DP Maths AA SL - Functions - Functions and Their Properties

DP Maths AA SL - Functions - Functions and Their Properties - Full notes

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DP Maths AA SL | Functions and Their Properties

Functions and Their Properties

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2.5 Composite and Inverse Functions

DP MATHEMATICS AA · TOPIC 2: FUNCTIONS

2.5

Composite and Inverse Functions

Full notes for SL · Syllabus SL 2.5

Contents

1	Composite functions
2	Domain and range of composite functions
3	Decomposing functions
4	The identity function and inverse functions
5	Finding an inverse algebraically
6	Graphs of inverse functions
7	Restricting domains to create inverses
8	Solving equations with inverses and composites
9	Evaluating composite functions
10	Domain and range of composite functions
11	Decomposing functions
12	Finding inverses algebraically
13	Restricting domains to make inverses
14	Graphs of inverse functions
15	Solving f(x) = f⁻¹(x)
16	Inverse of a composite
17	Self-inverse functions
18	Composite and inverse functions in context
19	Mixed review: worked problems
20	Additional exam-style practice
21	Graphical composition
22	Exam-style questions with markscheme
23	Common misconceptions and links
24	Check your understanding
25	Summary and topic map
26	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
form and evaluate composite functions (f ∘ g)(x) in the correct order
find the domain and range of a composite function
write a given function as a composite of simpler functions
explain why only one-to-one functions have inverses
find f−1(x) algebraically, state its domain and verify (f ∘ f−1)(x) = x
sketch y = f−1(x) as the reflection of y = f(x) in y = x
restrict a domain so that an inverse exists
solve equations involving composite and inverse functions

This booklet covers SL 2.5, which is common to SL and HL.

Guiding questions

How can functions be combined? How can a function be undone, and when is this impossible?

How this topic is examined

Composite and inverse functions appear on Paper 1 in most sessions, often in the same question. Typical tasks: find (f ∘ g)(x), solve (f ∘ g)(x) = k, find f−1(x) and its domain, sketch y = f−1(x), or find the largest domain for which an inverse exists.

The domain of f−1 is the range of f. Always state it, even if the question does not explicitly ask.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Composite functions

Apply one function, then another.

Think about it

A shop takes $5 off and then adds 10% tax. Another shop adds 10% tax first, then takes $5 off. Do customers pay the same?

The composite function (f ∘ g)(x) = f(g(x)) means: apply g first, then apply f to the result. The function written nearest to x acts first. In general, order matters: f(g(x)) ≠ g(f(x)).

In the shop example, let d(x) = x − 5 (discount) and t(x) = 1.1x (tax). Discount then tax: (t ∘ d)(x) = 1.1(x − 5) = 1.1x − 5.5. Tax then discount: (d ∘ t)(x) = 1.1x − 5. Discounting first saves the customer 50 cents.

A composite can be built from any number of functions: (f ∘ g ∘ h)(x) = f(g(h(x))), applying h, then g, then f. We also write f2(x) for (f ∘ f)(x), which is not the same as [f(x)]2.

WORKED EXAMPLE 1 Forming composites (basic)

Given f(x) = 2x + 1 and g(x) = x2 − 3, find (a) (f ∘ g)(x); (b) (g ∘ f)(x); (c) (f ∘ g)(2).

Solution

(a) f(x2 − 3) = 2(x2 − 3) + 1 = 2x2 − 5A1

(b) g(2x + 1) = (2x + 1)2 − 3 = 4x2 + 4x − 2A1

WORKED EXAMPLE 2 Composite with itself (standard)

Given h(x) = 1/(x + 1), find (h ∘ h)(x) and simplify.

Solution

h(h(x)) = 1/(1/(x + 1) + 1)M1

= 1/((1 + x + 1)/(x + 1))M1

= (x + 1)/(x + 2), x ≠ −1, −2A1

Common mistake

Reading (f ∘ g)(x) as 'apply f first'. The function nearest x is applied first.

