Problem Statement
Let $f(x) = \frac{2x + 1}{x - 3}, \quad x \neq 3$.
(a) Find $f^{-1}(x)$, stating its domain. [4]
(b) Solve $(f \circ f)(x) = x$. [2]
Verified Solution & Marking Scheme
(a) Interchange x and y and solve for y
Let $y = \frac{2x + 1}{x - 3}$. Interchange $x$ and $y$:
$$x = \frac{2y + 1}{y - 3} \implies x(y - 3) = 2y + 1$$
$$xy - 3x = 2y + 1 \implies xy - 2y = 3x + 1$$
$$y(x - 2) = 3x + 1 \implies y = \frac{3x + 1}{x - 2}$$
Domain of $f^{-1}$: $\{x \in \mathbb{R} : x \neq 2\}$.
(b) Solve (f ∘ f)(x) = x
Note that $(f \circ f)(x) = x \iff f(x) = f^{-1}(x)$:
$$\frac{2x + 1}{x - 3} = \frac{3x + 1}{x - 2}$$
Cross-multiplying:
$$(2x + 1)(x - 2) = (3x + 1)(x - 3)$$
$$2x^2 - 4x + x - 2 = 3x^2 - 9x + x - 3$$
$$2x^2 - 3x - 2 = 3x^2 - 8x - 3$$
$$x^2 - 5x - 1 = 0$$
Using quadratic formula:
$$x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-1)}}{2} = \frac{5 \pm \sqrt{29}}{2}$$