WAEC 2022 · Paper 2 · Q3

(a)
Two functions $f$ and $g$ are defined on the set of real numbers, $\mathbb{R}$ , by $f : x \to x^2 + 2$ and $g : x \to \dfrac{1}{x + 2}$ , $x \ne -2$ . Find the domain of $(g \circ f)^{-1}$ .
Show the answer
$\{x : 0 < x \le \frac14\}$

Worked solution (try it first)

$f$ acts first: $g \circ f(x) = g(x^2 + 2)$
$= \dfrac{1}{x^2 + 4}$ .
Write $y = \dfrac{1}{x^2 + 4}$ .
Turn both sides over: $x^2 + 4 = \dfrac1y$ .
Take 4 from both sides: $x^2 = \dfrac{1 - 4y}{y}$ , so $x = \pm\sqrt{\dfrac{1 - 4y}{y}}$ and $(g \circ f)^{-1}(x) = \pm\sqrt{\dfrac{1 - 4x}{x}}$ .
The inverse needs $x \ne 0$ and $\dfrac{1 - 4x}{x} \ge 0$ .
The top and bottom can't both be negative here, so both are positive: $0 < x \le \frac14$ .
So the domain is $\left\{x : 0 < x \le \frac14\right\}$ .
It is also the range of $g \circ f$ : $x^2 + 4 \ge 4$ , so $0 < \dfrac{1}{x^2 + 4} \le \frac14$ .