Practice exercise 1

1. Given f(x) = x + 4 and g(x) = 3x, find (f ∘ g)(x) and (g ∘ f)(x).

2. Given f(x) = x2 and g(x) = x − 2, find (f ∘ g)(5) and (g ∘ f)(5).

3. Given f(x) = 2x − 1, find (f ∘ f ∘ f)(x).

4. Given f(x) = ex and g(x) = 3 ln x, find (f ∘ g)(x).

5. Given f(x) = x/(x − 1), simplify (f ∘ f)(x).

2 Domain and range of composite functions

The output of g must be acceptable to f.

For f(g(x)) to be defined, x must be in the domain of g and g(x) must be in the domain of f. So the domain of f ∘ g may be smaller than the domain of g. To find the range, find the set of values g(x) takes on that domain, then apply f to that set.

WORKED EXAMPLE 3 Domain of a composite (standard)

Given f(x) = ln x and g(x) = 3 − x, find the domain of (f ∘ g)(x).

Solution

f(g(x)) = ln(3 − x); need 3 − x > 0M1

x < 3A1

WORKED EXAMPLE 4 Domain and range (exam level)

f(x) = √(x − 1), x ≥ 1, and g(x) = x2 + 2, x ∈ ℝ. Find (f ∘ g)(x) and state its domain and range.

Solution

f(g(x)) = √(x2 + 1)A1

g(x) ≥ 2 ≥ 1 for all x, so domain x ∈ ℝR1

x2 + 1 ≥ 1, so range (f ∘ g)(x) ≥ 1A1

Practice exercise 2

1. f(x) = √x and g(x) = x − 4. Find the domain of (f ∘ g)(x).

2. f(x) = 1/x and g(x) = x2 − 9. Find the domain of (f ∘ g)(x).

3. f(x) = ln x, g(x) = x2. State the domain and range of (f ∘ g)(x).

3 Decomposing functions

Spot the inner and outer functions.

It is often useful (especially for the chain rule in calculus) to write a complicated function as a composite of simpler ones. In h(x) = (3x − 1)5, the inner function is g(x) = 3x − 1 and the outer function is f(x) = x5, so h = f ∘ g.

Practice exercise 3

1. Write h(x) = √(x2 + 4) as f(g(x)).

2. Write h(x) = e2x+1 as f(g(x)).

3. Write h(x) = 1/(x − 3)2 as a composite of two functions.

4 The identity function and inverse functions

Undoing a function.

The identity function is I(x) = x: it leaves every input unchanged. The inverse f−1 of a function f reverses it: if f(a) = b then f−1(b) = a. Composing a function with its inverse in either order gives the identity: (f ∘ f−1)(x) = (f−1 ∘ f)(x) = x.

A function has an inverse only if it is one-to-one. If two inputs gave the same output, the inverse would need to send that output back to two different inputs, which is not allowed for a function. Graphically, f is one-to-one if it passes the horizontal line test: every horizontal line meets the graph at most once.

Inputs and outputs swap roles, so the domain of f−1 is the range of f, and the range of f−1 is the domain of f.

Common mistake

Confusing f−1(x) with 1/f(x). The inverse function undoes f; it is not the reciprocal.

5 Finding an inverse algebraically

Swap x and y, then rearrange.

Method: write y = f(x); interchange x and y; solve for y; write f−1(x) = ...; state its domain (the range of f).

WORKED EXAMPLE 5 Linear inverse (basic)

Find f−1(x) for f(x) = 4x − 7, and verify that (f ∘ f−1)(x) = x.

Solution

x = 4y − 7, so y = (x + 7)/4M1

f−1(x) = (x + 7)/4A1

f(f−1(x)) = 4(x + 7)/4 − 7 = xR1

WORKED EXAMPLE 6 Linear-over-linear inverse (standard)

Find f−1(x) for f(x) = (2x + 1)/(x − 3), x ≠ 3.

Solution

x = (2y + 1)/(y − 3)M1

xy − 3x = 2y + 1, so y(x − 2) = 3x + 1M1

f−1(x) = (3x + 1)/(x − 2), x ≠ 2A1

WORKED EXAMPLE 7 Exponential inverse (exam level)

Find f−1(x) for f(x) = e2x + 1, x ∈ ℝ, and state its domain.

