What is 2.8: Composite and inverse functions?

For a composite, substitute the whole of the inner function into the outer one and simplify: gf(x) with g(x)=3 x and f(x)=e^x is 3 e^x=3x, and leaving 3 e^x is not a finished answer.

How many marks are 2.8 questions worth?

2.8 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 2.8 appeared in?

We have not yet indexed a verified past-paper appearance for 2.8 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

2.8

Understand and use composite functions; inverse functions and their graphs.

Draft — not yet indexed

Composite and inverse functions

Worked answers and methods for 2.8 on Edexcel A-level Maths 9MA0.

Definition

Specification 2.8: understand and use composite functions; inverse functions and their graphs. $\mathrm{fg}(x)$ means $\mathrm f(\mathrm g(x))$ , the function $\mathrm g$ applied first; $\mathrm f^{-1}$ exists only when $\mathrm f$ is one-to-one, and its graph is the reflection of $y=\mathrm f(x)$ in the line $y=x$ .

Explanation

For a composite, substitute the whole of the inner function into the outer one and simplify: $\mathrm{gf}(x)$ with $\mathrm g(x)=3\ln x$ and $\mathrm f(x)=e^x$ is $3\ln e^x=3x$ , and leaving $3\ln e^x$ is not a finished answer.
Solving $\mathrm{fg}(x)=k$ means forming the equation, solving it, then rejecting any solution outside the domain of the inner function, with the reason stated.
For an inverse, write $y=\mathrm f(x)$ , make $x$ the subject, then swap the letters and write the answer as $\mathrm f^{-1}(x)=\ldots$ ; a rational function needs cross-multiplying and collecting the $x$ -terms on one side.
The domain of $\mathrm f^{-1}$ is the range of $\mathrm f$ and must be stated, found from a sketch, the minimum of a quadratic or the endpoint of the domain.
A many-to-one function such as a quadratic has no inverse unless its domain is restricted; a repeated output, $\mathrm g(0)=\mathrm g(3)$ , is the standard reason.

A function and its inverse are mirror images in the line

y=x

, so

(a,b)

becomes

(b,a)

Worked example

The function

\mathrm f

is defined by

\mathrm f(x)=\dfrac{3x+2}{x-1}

x\in\mathbb R

x\neq1

. Find

\mathrm f^{-1}(x)

and state its domain.

(4)

(Total for Question 1 is 4 marks)

Working

1.Let $y=\dfrac{3x+2}{x-1}$ and cross-multiply: $y(x-1)=3x+2$ , so $xy-y=3x+2$ .
2.Collect the $x$ -terms: $xy-3x=y+2$ , so $x(y-3)=y+2$ and $x=\dfrac{y+2}{y-3}$ .
3.Swap the letters: $\mathrm f^{-1}(x)=\dfrac{x+2}{x-3}$ .
4.The range of $\mathrm f$ is all real values except $3$ , so the domain of $\mathrm f^{-1}$ is $x\in\mathbb R$ , $x\neq3$ .

Answer: $\mathrm f^{-1}(x)=\dfrac{x+2}{x-3}$ , $x\in\mathbb R$ , $x\neq3$

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Cross-multiplies and attempts to collect the $x$ -terms on one side, eg $xy-3x=y+2$	M1
	Fully correct method to make $x$ the subject, eg $x=\dfrac{y+2}{y-3}$	M1
	$\mathrm f^{-1}(x)=\dfrac{x+2}{x-3}$	A1
	Domain $x\neq3$	B1
Notes
A1: In function notation, oe
B1: Must be stated

(4 marks)

Common mistakes

Don't evaluate $fg(x)$ by applying $f$ first, even though $fg(x)=f(g(x))$ .
Don't use both square-root branches for the inverse of a restricted quadratic instead of choosing the branch fixed by the stated domain.

