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Cambridge IGCSE Mathematics — 0580 Extended

Topic 2.13: Algebra and graphs — Functions

Function notation, domain and range

\(f(x)\) means “the output of \(f\) when the input is \(x\)”. The domain is the set of allowed inputs; the range is the set of possible outputs. Mapping diagrams are allowed.

You are not expected to find the domain or range of a composite function.

Method

  1. To evaluate \(f(a)\), replace every \(x\) in the rule by \(a\) (use brackets if \(a\) is negative).
  2. Domain: exclude values that make a denominator zero (and, later, a square root negative).
  3. Range: ask what \(y\)-values the rule can actually produce. For \(y = x^2\), that is \(y \ge 0\).

If \(f(x) = 3x - 5\), find \(f(2)\). If \(h(x) = 2x^2 + 3\), find \(h(-1)\).

\(f(2) = 3(2) - 5 = 1\). \(h(-1) = 2(-1)^2 + 3 = 2(1) + 3 = 5\).

Mapping diagram sending 0, 1 and 2 to negative 5, negative 2 and 1 under f of x equals 3x minus 5
Each input has exactly one output: \(0 \mapsto -5\), \(1 \mapsto -2\), \(2 \mapsto 1\).

State the domain of \(\dfrac{1}{x-2}\) and the range of \(y = x^2\).

Number line with x equals 2 excluded, and range of x squared is y greater than or equal to 0
Division by zero is undefined, so \(x \ne 2\). Squares are never negative, so \(y \ge 0\).

Inverse functions

The inverse \(f^{-1}\) undoes \(f\). It is not the reciprocal \(1/f(x)\) unless \(f\) itself is of the form \(k/x\).

Method

  1. Write \(y = f(x)\).
  2. Rearrange to make \(x\) the subject.
  3. Replace \(y\) by \(x\) to write \(f^{-1}(x)\).
  4. Check: \(f\big(f^{-1}(x)\big) = x\) (and \(f^{-1}\big(f(x)\big) = x\)).

Find \(f^{-1}(x)\) if \(f(x) = 3x - 5\).

Inverse of 3x minus 5 is x plus 5 over 3, checked by composition
\(y = 3x - 5 \Rightarrow x = \dfrac{y+5}{3}\), so \(f^{-1}(x) = \dfrac{x+5}{3}\).

Composite functions

Cambridge writes \(gf(x) = g\big(f(x)\big)\): apply \(f\) first, then \(g\). In general \(fg \ne gf\).

Method

  1. Read the order from the right: \(gf\) means \(f\) first.
  2. Replace the input of the outer function by the whole inner expression, using a bracket.
  3. Simplify only as far as the question asks. Do not multiply \(g\) by \(f\).

Given \(f(x) = 3x + 2\) and \(g(x) = (3x + 5)^2\), find \(gf(x)\) and \(fg(x)\).

gf of x equals (9x plus 11) squared, while fg of x equals 3 times (3x plus 5) squared plus 2
\(gf(x) = (9x+11)^2\), but \(fg(x) = 3(3x+5)^2 + 2\).

Paper 2 (non-calculator)

\(fg \ne gf\) in general. Read the order carefully: \(gf\) means \(g\) after \(f\). Substituting a number is often safer than expanding: \(f(2) = 8\), then \(g(8) = (3\cdot 8 + 5)^2 = 29^2\) if you only need a value.

Try this

If \(f(x) = 2x + 1\), find \(f^{-1}(x)\) and \(f^{-1}(7)\).

Show answer
Answer

  1. Set \(y = 2x + 1\) and rearrange

    \[ y - 1 = 2x \Rightarrow x = \dfrac{y-1}{2} \]

  2. Write the inverse in \(x\)

    \[ f^{-1}(x) = \dfrac{x-1}{2} \]

  3. Evaluate at \(7\)

    \[ f^{-1}(7) = \dfrac{7-1}{2} = 3 \]

Exam Traps

  • \(gf(x) = g\big(f(x)\big)\), not \(g \times f\).
  • \(f^{-1}\) undoes \(f\). Do not take the reciprocal of \(f(x)\) unless \(f\) is a \(1/x\) type. \(f^{-1}\) is not \(1/f\).

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