Problem Statement
Let $f(x) = 2x - 3$ and $g(x) = \frac{1}{\sqrt{x + 2}}$ for $x > -2$.
**(a)** [2 marks] Find $(g \circ f)(x)$ and state its domain.
**(b)** [4 marks] Find an expression for $f^{-1}(x)$ and solve the equation $(f^{-1} \circ f)(x) = f(x)$.
Verified Solution & Marking Scheme
Part (a) [M1]
$(g \circ f)(x) = g(f(x)) = \frac{1}{\sqrt{(2x - 3) + 2}} = \frac{1}{\sqrt{2x - 1}}$
Part (a) [A1]
Domain: require $2x - 1 > 0 \implies x > \frac{1}{2}$
Part (b) [M1]
Let $y = 2x - 3 \implies 2x = y + 3 \implies x = \frac{y + 3}{2}$. Hence $f^{-1}(x) = \frac{x + 3}{2}$
Part (b) [M1]
Note $(f^{-1} \circ f)(x) = x$ by definition of inverse
Part (b) [M1]
Set $x = f(x) \implies x = 2x - 3$
Part (b) [A1]
$x = 3$