WAEC 2009 · Paper 2 · Q2

Functions

Given that $f(x) = \dfrac{x}{2}$ , $g(x) = \dfrac{5x}{x^2 - 4}$ , $x \ne \pm 2$ , and $h(x) = (x - 1)^{\frac12}$ , where $x \in \mathbb R$ , the set of real numbers, find:

(a)
$g \circ f(x)$ ;
(b)
the inverse of $h(x)$ .

Worked solution (try it first)

(a)

$g \circ f(x)$ means $g(f(x))$ : put $f(x) = \frac{x}{2}$ in place of $x$ in $g$ .
$g\left(\frac{x}{2}\right) = \dfrac{5 \cdot \frac{x}{2}}{\left(\frac{x}{2}\right)^2 - 4}$
$= \dfrac{\frac{5x}{2}}{\frac{x^2}{4} - 4}$ .
Multiply the top and bottom by 4: $\dfrac{10x}{x^2 - 16}$ .
So $g \circ f(x) = \dfrac{10x}{x^2 - 16}$ , $x \ne \pm 4$ .

(b)

Let $y = (x - 1)^{\frac12}$ and make $x$ the subject.
Square both sides: $y^2 = x - 1$ , so $x = y^2 + 1$ .
Swap the letters: $h^{-1}(x) = x^2 + 1$ (for $x \ge 0$ ).

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