NECO 2023 · Paper 1 · Q19

Let $f$ and $g$ on the set of real numbers be defined by $f(x) = 3x^2 - 4$ and $g(x) = 2x + 1$ . Find $fg(x)$ .

Worked solution (try it first)

$fg(x)$ means $f(g(x))$ : $g$ acts first.
So $fg(x) = f(2x + 1)$ .
Replace $x$ in $f$ by $(2x + 1)$ : $3(2x + 1)^2 - 4$ .
Expand the square: $(2x + 1)^2 = 4x^2 + 4x + 1$ , so $3(2x + 1)^2 = 12x^2 + 12x + 3$ .
Take away 4: $fg(x) = 12x^2 + 12x - 1$ , option D.