Mark-scheme wording to use

“M1 for cross-multiplying and attempting to collect the x-terms, or the swapped y-terms, on one side. M1 for a fully correct method to isolate the original input. A1 for f^(-1)(x) = (x + 5)/(3 - x), stated in function notation with domain x != 3” — Pearson 9MA0, Specimen Paper 1, Q10(a)
“M1 for applying the functions in the correct order. A1 for the simplified result gf(x) = 3x … leaving 3ln(e^x) is not sufficient for the accuracy mark” — Pearson 9MA0, Sample Paper 2, Q4(a)
“B1 for explaining that g is many-one or not one-to-one, or giving a valid repeated-value example such as g(0) = g(3) = 0” — Pearson 9MA0, Specimen Paper 1, Q10(e)

What the examiner wants

The domain, every time. The June 2024 Paper 1 report repeats a point from earlier reports that continues to be overlooked: the requirement for the domain to be stated when an inverse function is asked for, and the need to avoid incorrect statements such as “square numbers are always positive” when justifying a range.

Exam tip

To find an inverse, rearrange $y=f(x)$ for $x$ and state the inverse domain obtained from the original range.

Worked practice

Tier 1 · Easy

Let

f(x)=3x-1

and

g(x)=x^2+2

. Calculate

fg(2)

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$17$	2
Notes
Apply $g$ first: $g(2)=2^2+2=6$ . Then $f(6)=3(6)-1=17$ .

(2 marks)

Tier 2 · Standard

Let

f(x)=2x-3

and let

g(x)=\dfrac1{x+1}

. Find

fg(x)

, state its domain and solve

fg(x)=5

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$fg(x)=\dfrac2{x+1}-3$ Domain: $x\ne-1$ $x=-\dfrac34$	4
Notes
Applying $g$ first gives $fg(x)=f(g(x))=2\left(1/(x+1)\right)-3=2/(x+1)-3$ , with $x\ne-1$ . Solving $2/(x+1)-3=5$ gives $2/(x+1)=8$ , so $x+1=1/4$ and $x=-3/4$ .

(4 marks)

Tier 3 · Hard

The function

f(x)=(x-3)^2+1

has domain

x\geq3

. Find

f^{-1}(x)

, state its domain, and solve

f^{-1}(2x+1)=x+3

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$f^{-1}(x)=3+\sqrt{x-1}$ with domain $x\geq1$ $x=0$ or $x=2$	6
Notes
From $y=(x-3)^2+1$ and $x\geq3$ , take the positive root: $x=3+\sqrt{y-1}$ . Thus $f^{-1}(x)=3+\sqrt{x-1}$ , whose domain is $x\geq1$ . The equation becomes $3+\sqrt{2x}=x+3$ , so $\sqrt{2x}=x$ with $x\geq0$ . Squaring gives $x^2=2x$ , hence $x=0$ or $x=2$ ; both satisfy the unsquared equation.

(6 marks)

Tier 1 · Easy

Given

f(x)=5x+4

, find

f^{-1}(x)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$f^{-1}(x)=\dfrac{x-4}{5}$	2
Notes
Write $y=5x+4$ and rearrange to $x=(y-4)/5$ . Exchanging the variable labels gives $f^{-1}(x)=(x-4)/5$ .

(2 marks)

Tier 2 · Standard

Let

f(x)=4x+1

and

g(x)=\sqrt{x+2}

. Find

f^{-1}g(7)

. Then find

gf^{-1}(x)

and state its domain.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$f^{-1}g(7)=\dfrac12$ $gf^{-1}(x)=\dfrac12\sqrt{x+7}$ with domain $x\geq-7$	4
Notes
Since $f^{-1}(x)=(x-1)/4$ and $g(7)=3$ , $f^{-1}g(7)=f^{-1}(3)=1/2$ . Also $gf^{-1}(x)=g((x-1)/4)=\sqrt{(x-1)/4+2}=\frac12\sqrt{x+7}$ . The square-root input requires $x+7\geq0$ , so the domain is $x\geq-7$ .

(4 marks)

Tier 3 · Hard

Let

g(x)=x^2

and let

f

be a linear function. Given that

fg(x)=6x^2-5

, find

f(x)

. Hence solve

fg(x)=gf(x)

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$f(x)=6x-5$ $x=1$	5
Notes
Because $fg(x)=f(x^2)=6x^2-5$ , the linear function is $f(u)=6u-5$ , so $f(x)=6x-5$ . Then $gf(x)=g(6x-5)=(6x-5)^2$ . Equating the composites gives $6x^2-5=(6x-5)^2$ , so $30x^2-60x+30=0$ , or $30(x-1)^2=0$ . Hence $x=1$ .

