a Function Is Self-Inverse

A self-inverse function is its own inverse: $f^{-1}(x) = f(x)$. To prove it, you find the inverse the usual way and show the result is identical to the original. Rational functions are the common examples.

The function $f$ is defined by $f(x) = \dfrac{x + 2}{x - 1}$ for $x \neq 1$. Show that $f$ is self-inverse. [4]

The working

Step 1, set $y = f(x)$ and make $x$ the subject: $y = \frac{x + 2}{x - 1}$

Multiply up: $y(x - 1) = x + 2 \;\Rightarrow\; yx - y = x + 2 \quad \text{(M1)}$

Step 2, collect the $x$-terms on one side: $yx - x = y + 2 \;\Rightarrow\; x(y - 1) = y + 2 \quad \text{(M1)}$

Step 3, solve for $x$: $x = \frac{y + 2}{y - 1} \quad \text{(A1)}$

Step 4, rename ($y \to x$) to get the inverse: $f^{-1}(x) = \frac{x + 2}{x - 1} = f(x) \quad \text{(A1)}$

Since $f^{-1}(x)$ is identical to $f(x)$, the function is self-inverse. $\blacksquare$

Where the marks are won and lost

• The key algebra is collecting all $x$-terms on one side and factoring out $x$, that’s what isolates it in a rational inverse.
• The conclusion must be explicit: state that $f^{-1}(x) = f(x)$, therefore self-inverse. Reaching the expression without the concluding sentence can cost the final mark.
• A self-inverse function satisfies $ff(x) = x$; you could alternatively verify by computing the composite, but the swap-and-rearrange route is cleaner here.

Common mistakes

• Sign errors moving terms across ($yx - y = x + 2$).
• Failing to factor out $x$ from $yx - x$.
• Stopping at the expression without stating the self-inverse conclusion.

Full method: Inverse Functions notes. Topic home: Functions pillar.