2.5 Composite and inverse functions (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

Composite functions

A composite function applies one function and then another. $(f \circ g)(x) = f(g(x))$ : apply $g$ first, then $f$ to the result.

Numerically, work from the inside out: with $f(x) = 3x - 1$ , $g(x) = x^2 + 2$ , $(f \circ g)(2) = f(6) = 17$ .
Algebraically, replace every $x$ in the outer function by the whole inner expression: $(g \circ f)(x) = (3x - 1)^2 + 2 = 9x^2 - 6x + 3$ .
Order matters: in general $f \circ g \ne g \circ f$ . Here $(f \circ g)(x) = 3x^2 + 5$ .

Key termscomposite function

Common mistake

$(f \circ g)(x)$ is not $f(x) \times g(x)$ . The circle means 'apply one after the other', not 'multiply'.

Exam tip

Read $f \circ g$ right to left: $g$ happens first.

Section 2

The domain of a composite function

For $f(g(x))$ to exist, $x$ must be in the domain of $g$ and $g(x)$ must be in the domain of $f$ .

With $f(x) = \sqrt{x + 5}$ , $x \ge -5$ , and $g(x) = x^2 - 9$ :

$(f \circ g)(x) = \sqrt{x^2 - 4}$ needs $x^2 - 4 \ge 0$ , so $x \le -2$ or $x \ge 2$ .
$(g \circ f)(x) = x - 4$ , but only for $x \ge -5$ , because $x$ must go into $f$ first.

Two functions are the same only if they have the same rule and the same domain, so $g \circ f$ is not the same as $x - 4$ on $\mathbb{R}$ .

Key termsdomain of a composite

Common mistake

Simplifying first and reading the domain from the simplified rule. Always carry the restriction from the inner function.

Section 3

Composite functions in context

Many processes happen in stages, and the order changes the result. Bank A deducts a 15 AED fee and then converts at 0.24 euros per dirham: with $f(x) = x - 15$ and $g(x) = 0.24x$ this is $(g \circ f)(x) = 0.24x - 3.6$ .

The other order, $(f \circ g)(x) = 0.24x - 15$ , would convert first and then take off 15 euros — a different model. When you form a composite in context, check the units at each stage.

Exam tip

Write each stage as its own function, then compose them in the order they happen: the first stage goes on the inside.

Section 4

The identity function and inverses

The identity function is $I(x) = x$ : it leaves every input unchanged.

An inverse undoes a function, so composing them in either order gives the identity: $(f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x.$

You can use this to check an inverse. With $f(x) = 3x - 1$ and $f^{-1}(x) = \frac{x + 1}{3}$ : $f(f^{-1}(x)) = 3 \cdot \frac{x + 1}{3} - 1 = x$ .

If $(h \circ h)(x) = x$ , then $h$ is its own inverse (self-inverse): $h^{-1} = h$ . An example is $h(x) = \frac{2x + 3}{x - 2}$ .

Key termsidentity functionself-inverse

Example

$h(4) = 5.5$ and $h(5.5) = 4$ , so $(h \circ h)(4) = 4$ .

Section 5

Finding an inverse function

An inverse function exists only when $f$ is one-to-one. To find $f^{-1}(x)$ :

Write $y = f(x)$ .
Interchange $x$ and $y$ .
Rearrange to make $y$ the subject; this is $f^{-1}(x)$ .

Example: $y = 0.24x - 3.6 \to x = 0.24y - 3.6 \to f^{-1}(x) = \frac{x + 3.6}{0.24}$ . The domain of $f^{-1}$ is the range of $f$ .

In context the inverse answers the reverse question: 'how many dirhams give 300 euros?' is $(g \circ f)^{-1}(300) = 1265$ .

Key termsone-to-oneinverse function

Common mistake

Leaving the answer in terms of $y$ . The final inverse must be written as a function of $x$ .

Must know

$(f \circ g)(x) = f(g(x))$ : $g$ first. Usually $f \circ g \ne g \circ f$ .
The domain of $f \circ g$ is restricted by the domain of $g$ and by what $f$ can accept.
Identity: $I(x) = x$ ; $(f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x$ .
Find $f^{-1}$ by interchanging $x$ and $y$ and rearranging; only one-to-one functions have inverses.
If $(h \circ h)(x) = x$ then $h$ is self-inverse.

That's the notes covered.

Carry on to the next subtopic.