Home / IGCSE Maths 0580 / Algebra and graphs / Functions: notation, inverse and composite
Cambridge IGCSE · Topic 2.13
IGCSE Maths 0580 / Topic 2 / IGCSE

Input, rule, output.

Extended. A function is a machine: put a number in, apply the rule, get one number out. Run the machine backwards for the inverse, or feed one machine into another for a composite.

Full-screen labFunctions and differentiation lab → Topic 2All algebra and graphs lessons →
The key idea

\(f(a)\) means substitute \(a\) into the rule. \(f^{-1}\) undoes \(f\): write \(y = f(x)\) and rearrange to make \(x\) the subject. \(gf(x) = g(f(x))\): apply \(f\) FIRST, then \(g\).

x y −224 −224 (1, 3) (3, 1) f(x) = 2x + 1 y = f⁻¹(x) y = x
Fig. 1: f(x) = 2x + 1 and its inverse f⁻¹(x) = (x − 1)/2 are reflections of each other in the line y = x; f(1) = 3, so f⁻¹(3) = 1
Section 01

Function notation.

In the lab's functions tab, type inputs into the function machine and watch the mapping diagram fill in, each arrow joining an input to its output. Change the domain and see the range change with it.

Worked example Evaluating and solving
\(f(x) = 3x - 5\). Find (a) \(f(-2)\) (b) the value of \(x\) for which \(f(x) = 10\).
  1. (a) \(f(-2) = 3(-2) - 5 = -6 - 5 = -11\).
  2. (b) Set the rule equal to 10: \(3x - 5 = 10\).
  3. \(3x = 15\), so \(x = 5\).
Answer: (a) \(-11\)   (b) \(x = 5\)
Worked example Domain and range
\(h(x) = 2x^2 + 3\) has domain \(\{-2, -1, 0, 1, 2\}\). Find the range.
  1. \(h(-2) = 2(4) + 3 = 11\) and \(h(2) = 11\).
  2. \(h(-1) = 2(1) + 3 = 5\) and \(h(1) = 5\).
  3. \(h(0) = 3\).
  4. List each output once.
Answer: Range \(= \{3, 5, 11\}\). (With all real numbers as the domain, the range would be \(h(x) \ge 3\).)
Examiner trap

In \(2x^2\) only the \(x\) is squared: \(h(-2) = 2 \times (-2)^2 + 3 = 2 \times 4 + 3 = 11\), not \((2 \times -2)^2 + 3 = 19\).

Section 02

Inverse functions.

Open the inverse functions tab and drag the blue point along \(y = f(x)\). Its reflection in the line \(y = x\) traces out \(y = f^{-1}(x)\): every input and output swap places.

Worked example A linear inverse
\(f(x) = 3x - 5\). Find \(f^{-1}(x)\).
  1. Write \(y = 3x - 5\).
  2. Make \(x\) the subject: \(y + 5 = 3x\), so \(x = \dfrac{y + 5}{3}\).
  3. Write the answer as a function of \(x\).
Answer: \(f^{-1}(x) = \dfrac{x + 5}{3}\)
Worked example An inverse with a fraction
\(g(x) = \dfrac{3(x + 4)}{5}\). Find \(g^{-1}(x)\).
  1. Write \(y = \dfrac{3(x + 4)}{5}\).
  2. Multiply by 5: \(5y = 3x + 12\).
  3. Subtract 12 and divide by 3: \(x = \dfrac{5y - 12}{3}\).
  4. Check: \(g(1) = \dfrac{3 \times 5}{5} = 3\), and \(\dfrac{5 \times 3 - 12}{3} = 1\). It undoes \(g\).
Answer: \(g^{-1}(x) = \dfrac{5x - 12}{3}\)
Your turn
\(f(x) = 3x - 5\). Find \(f^{-1}(7)\) without finding \(f^{-1}(x)\).
Examiner trap

\(f^{-1}(x)\) is NOT \(\dfrac{1}{f(x)}\). The \(-1\) means 'inverse', not a power: for \(f(x) = 3x - 5\), \(f^{-1}(x) = \dfrac{x + 5}{3}\), not \(\dfrac{1}{3x - 5}\).

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

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\(f(x) = 3x - 5\). Find \(f(4)\).

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What is the domain of a function?

Quick check+10 XP

\(f(x) = x + 6\). What is \(f^{-1}(x)\)?

Quick check+10 XP

\(f(x) = 2x\) and \(g(x) = x + 1\). Find \(gf(3)\).

Section 03

Composite functions.

In the composite functions tab, chain two machines together and swap their order. The output of the first machine becomes the input of the second, and changing the order usually changes the answer.

Worked example Numbers through two functions
\(f(x) = 3x - 5\) and \(h(x) = 2x^2 + 3\). Find (a) \(hf(2)\) (b) \(fh(2)\).
  1. (a) \(hf(2) = h(f(2))\): work out \(f(2)\) first: \(6 - 5 = 1\).
  2. Then \(h(1) = 2 + 3 = 5\).
  3. (b) \(fh(2) = f(h(2))\): \(h(2) = 8 + 3 = 11\).
  4. Then \(f(11) = 33 - 5 = 28\).
Answer: (a) \(5\)   (b) \(28\): the order matters
Worked example An algebraic composite
\(f(x) = \dfrac{2}{x - 1}\) and \(g(x) = (x + 3)^2\). Find \(fg(x)\), giving your answer as a fraction in its simplest form.
  1. \(fg(x) = f(g(x))\): replace every \(x\) in \(f\) with \((x + 3)^2\).
  2. \(fg(x) = \dfrac{2}{(x + 3)^2 - 1}\).
  3. Expand: \((x + 3)^2 = x^2 + 6x + 9\), so the denominator is \(x^2 + 6x + 9 - 1\).
Answer: \(fg(x) = \dfrac{2}{x^2 + 6x + 8}\)
Your turn
\(f(x) = 2x + 1\). Find \(ff(x)\) in its simplest form.
Examiner trap

\(gf(x)\) means \(f\) first, even though \(g\) is written first. It is also not a product: \(gf(x) \ne g(x) \times f(x)\).

Stage 2 · Exam

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\(g(x) = \dfrac{3(x + 4)}{5}\). Find \(g^{-1}(x)\).

Exam style+20 XP

\(f(x) = 3x - 5\) and \(h(x) = 2x^2 + 3\). Find \(fh(-1)\).

Exam style+20 XP

\(f(x) = \dfrac{2}{x - 1}\) and \(g(x) = (x + 3)^2\). Find \(fg(x)\) in its simplest form.

Exam style+20 XP

\(f(x) = 3x - 5\). Solve \(f^{-1}(x) = 4\).

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Six original Cambridge-style questions
Evaluating and solving with function notation, range from a domain, inverses, composites in both orders and a temperature-conversion problem.