Range of a Composite Function on an Interval

This question chains two skills: forming a composite function, then finding its range over an interval, which means weighing the endpoints against any turning point inside the interval, exactly as with a restricted-domain quadratic.

The functions $f$ and $g$ are defined by $f(x) = x^2$ and $g(x) = x - 3$. Find the range of the composite function $gf$ for the domain $-2 \le x \le 2$. [5]

The working

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

Step 2, find the values of $x^2$ on $[-2, 2]$. The square is smallest at $x = 0$ (giving $0$) and largest at the endpoints $x = \pm 2$ (giving $4$): $0 \le x^2 \le 4 \quad \text{(M1)}$

Step 3, apply the "$-3$" to get the range of $gf$: $0 - 3 \le x^2 - 3 \le 4 - 3 \;\Rightarrow\; -3 \le gf(x) \le 1 \quad \text{(A1, A1)}$

Where the marks are won and lost

• Composite order: $gf$ means $f$ first, so $g$ acts on $x^2$, giving $x^2 - 3$, not $(x - 3)^2$.
• The minimum comes from $x^2 = 0$ at $x = 0$, which is inside $[-2, 2]$. If the interval didn’t include $0$, the minimum would move to the nearer endpoint, always check whether the turning point is in range.
• Both endpoints give $x^2 = 4$ here (symmetry), so the maximum is $4 - 3 = 1$.

Common mistakes

• Computing $fg$ instead of $gf$ (wrong order).
• Assuming the endpoints give both extremes and missing the minimum at $x = 0$.
• Reading the range as $-3 \le gf \le 1$ but writing it back-to-front.

Full method: Composite Functions notes. See also Domain & Range. Topic home: Functions pillar.