Pure 2.8 - Composite and inverse functions | Edexcel A-Level Mathematics

Functions describe processes with one output for each allowed input. Learn to combine those processes, undo them and use their domains and graphs to decide which answers make sense. Keeping track of the direction of each mapping is the central skill.

Functions as mappings

A machine squares a number and adds one. If its display reads $5$ , can you recover the input? Both $2$ and $-2$ produce $5$ , so the display alone cannot tell you. We will find the condition that makes reversal possible.

You already know

Substitution evaluates a rule: if $p(x)=x^2+1$ , then $p(2)=5$ . Solving $x^2=4$ gives two real values, $x=2$ and $x=-2$ .

A function must say which inputs are allowed. For $p:x\mapsto x^2+1$ , $x\in\mathbb R$ , every real input is allowed and the outputs are all at least $1$ .

Function, domain and range

A function assigns each input in its domain exactly one output. Its range is the set of outputs actually produced. A one-one function never gives the same output from two different inputs; a many-one function does.

For example, $x\mapsto2x+3$ on $\mathbb R$ is one-one; $x\mapsto x^2$ on $\mathbb R$ is many-one. A rule assigning both $2$ and $-2$ to the single input $4$ is not a function.

The notation $p:x\mapsto x^2+1$ gives the rule as a mapping; $p(x)=x^2+1$ gives the same rule in function notation. Its range can be written $[1,\infty)$ or $\{y\in\mathbb R:y\ge1\}$ . Square brackets include an endpoint; the round bracket at infinity never includes infinity as a number.

[DIAGRAM: asset_name: One-one and many-one mappings; asset_slug: p1_2_8_understand_and_use_composite_functions_inverse_functions_and_their_graphs__diagram_01; caption: Each input has exactly one output in both mappings; description: A one-one mapping sends 1, 2, 3 to 4, 5, 6 respectively. A many-one mapping sends -2 and 2 to 4 and 3 to 9.]

There is a neat surprise: restricting $x^2+1$ to $x\ge0$ makes it one-one without changing its range at all. Every output at least $1$ still occurs, but its duplicate negative input has disappeared. Restricting to $x\le0$ works too, and gives a different inverse later.

Key idea

Many-one is allowed for a function; it is the obstacle to reversing that function uniquely.

Use the domain as well as the formula in this check.

Composite functions

A composite connects the output of one function to the input of another. With $f(x)=2x+3$ and $g(x)=x^2-1$ , input $2$ goes through $g$ to $3$ , then through $f$ to $9$ .

Composite function

$fg(x)=f(g(x))$ means do $g$ first, then $f$ . The input must belong to the domain of $g$ , and $g(x)$ must belong to the domain of $f$ .

The nearest function to the input acts first. Composition generally changes when its order changes.

[DIAGRAM: asset_name: Function composition pipeline; asset_slug: p1_2_8_understand_and_use_composite_functions_inverse_functions_and_their_graphs__diagram_02; caption: The inner function acts first; description: Input x enters g, output g(x) enters f, and the final output is f(g(x)) or fg(x). The intermediate output must be in the domain of f.]

The pipeline notation is not a multiplication rule; replacing a whole input by brackets keeps that distinction clear.

Worked example 1: compose two rules

Question. For $f(x)=2x+3$ and $g(x)=x^2-1$ , both on $\mathbb R$ , find $fg(x)$ .

Plan. Substitute the complete expression $g(x)$ into the input position of $f$ .

\begin{aligned} fg(x)&=f(x^2-1) &&\text{do }g\text{ first}\\ &=2(x^2-1)+3 &&\text{use the rule for }f\\ &=2x^2+1 &&\text{expand and collect} \end{aligned}

Why the brackets? Everything produced by $g$ is the input multiplied by $2$ .

Answer. $fg(x)=2x^2+1$ , for $x\in\mathbb R$ .

Check. At $x=2$ , $g(2)=3$ , $f(3)=9$ , and $2(2)^2+1=9$ .

Change only the order. The same two rules now carry out the operations in reverse sequence.

Worked example 2: reverse the composition order

Question. For the same $f$ and $g$ , find $gf(x)$ .

Plan. Put $2x+3$ into the rule for $g$ .

\begin{aligned} gf(x)&=g(2x+3) &&\text{do }f\text{ first}\\ &=(2x+3)^2-1 &&\text{square the whole input}\\ &=4x^2+12x+8 &&\text{expand and collect} \end{aligned}

Why is it different? Squaring after adding $3$ creates a cross term; adding after squaring does not.

