Composite Functions and Solving fg(x) = k

Composite functions test one idea: the inner function acts first. Write $fg(x)$ as $f(g(x))$, substitute the whole of $g$ into $f$, and the algebra follows. The examiner usually adds a “solve” part to check you can then treat the result as an ordinary equation.

The functions $f$ and $g$ are defined by $f(x) = 2x + 3$ and $g(x) = x^2 - 1$. (i) Find $fg(x)$ and $gf(x)$, simplifying each. [4] (ii) Solve the equation $fg(x) = 21$. [2]

The working

(i) $fg(x) = f\big(g(x)\big)$: put $g(x) = x^2 - 1$ into $f$ wherever $x$ appears: $fg(x) = 2(x^2 - 1) + 3 = 2x^2 - 2 + 3 = 2x^2 + 1 \quad \text{(M1, A1)}$

$gf(x) = g\big(f(x)\big)$: put $f(x) = 2x + 3$ into $g$: $gf(x) = (2x + 3)^2 - 1 = 4x^2 + 12x + 9 - 1 = 4x^2 + 12x + 8 \quad \text{(M1, A1)}$

The two are different, order matters.

(ii) Solve $fg(x) = 21$ using the simplified $fg(x) = 2x^2 + 1$: $2x^2 + 1 = 21 \;\Rightarrow\; 2x^2 = 20 \;\Rightarrow\; x^2 = 10 \;\Rightarrow\; x = \pm\sqrt{10} \quad \text{(M1, A1)}$

Both roots are valid ($g$ and $f$ are defined for all real $x$), so $x = \sqrt{10}$ or $x = -\sqrt{10}$.

Where the marks are won and lost

• The substitution direction is the whole topic. $fg = f(g)$, so $g$ goes inside $f$. Writing $g(f)$ for $fg$ loses both marks in (i).
• Expanding $(2x + 3)^2$ needs the middle term $12x$. Dropping it gives $4x^2 + 8$ and a wrong $gf$.
• In (ii), $x^2 = 10$ has two roots. Giving only $+\sqrt{10}$ concedes the accuracy mark unless a domain rules the negative out (here it does not).

Common mistakes

• Computing $gf$ when $fg$ was asked (reversed order).
• Expanding $(2x+3)^2$ as $4x^2 + 9$.
• Forgetting the $\pm$ when square-rooting.

Full method: Composite Functions notes. Topic home: Functions pillar.