Finding an Inverse Function and Its Domain

Finding an inverse is a fixed three-step routine: write $y =$, swap the roles by making $x$ the subject, then rename. The part students forget is the domain of the inverse, which equals the range of the original.

The function $f$ is defined by $f(x) = \dfrac{3}{x - 2}$ for $x > 2$. (i) Find $f^{-1}(x)$. [3] (ii) State the domain of $f^{-1}$. [2]

The working

(i) Set $y = f(x)$ and make $x$ the subject: $y = \frac{3}{x - 2}$

Multiply up, then isolate $x$: $y(x - 2) = 3 \;\Rightarrow\; x - 2 = \frac{3}{y} \;\Rightarrow\; x = 2 + \frac{3}{y} \quad \text{(M1, M1)}$

Rename ($y \to x$) to write the inverse as a function of $x$: $f^{-1}(x) = 2 + \frac{3}{x} \quad \text{(A1)}$

(ii) The domain of $f^{-1}$ is the range of $f$. For $x > 2$, the denominator $x - 2$ is positive and can be any positive number, so $\frac{3}{x-2}$ takes any positive value: $\text{range of } f: \ f(x) > 0 \;\Rightarrow\; \text{domain of } f^{-1}: \ x > 0 \quad \text{(M1, A1)}$

Where the marks are won and lost

• Making $x$ the subject is the method. Multiplying by $(x - 2)$ first, then dividing by $y$, keeps the algebra clean, trying to invert the fraction in one leap invites slips.
• The domain of the inverse is a separate two-mark part. It comes from the range of $f$, which here is $f(x) > 0$ because the domain was restricted to $x > 2$ (denominator positive).
• Rename at the end. Leaving the answer as $x = 2 + \frac{3}{y}$ is not yet "$f^{-1}(x)$".

Common mistakes

• Giving the domain of $f^{-1}$ as $x > 2$ (that is $f$‘s domain, not its range).
• Forgetting to swap back to $x$, so the inverse is left in terms of $y$.
• Sign or reciprocal errors when isolating $x$ from $y(x - 2) = 3$.

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