Answer. $gf(x)=4x^2+12x+8$ , for $x\in\mathbb R$ .

Check. At $x=2$ , $f(2)=7$ , $g(7)=48$ , agreeing with the expression.

Domains can limit a composite even when its algebra looks simple. For $s(x)=\sqrt x$ and $r(x)=x-4$ , $sr(x)=\sqrt{x-4}$ needs $x\ge4$ ; $rs(x)=\sqrt x-4$ needs $x\ge0$ .

Watch out

The wrong line $fg(x)=f(x)g(x)$ treats composition as multiplication. The correct line is $fg(x)=f(g(x))$ .

Now change the linear rule and evaluate both orders at one input.

Inverse functions

An inverse reverses the direction of a one-one function. If $f$ sends $a$ to $b$ , then $f^{-1}$ sends $b$ back to $a$ . The inverse's inputs therefore come from the original range.

Inverse identities

For a one-one function $f$ ,

f^{-1}f(x)=x\quad\text{on the domain of }f, \qquad ff^{-1}(x)=x\quad\text{on the range of }f.

Not in the formula booklet: learn it.

To find the reverse rule, solve $y=f(x)$ for the original input $x$ , then name the inverse's input $x$ again.

Worked example 3: invert a rational function

Question. Find the inverse of $f(x)=\frac{2x+1}{x-3}$ , $x\ne3$ .

Plan. Rearrange $y=f(x)$ , collecting all terms containing the original input.

\begin{aligned} y(x-3)&=2x+1 &&\text{clear the denominator}\\ xy-2x&=3y+1 &&\text{collect }x\text{ terms}\\ x(y-2)&=3y+1 &&\text{factorise}\\ x&=\frac{3y+1}{y-2} &&\text{divide for }y\ne2 \end{aligned}

Why collect before dividing? The input occurs in both numerator and denominator; it must be isolated as one factor.

Answer. $f^{-1}(x)=\frac{3x+1}{x-2}$ , $x\ne2$ . The range of $f$ excludes $2$ , since $f(x)=2+\frac7{x-3}$ .

Check. $f(4)=9$ and $f^{-1}(9)=\frac{28}7=4$ .

Change only the numerator's constant, from $1$ to $5$ . The rearrangement is unchanged, but the inverse's rule changes.

Worked example 4: track a changed constant

Question. Find the inverse of $h(x)=\frac{2x+5}{x-3}$ , $x\ne3$ .

Plan. Clear the denominator and factorise the input terms.

\begin{aligned} y(x-3)&=2x+5 &&\text{multiply through}\\ x(y-2)&=3y+5 &&\text{collect and factorise}\\ x&=\frac{3y+5}{y-2} &&\text{isolate the input} \end{aligned}

Why exclude $2$ again? $h(x)=2+\frac{11}{x-3}$ never equals $2$ .

Answer. $h^{-1}(x)=\frac{3x+5}{x-2}$ , $x\ne2$ .

Check. $h(4)=13$ and $h^{-1}(13)=\frac{44}{11}=4$ .

The superscript is notation for reversal, not an instruction to take a reciprocal.

Watch out

$f^{-1}(x)\ne\frac1{f(x)}$ in general. For $f(x)=4x-7$ , the inverse is $\frac{x+7}4$ , whereas the reciprocal is $\frac1{4x-7}$ .

Change the numerator's constant once more and show the reverse rule.

Domains, ranges, and graphs of inverses

Swapping input and output swaps domain and range. It also swaps coordinates: a point $(a,b)$ on $y=f(x)$ becomes $(b,a)$ on $y=f^{-1}(x)$ . Geometrically, that is reflection in $y=x$ .

[DIAGRAM: asset_name: Reflection of an inverse graph; asset_slug: p1_2_8_understand_and_use_composite_functions_inverse_functions_and_their_graphs__diagram_03; caption: Reflect the non-negative branch of the square graph in y equals x; description: Equal-scale axes show y=x squared for x non-negative and its inverse y=square root of x, with reflection line y=x and swapped points (2,4) and (4,2). The excluded negative branch is faint and dashed.]

A quadratic must be restricted to one branch before it can be reversed. The branch determines which square root to choose.

Worked example 5: the increasing quadratic branch

Question. Find the inverse of $f(x)=x^2+1$ , $x\ge0$ , stating its domain and range.

