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Mathematics: Analysis and Approaches

SL 2.5 Composite, identity and inverse functions

Composite functions chain one function into another, and an inverse function undoes a function so that composing the two gives the identity. Study guide, 2 practice questions with markschemes and 2 flashcards. Free.

Forming composite functions · Domain of a composite · Identity function · Finding an inverse algebraically · Verifying inverses by composition

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Study guide

Composition: functions in a chain

\((f \circ g)(x) = f(g(x))\) means "do \(g\) first, then \(f\)". The function written closest to \(x\) acts first. Picture two machines in a row: the output of \(g\) is fed straight into \(f\).

Order matters. With \(f(x) = 2x + 1\) and \(g(x) = x^2\):

These are different functions. At \(x = 1\): \(f(g(1)) = 3\) but \(g(f(1)) = 9\).

Domain of a composite

For \(f(g(x))\) to exist, two things must hold: \(x\) must be in the domain of \(g\), and \(g(x)\) must be in the domain of \(f\). Example: \(f(x) = x - 5\) and \(g(x) = \sqrt{x}\). Then \((g \circ f)(x) = \sqrt{x - 5}\), which needs \(x \ge 5\), even though \(f\) itself accepts every real number.

The identity function

The identity is \(I(x) = x\): it changes nothing. It plays the role that 1 plays in multiplication. A function and its inverse compose to give the identity:

\[f(f^{-1}(x)) = x \quad\text{and}\quad f^{-1}(f(x)) = x.\]

Finding an inverse algebraically

  1. Write \(y = f(x)\).
  2. Swap \(x\) and \(y\).
  3. Rearrange to make \(y\) the subject.
  4. Write \(f^{-1}(x) = \dots\) and state its domain (the range of \(f\)).

Worked example

Find the inverse of \(f(x) = \dfrac{x + 4}{3}\) and verify it.

A second example with a fraction: \(h(x) = \frac{2x}{x - 1}\), \(x \ne 1\). Swap: \(x = \frac{2y}{y - 1}\), so \(xy - x = 2y\), giving \(y(x - 2) = x\) and \(h^{-1}(x) = \frac{x}{x - 2}\), \(x \ne 2\). The excluded value is the horizontal asymptote of \(h\), which matches the idea that the range of \(h\) becomes the domain of \(h^{-1}\).

Solving with composites

Questions like "solve \(f(g(x)) = 7\)" just need you to form the composite and solve. Questions like "find \(f^{-1}(5)\)" can skip finding the inverse formula: solve \(f(x) = 5\) instead.

Common mistakes

Exam tips

Cheat sheet

Key definitions

Composite function

The function \(f \circ g\) defined by \((f \circ g)(x) = f(g(x))\), applying \(g\) first and then \(f\).

Identity function

The function \(I(x) = x\) that leaves every input unchanged.

Inverse function

The function \(f^{-1}\) satisfying \(f(f^{-1}(x)) = f^{-1}(f(x)) = x\).

Self-inverse function

A function that is its own inverse, so \(f(f(x)) = x\), for example \(f(x) = \frac{1}{x}\).

Domain of a composite

The inputs \(x\) in the domain of \(g\) for which \(g(x)\) lies in the domain of \(f\).

Practice questions

SLPaper 1Foundation

Question 1

Let \(f(x) = 2x - 3\) and \(g(x) = x^2 + 1\).

(a) [2]

Find \((g \circ f)(2)\).

Markscheme
  • 1 \(f(2) = 1\)
  • 1 \(g(1) = 2\)
(b) [2]

Find \((f \circ g)(x)\).

Markscheme
  • 1 Substitutes \(g\) into \(f\): \(2(x^2 + 1) - 3\)
  • 1 \(2x^2 - 1\)
(c) [2]

Solve \((f \circ g)(x) = 17\).

Markscheme
  • 1 \(x^2 = 9\)
  • 1 \(x = \pm 3\)
SLPaper 1Challenging

Question 2

Let \(f(x) = 3x + k\) and \(g(x) = \frac{x}{2} - 1\), where \(k\) is a constant.

(a) [3]

Find \((f \circ g)(x)\) and \((g \circ f)(x)\) in terms of \(x\) and \(k\).

Markscheme
  • 1 \((f \circ g)(x) = 3\left(\frac{x}{2} - 1\right) + k = \frac{3x}{2} - 3 + k\)
  • 1 Substitutes \(f\) into \(g\): \(\frac{3x + k}{2} - 1\)
  • 1 \((g \circ f)(x) = \frac{3x}{2} + \frac{k}{2} - 1\)
(b) [2]

Find the value of \(k\) for which \((f \circ g)(x) = (g \circ f)(x)\) for all \(x\).

Markscheme
  • 1 Equates constant terms: \(k - 3 = \frac{k}{2} - 1\)
  • 1 \(k = 4\)
(c) [2]

Using this value of \(k\), find \(f^{-1}(x)\).

Markscheme
  • 1 Rearranges \(y = 3x + 4\), e.g. \(x = \frac{y - 4}{3}\)
  • 1 \(f^{-1}(x) = \frac{x - 4}{3}\)

Key facts

What does \((f \circ g)(x)\) mean, and which function acts first?

\((f \circ g)(x) = f(g(x))\). Apply \(g\) first, then \(f\). In general \(f \circ g \ne g \circ f\).

Steps to find an inverse function \(f^{-1}(x)\)

Write \(y = f(x)\), swap \(x\) and \(y\), rearrange to make \(y\) the subject. Check: \((f \circ f^{-1})(x) = x\).