IGCSE · Practice questions · Composite functions

Inside first, outside second.

Six original Cambridge-style questions on composite functions: evaluating and building composites, solving equations exactly without a calculator, and finding domains and ranges, ending with a chain of temperature scales.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Inside out, then simplify.

Order: \(fg(x) = f\big(g(x)\big)\), so \(g\) acts first; \(f^{2}(x) = f\big(f(x)\big)\).
Substitute the whole inner expression in brackets wherever \(x\) appears in the outer function.
Solve: write the composite as one expression first; take both square roots; use \(\ln\) to undo \(\mathrm{e}\).
Exist? \(gf\) needs every output of \(f\) to be an allowed input of \(g\); restrict the domain of \(f\) if not.
Range: feed the range of \(f\) (not its domain) into \(g\).

Recall

[2 marks]

The functions f and g are defined by \(f(x) = 4x - 1\) and \(g(x) = x^{2} + 2\) for \(x \in \mathbb{R}\). Find (a) \(fg(3)\) (b) \(gf(3)\).

(a) \(g(3) = 11\), so \(fg(3) = f(11) = 43\). ✓
(b) \(f(3) = 11\), so \(gf(3) = g(11) = 123\). ✓

Calculation

[4 marks]

The functions f and g are defined by \(f(x) = 2x + 5\) for \(x \in \mathbb{R}\) and \(g(x) = \dfrac{1}{x - 1}\) for \(x \ne 1\).

(a) Find \(fg(x)\), giving your answer as a single fraction.

(b) Find \(gf(x)\), stating the value of \(x\) for which it is not defined.

(a) \(fg(x) = \dfrac{2}{x - 1} + 5 = \dfrac{2 + 5(x - 1)}{x - 1} = \dfrac{5x - 3}{x - 1}\). ✓
(b) \(gf(x) = \dfrac{1}{(2x + 5) - 1} = \dfrac{1}{2x + 4}\). ✓
It is not defined when \(2x + 4 = 0\), that is \(x = -2\). ✓
(c) \(f^{2}(x) = 2(2x + 5) + 5 = 4x + 15\). ✓

Calculation

[4 marks]

The functions f and g are defined by \(f(x) = \mathrm{e}^{x}\) and \(g(x) = 3x + 2\) for \(x \in \mathbb{R}\). Giving exact answers, solve

(a) \(gf(x) = 14\)

(b) \(fg(x) = 4\).

(a) \(3\mathrm{e}^{x} + 2 = 14 \Rightarrow \mathrm{e}^{x} = 4\). ✓
\(x = \ln 4\) (or \(2\ln 2\)). ✓
(b) \(\mathrm{e}^{3x + 2} = 4 \Rightarrow 3x + 2 = \ln 4\). ✓
\(x = \dfrac{\ln 4 - 2}{3}\). ✓

Analysis

[5 marks]

The functions f and g are defined by \(f(x) = x^{2} - 2x\) and \(g(x) = 2x + 3\) for \(x \in \mathbb{R}\).

(a) Find \(fg(x)\) and \(gf(x)\).

(b) Solve \(fg(x) = gf(x)\).

(a) \(fg(x) = (2x + 3)^{2} - 2(2x + 3) = 4x^{2} + 12x + 9 - 4x - 6 = 4x^{2} + 8x + 3\). ✓
\(gf(x) = 2(x^{2} - 2x) + 3 = 2x^{2} - 4x + 3\). ✓
(b) \(4x^{2} + 8x + 3 = 2x^{2} - 4x + 3 \Rightarrow 2x^{2} + 12x = 0 \Rightarrow 2x(x + 6) = 0\), so \(x = 0\) or \(x = -6\). ✓
(c) \(f(x) = (x - 1)^{2} - 1\), so the range of f is \(f(x) \ge -1\). ✓
\(g\) is increasing, so \(gf(x) \ge 2(-1) + 3\), giving \(gf(x) \ge 1\). ✓

Proof

[4 marks]

The functions f and g are defined by \(f(x) = \ln x\) for \(x > 0\) and \(g(x) = 5 - x\) for \(x \in \mathbb{R}\).

(a) Explain why the composite function \(fg\) does not exist.

(b) The domain of g is changed to \(x < k\) so that \(fg\) exists. State the greatest value of \(k\), find \(fg(x)\) and solve \(fg(x) = 0\).

(a) The range of g is all real numbers, including values such as \(g(6) = -1\), but f only accepts positive inputs (\(\ln(-1)\) is undefined). So the range of g is not within the domain of f. ✓
(b) We need \(5 - x > 0\), so \(x < 5\): the greatest value is \(k = 5\). ✓
\(fg(x) = \ln(5 - x)\). ✓
\(\ln(5 - x) = 0 \Rightarrow 5 - x = 1 \Rightarrow x = 4\), which satisfies \(x < 5\). ✓

Analysis

[5 marks]

A temperature of \(x\) kelvin is converted to degrees Celsius by \(g(x) = x - 273\), and a temperature of \(x\) degrees Celsius is converted to degrees Fahrenheit by \(f(x) = 1.8x + 32\).

(a) Find \(fg(x)\) and explain what it represents.

(b) Find the temperature in kelvin that is equal to 50 degrees Fahrenheit.

(d) Explain why \(gf\) has no sensible meaning here.

(a) \(fg(x) = 1.8(x - 273) + 32 = 1.8x - 459.4\). It converts kelvin directly to degrees Fahrenheit. ✓
(b) \(1.8x - 459.4 = 50 \Rightarrow 1.8x = 509.4 \Rightarrow x = 283\): 283 kelvin. ✓
(c) \(fg\) is increasing and \(fg(0) = -459.4\), so the range is \(fg(x) \ge -459.4\). ✓
(d) \(gf\) would convert Celsius to Fahrenheit and then subtract 273 as if the Fahrenheit value were in kelvin, which gives no real temperature scale. The order of the functions matters. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.