3.2 - Composite and inverse functions

Learn to combine function rules in the right order and find a function that reverses a mapping. You will keep track of allowed inputs and choose the correct square-root branch when an inverse needs a domain restriction. This lesson is for Higher students.

The order of a composite function (Higher)

Two machines use the rules “double, then subtract $1$ ” and “square, then add $3$ ”. Starting with $0$ , can reversing their order really change the final answer? We will get $5$ one way and $4$ the other.

You already know

A function gives one output for each allowed input. In $f(t)$ , replace every input variable in the rule for $f$ by the whole expression $t$ , using brackets where needed.

A second machine can act on the first machine's output. Read a nested expression from the innermost brackets outwards.

Composite function

A composite function applies one function, then another. Pearson's notation is

fg(x)=f(g(x))

It means do $g$ first, then $f$ . It does not mean multiply the two outputs. This notation is not supplied on the Higher formulae sheet: learn it.

For the opening machines, let $f(x)=2x-1$ and $g(x)=x^2+3$ , both with all real inputs. When forming a rule, insert the complete inner expression into the outer rule before expanding.

Worked example 1: apply the rightmost function first

Question. For $f(x)=2x-1$ and $g(x)=x^2+3$ , find $fg(x)$ and $fg(0)$ .

Plan. Apply $g$ , then substitute its output into $f$ .

\begin{aligned} fg(x)&=f(x^2+3) &&\text{apply }g\text{ first}\\ &=2(x^2+3)-1 &&\text{replace the whole input of }f\\ &=2x^2+5 &&\text{expand and simplify}\\ fg(0)&=5 &&\text{substitute zero} \end{aligned}

Why double the whole bracket? The input to $f$ is $x^2+3$ , including its constant.

Answer. $fg(x)=2x^2+5$ , and $fg(0)=5$ .

Check. Directly, $0$ goes to $g(0)=3$ , then $f(3)=5$ .

Now change only the order. The same functions and same starting input will give a different answer.

Worked example 2: reverse the order

Question. For $f(x)=2x-1$ and $g(x)=x^2+3$ , find $gf(x)$ and $gf(0)$ .

Plan. Apply $f$ , then substitute its output into $g$ .

\begin{aligned} gf(x)&=g(2x-1) &&\text{apply }f\text{ first}\\ &=(2x-1)^2+3 &&\text{square the whole input}\\ &=4x^2-4x+4 &&\text{expand and simplify}\\ gf(0)&=4 &&\text{substitute zero} \end{aligned}

Why retain the bracket before squaring? The output of $f$ is the entire expression $2x-1$ .

Answer. $gf(x)=4x^2-4x+4$ , and $gf(0)=4$ .

Check. Directly, $f(0)=-1$ , then $g(-1)=(-1)^2+3=4$ .

The opening puzzle shows that order matters: $fg$ and $gf$ are usually different functions, although some special pairs do give the same rule.

Watch out

The wrong line $fg(x)=(2x-1)^2+3$ applies $f$ first. That is $gf(x)$ . For $fg$ , start with $g$ : $fg(x)=2(x^2+3)-1$ .

Write the nested form first when unsure of the order. It makes the first operation visible.

Allowed inputs of a composite (Higher)

A composite is a journey through two domains. The starting input must be allowed by the first function; its output must then be allowed by the second. A restriction on an intermediate value can therefore exclude a different starting input.

For $fg$ , first check the domain of $g$ , then require $g(x)$ to lie in the domain of $f$ . The Higher formulae sheet does not supply these domain rules. Use the original functions, even if later simplification hides a restriction.

Worked example 3: an intermediate value is forbidden

Question. Let $f(x)=\frac{1}{x-2}$ and $g(x)=x+5$ , each with its largest real domain. Find $fg(x)$ and its domain.

Plan. Substitute $g(x)$ into $f$ , then exclude an intermediate input of $2$ .

\begin{aligned} fg(x)&=\frac{1}{(x+5)-2}=\frac{1}{x+3} &&\text{substitute the inner rule}\\ x+5&\ne2 &&\text{input to }f\text{ must be allowed}\\ x&\ne-3 &&\text{solve the restriction} \end{aligned}

Why exclude $-3$ , rather than $2$ ? Starting at $-3$ makes $g$ output $2$ , which $f$ cannot accept.

Answer. $fg(x)=1/(x+3)$ , with domain all real $x\ne-3$ .

Check. At $x=2$ , the intermediate value is $7$ , and $f(7)=1/5$ , so $2$ is allowed here.

Change only the outer operation: replace the reciprocal by a square root, so $f(x)=\sqrt{x-2}$ , and keep $g(x)=x+5$ . Now the intermediate value must be at least $2$ , rather than merely different from $2$ .

Worked example 4: a square-root restriction

Question. Let $f(x)=\sqrt{x-2}$ and $g(x)=x+5$ , each with its largest real domain. Find $fg(x)$ and its domain.

Plan. Substitute into the root and require its inside expression to be nonnegative.

\begin{aligned} fg(x)&=\sqrt{(x+5)-2}=\sqrt{x+3} &&\text{substitute the inner rule}\\ x+3&\ge0 &&\text{real square-root condition}\\ x&\ge-3 &&\text{solve the inequality} \end{aligned}

Why include $-3$ ? The intermediate value is $2$ , giving output $\sqrt0=0$ , which is defined.

Answer. $fg(x)=\sqrt{x+3}$ , with domain $x\ge-3$ .

Check. Starting at $-2$ gives $g(-2)=3$ , then $f(3)=1$ ; the composite also gives $\sqrt1=1$ .

Order affects the domain as well as the expression. For the original reciprocal $f(x)=1/(x-2)$ and $g(x)=x+5$ , reversing the order gives

gf(x)=\frac{1}{x-2}+5\qquad (x\ne2)

Here $f$ acts first, and adding $5$ creates no further exclusion. Also retain a restriction if cancellation hides it: composing $f(x)=1/x$ with itself simplifies to $x$ , but input $0$ remains forbidden.

Watch out

The wrong restriction “ $x\ne2$ ” for $fg(x)=1/(x+3)$ copies the outer function's restriction without checking what reaches it. Solve $g(x)\ne2$ , giving $x\ne-3$ .

Check the root condition on the complete substituted expression in the next problem.

Finding an inverse algebraically (Higher)

If a machine triples its input and adds $4$ , undo it by subtracting $4$ and dividing by $3$ . Reverse both the operations and their order.

Inverse function

An inverse function, written $f^{-1}$ , reverses a function's input-output pairs. Its domain is the original range, and its range is the original domain. The notation $f^{-1}(x)$ does not mean the reciprocal $1/f(x)$ .

To find a rule, write $y=f(x)$ , solve for the old input $x$ , then use $x$ as the input name in the inverse rule. These inverse methods are not supplied on the Higher formulae sheet.

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Algebra

Shape, space and measure

Handling data

3.2 - Composite and inverse functions

The order of a composite function (Higher)

Allowed inputs of a composite (Higher)

Finding an inverse algebraically (Higher)

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