Inverse Functions

The inverse function $f^{-1}$ undoes $f$: if $f$ sends 3 to 11, $f^{-1}$ sends 11 back to 3. It exists only when $f$ is one-one, and it is not the reciprocal, $f^{-1}(x) \ne \frac{1}{f(x)}$, a confusion 0606 punishes.

The four-step routine

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

1. Set $y = \frac{2x + 1}{x - 3}$
2. Make $x$ the subject: $y(x - 3) = 2x + 1 \to xy - 3y = 2x + 1 \to x(y - 2) = 3y + 1 \to x = \frac{3y + 1}{y - 2}$
3. Swap letters: $f^{-1}(x) = \frac{3x + 1}{x - 2}$
4. State the domain if asked: $x \ne 2$

Each rearrangement line is visible method, multiply up, collect $x$-terms, factorise $x$ out. Cramming them into one line risks the M marks that survive an algebra slip.

The square-root case: justify the sign

When inverting anything squared, a $\pm$ appears and one sign must be chosen with a written reason:

$f(x) = (x - 2)^2 + 3$ for $x \ge 2$: $y - 3 = (x - 2)^2 \to x - 2 = +\sqrt{y - 3}$ (positive root, since $x \ge 2$) $\to f^{-1}(x) = 2 + \sqrt{x - 3}$

That bracketed reason is the most-missed mark in this subtopic. No reason, no mark, even with the right answer.

Domain and range swap

Domain of $f^{-1} =$ range of $f$; range of $f^{-1} =$ domain of $f$. So “state the domain of $f^{-1}$” is secretly a range question about $f$, sketch $f$ first.

The graph: reflection in $y = x$

$y = f^{-1}(x)$ is $y = f(x)$ reflected in the line $y = x$. Sketch questions award a B mark for drawing the dashed $y = x$ line and showing the mirror symmetry; intersections of $f$ and $f^{-1}$ lie on $y = x$ (so solving $f(x) = f^{-1}(x)$ can collapse to solving $f(x) = x$, a slick “hence” route 0606 likes).

Common mistakes

• $f^{-1}$ computed as $\frac{1}{f}$
• $\pm$ unresolved, or resolved silently
• Domain of $f^{-1}$ stated as the domain of $f$ (it’s the range)
• Verification skipped: $ff^{-1}(x) = x$ is a ten-second self-check that catches most slips

Full topic context: Functions notes · pairs with composite functions.