Solving a Composite Equation gf(x) = k

Forming a composite that contains a square and setting it to a value produces a squared equation, so square-rooting gives two solutions. This tests both composite formation and remembering the $\pm$.

The functions $f$ and $g$ are defined by $f(x) = 3x - 1$ and $g(x) = x^2$. Solve the equation $gf(x) = 16$. [5]

The working

Step 1, form $gf(x) = g\big(f(x)\big)$ (apply $f$ first, then $g$ squares it): $gf(x) = (3x - 1)^2 \quad \text{(M1, A1)}$

Step 2, set equal to $16$: $(3x - 1)^2 = 16 \quad \text{(M1)}$

Step 3, square-root both sides (remember the $\pm$): $3x - 1 = \pm 4$

Step 4, solve each case: $3x - 1 = 4 \Rightarrow 3x = 5 \Rightarrow x = \frac{5}{3}$ $3x - 1 = -4 \Rightarrow 3x = -3 \Rightarrow x = -1 \quad \text{(A1, A1)}$

So $x = \frac{5}{3}$ or $x = -1$.

Where the marks are won and lost

• Composite order: $gf$ means $f$ first, so $g$ squares $(3x - 1)$, giving $(3x - 1)^2$, not $3x^2 - 1$.
• The $\pm$ on square-rooting is essential: $\sqrt{16} = \pm 4$ leads to two equations. Taking only $+4$ loses $x = -1$.
• Both solutions are valid (no domain restriction rules either out).

Common mistakes

• Computing $fg$ instead of $gf$.
• Expanding $(3x-1)^2$ and forming a quadratic (valid but slower; the $\pm$ route is cleaner) then mishandling it.
• Forgetting the negative square root.

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