(5 marks)

Tier 2 · Standard

Let

f(x)=3x-4

and

g(x)=ax+b

. Given that

fg(x)=x

for every real

x

, find

a

and

b

, and verify that

gf(x)=x

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$a=\dfrac13$ , $b=\dfrac43$ $gf(x)=x$	4
Notes
Since $fg(x)=3(ax+b)-4=3ax+3b-4$ , comparison with $x$ gives $3a=1$ and $3b-4=0$ . Thus $a=1/3$ and $b=4/3$ . Then $gf(x)=((3x-4)/3)+4/3=x$ , as required.

(4 marks)

Tier 3 · Hard

The function

f(x)=x+\dfrac4x

has domain

x\geq2

. Find

f^{-1}(x)

and state its domain. Hence find the exact value of

f^{-1}\left(\dfrac{13}{3}\right)

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$f^{-1}(x)=\dfrac{x+\sqrt{x^2-16}}2$ with domain $x\geq4$ $f^{-1}\left(\dfrac{13}{3}\right)=3$	6
Notes
For $x\geq2$ , $f(x)-4=(x-2)^2/x\geq0$ , so the range of $f$ , and hence the domain of its inverse, is $[4,\infty)$ . Write $y=x+4/x$ and rearrange to $x^2-yx+4=0$ . Thus $x=(y\pm\sqrt{y^2-16})/2$ . The two roots of this quadratic multiply to $4$ , so they are $x$ and $4/x$ ; since $x\geq2$ gives $x\geq4/x$ , the required value is the larger root, so $f^{-1}(x)=(x+\sqrt{x^2-16})/2$ . At $x=13/3$ , the square root is $5/3$ , giving $f^{-1}(13/3)=((13/3)+(5/3))/2=3$ .

(6 marks)

Tier 3 · Hard

Let

f(x)=x^3+x+1

. Prove that the graphs of

y=f(x)

and

y=f^{-1}(x)

have exactly one point of intersection, and find its coordinates.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	The unique point of intersection is $(-1,-1)$ .	6
Notes
For $v>u$ , $f(v)-f(u)=(v-u)(v^2+uv+u^2+1)>0$ , because $v^2+uv+u^2=(v+u/2)^2+3u^2/4\geq0$ . Thus $f$ is strictly increasing; its cubic end behaviour also gives range $\mathbb R$ , so it has an inverse on $\mathbb R$ . If $(a,b)$ lies on both graphs, then $f(a)=b$ and $f(b)=a$ . If $a<b$ , strict increase would give $b=f(a)<f(b)=a$ , a contradiction; the case $a>b$ is similar. Hence $a=b$ . An intersection must therefore satisfy $f(a)=a$ , so $a^3+a+1=a$ , giving $a^3=-1$ and $a=-1$ . Substitution gives the unique point $(-1,-1)$ .

(6 marks)

Q10

Tier 3 · Hard

10.

Let

f(x)=\sqrt{x-1}

and

g(x)=\dfrac3{x-2}

. Find

fg(x)

and

gf(x)

, stating the domain of each composite. Hence solve

fg(x)=1

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$fg(x)=\sqrt{\dfrac{5-x}{x-2}}$ with domain $2<x\leq5$ $gf(x)=\dfrac3{\sqrt{x-1}-2}$ with domain $x\geq1$ , $x\ne5$ $x=\dfrac72$	6
Notes
Applying $g$ first gives $fg(x)=\sqrt{3/(x-2)-1}=\sqrt{(5-x)/(x-2)}$ . Its radicand is non-negative and $x\ne2$ , giving $2<x\leq5$ . Applying $f$ first gives $gf(x)=3/(\sqrt{x-1}-2)$ . This requires $x\geq1$ and excludes $\sqrt{x-1}=2$ , so $x\ne5$ . Finally, $fg(x)=1$ gives $(5-x)/(x-2)=1$ , hence $5-x=x-2$ and $x=7/2$ , which lies in the domain.

(6 marks)

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Other points in 2 Algebra and functions

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