The function f is defined by $f(x) = \frac{3x - 7}{x - 2} \quad (x \in \mathbb{R}, x \neq 2)$ (a) Find $f^{-1}(7)$ (b) Show that $f(f(x)) = \frac{ax + b}{x - 3}$ where a and b are integers to be found. - Edexcel - A-Level Maths Pure - Question 6 - 2020 - Paper 1

We start by substituting $f(x)$ into itself:

$f(f(x)) = f\left(\frac{3x - 7}{x - 2}\right)$

Substituting into the function:

$= \frac{3\left(\frac{3x - 7}{x - 2}\right) - 7}{\left(\frac{3x - 7}{x - 2}\right) - 2}$

Calculating the numerator:

$= \frac{\frac{9x - 21}{x - 2} - 7}{\frac{3x - 7 - 2(x - 2)}{x - 2}}$

First, simplify the numerator:

$= \frac{\frac{9x - 21 - 7(x - 2)}{x - 2}}{\frac{3x - 7 - 2(x - 2)}{x - 2}}$

Expanding and combining terms in the numerator:

$= \frac{9x - 21 - 7x + 14}{3x - 7 - 2x + 4}$

$= \frac{2x - 7}{x - 3}$

Thus, $a = 2$ and $b = -7$, and we verify:

$f(f(x)) = \frac{2x - 7}{x - 3}$