Plan. Rearrange, choosing the square root consistent with the original domain.

\begin{aligned} y&=x^2+1 &&\text{write the original rule}\\ x^2&=y-1 &&\text{subtract }1\\ x&=\sqrt{y-1} &&\text{use }x\ge0 \end{aligned}

Why the positive root? The original function accepts no negative inputs.

Answer. $f^{-1}(x)=\sqrt{x-1}$ , with domain $[1,\infty)$ and range $[0,\infty)$ . Its graph reflects the original branch in $y=x$ .

Check. The original point $(2,5)$ becomes $(5,2)$ , since $\sqrt{5-1}=2$ .

Change only the domain to $x\le0$ . The formula and original range stay the same, but the inverse must return negative inputs.

Worked example 6: the decreasing quadratic branch

Question. Find the inverse of $g(x)=x^2+1$ , $x\le0$ , stating its domain and range.

Plan. Use the negative square root when reversing the rule.

\begin{aligned} y&=x^2+1 &&\text{original rule}\\ x^2&=y-1 &&\text{rearrange}\\ x&=-\sqrt{y-1} &&\text{use }x\le0 \end{aligned}

Why not write both signs? An inverse function must produce exactly one output for each input.

Answer. $g^{-1}(x)=-\sqrt{x-1}$ , domain $[1,\infty)$ , range $(-\infty,0]$ .

Check. $g(-2)=5$ , and $g^{-1}(5)=-2$ . Also $g^{-1}g(x)=-\sqrt{x^2}=x$ when $x\le0$ .

For a shifted quadratic, apply the same sign decision to the bracket rather than to $x$ alone. For example, $(x-2)^2+3$ on $x\ge2$ has inverse $2+\sqrt{x-3}$ , domain $x\ge3$ .

Exam tip

State the inverse's domain as the original range. A correct rearrangement with the wrong branch is not a correct inverse.

Keep the negative branch and change the vertical shift.

Using composites and inverses together

Read a mixed expression from the input outwards. In $fg^{-1}(10)$ , first undo $g$ , then apply $f$ . You do not undo the entire composite unless brackets say so.

Worked example 7: undo then apply

Question. Let $f(x)=2x-5$ , $x\in\mathbb R$ , and $g(x)=x^2+1$ , $x\ge0$ . Find $fg^{-1}(10)$ .

Plan. Find the permitted inverse of $g$ , then feed its output into $f$ .

\begin{aligned} g^{-1}(x)&=\sqrt{x-1},\quad x\ge1 &&\text{reverse }g\\ g^{-1}(10)&=\sqrt9=3 &&\text{apply the inner rule}\\ f(3)&=2(3)-5=1 &&\text{apply the outer rule} \end{aligned}

Why start with the inverse? It is nearest to the input $10$ .

Answer. $fg^{-1}(10)=1$ .

Check. $g(3)=10$ , so $3$ is the correct recovered input; it lies in the domain of $f$ .

Change only the input from $10$ to $26$ . This lets you follow the same route without rebuilding it.

Worked example 8: a different input through the same pipeline

Question. For the same functions, find $fg^{-1}(26)$ .

Plan. Undo the square-and-add rule first.

\begin{aligned} g^{-1}(26)&=\sqrt{26-1}=5 &&\text{recover the non-negative input}\\ f(5)&=2(5)-5=5 &&\text{apply }f \end{aligned}

Why is the input allowed? $26\ge1$ , so it belongs to the domain of $g^{-1}$ .

Answer. $fg^{-1}(26)=5$ .

Check. $g(5)=26$ , confirming the inverse step.

The cancellation identities also explain the opening puzzle. The machine $x^2+1$ can be undone only after a one-one domain is chosen: $5$ returns $2$ on $x\ge0$ , and $-2$ on $x\le0$ .

Problem-solving

Write a mixed expression as nested brackets, check the inner domain, and work outwards one function at a time.

A final input gives the same sequence with new numbers.

Quick Recap

A function gives exactly one output for each allowed input; it may be many-one.
$fg$ means do $g$ first, then $f$ , checking both domains.
A one-one function has an inverse that swaps its domain and range.
Inverse graphs reflect in $y=x$ , so corresponding coordinates swap.

The machine displaying $5$ reveals its input only after we choose a one-one domain: non-negative inputs give $2$ ; non-positive inputs give $-2$ .

Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 2.8 - Composite and inverse functions