What is 2.8: Composite and inverse functions?

The composite fg(x) means f(g(x)): apply the right-hand function first and keep its whole output inside the outer function. A one-to-one mapping has no repeated outputs and can have an inverse; a many-to-one mapping needs a restricted domain first.

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.

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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.

Explanation

The composite $fg(x)$ means $f(g(x))$ : apply the right-hand function first and keep its whole output inside the outer function. A one-to-one mapping has no repeated outputs and can have an inverse; a many-to-one mapping needs a restricted domain first.
To find an inverse, write $y=f(x)$ , rearrange for $x$ , then exchange the variable labels and use the notation $f^{-1}$ . A function and its inverse have graphs reflected in $y=x$ , with domain and range interchanged.
In general $fg\ne gf$, whereas $f^{-1}f(x)=ff^{-1}(x)=x$ on the appropriate domains.
For inverse quadratics, use the stated domain to choose the correct square-root branch.
Check a proposed inverse by composing the functions on the permitted domain.

A function and its inverse are mirror images in the line

y=x

, so

(a,b)

becomes

(b,a)

Worked example

For $f(x)=\dfrac{3x+2}{x-1}$ , find $f^{-1}(x)$ and state its domain.

1.Let $y=(3x+2)/(x-1)$ .
2.Then $yx-y=3x+2$ , so $x(y-3)=y+2$ and $x=(y+2)/(y-3)$ .
3.Exchanging the labels gives $f^{-1}(x)=(x+2)/(x-3)$ .
4.The inverse denominator requires $x\ne3$ .

Answer: $f^{-1}(x)=\dfrac{x+2}{x-3}$ Domain: $x\ne3$

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.

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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