Solution

x = e2y + 1, so e2y = x − 1M1

f−1(x) = ½ ln(x − 1)A1

Domain x > 1 (the range of f)A1

Practice exercise 4

1. Find the inverse of f(x) = (x − 5)/3.

2. Find the inverse of f(x) = x3 + 2.

3. Find the inverse of f(x) = 3/(x + 1), x ≠ −1, and state its domain.

4. Find the inverse of f(x) = ln(2x − 4), x > 2.

5. If f(3) = 10, find f−1(10).

6 Graphs of inverse functions

Reflect in y = x.

The graph of y = f−1(x) is the reflection of y = f(x) in the line y = x, because swapping x and y swaps the coordinates of every point: (a, b) on f becomes (b, a) on f−1.

Consequences: intercepts swap axes (a y-intercept (0, c) of f becomes an x-intercept (c, 0) of f−1); vertical asymptotes become horizontal and vice versa; any intersection of y = f(x) with y = f−1(x) on the line y = x satisfies f(x) = x. A self-inverse function (f−1 = f) is symmetric about y = x.

Figure 1. A function and its inverse are reflections in y = x.

Practice exercise 5

1. The point (2, 7) lies on y = f(x). Which point lies on y = f−1(x)?

2. f has asymptotes x = 1 and y = 3. State the asymptotes of f−1.

3. Sketch y = 2x and its inverse on the same axes, labelling intercepts.

7 Restricting domains to create inverses

Make it one-to-one.

A many-to-one function can be given an inverse by restricting its domain so that it becomes one-to-one. For a quadratic, restrict to one side of the vertex. For example, f(x) = (x − 2)2 + 1 is one-to-one on x ≥ 2, and then f−1(x) = 2 + √(x − 1), x ≥ 1.

When solving for y after swapping, you will get ±. Choose the sign that matches the range of f−1, which is the restricted domain of f.

WORKED EXAMPLE 8 Largest domain (exam level)

f(x) = x2 − 6x + 4, x ≤ k. Find the largest k for which f−1 exists, and find f−1(x).

Solution

Vertex at x = 3; f is one-to-one for x ≤ 3: k = 3A1

y = (x − 3)2 − 5; swap: x + 5 = (y − 3)2, y = 3 ± √(x + 5)M1

Range of f−1 is y ≤ 3, so f−1(x) = 3 − √(x + 5), x ≥ −5A1 R1

Practice exercise 6

1. f(x) = (x + 1)2, x ≥ −1. Find f−1(x).

2. f(x) = 4 − x2, x ≤ 0. Find f−1(x) and its domain.

3. Find the largest domain of the form x ≥ a for which f(x) = x2 + 4x has an inverse.

8 Solving equations with inverses and composites

Use the definitions directly.

WORKED EXAMPLE 9 Solving a composite equation (standard)

f(x) = 3x − 2 and g(x) = 2/(x + 1). Solve (f ∘ g)(x) = 1.

Solution

6/(x + 1) − 2 = 1M1

x + 1 = 2A1

x = 1A1

WORKED EXAMPLE 10 Using an inverse value (standard)

f(x) = 2x3 + 3. Find f−1(19) without finding f−1(x).

Solution

Solve f(x) = 19: 2x3 = 16M1

x = 2A1

WORKED EXAMPLE 11 Intersection with the inverse (exam level)

f(x) = (x2 + 4)/5, x ≥ 0. Find the points where y = f(x) meets y = f−1(x).

Solution

They meet on y = x: (x2 + 4)/5 = xM1

x2 − 5x + 4 = 0, x = 1 or 4A1

(1, 1) and (4, 4)A1

Practice exercise 7

1. f(x) = x + 3 and g(x) = x2. Solve (g ∘ f)(x) = 25.

2. f(x) = ex − 1. Find f−1(e4).

3. f(x) = 2x − 3 and g(x) = (x + 3)/2. Show that g = f−1.

9 Evaluating composite functions

Inside first.

(f ∘ g)(x) = f(g(x)): apply g first, then f. To evaluate at a number, work from the inside out; to find a formula, substitute the whole expression g(x) into f.

WORKED EXAMPLE 12 Numerical (basic)

f(x) = 3x − 1, g(x) = x2 + 2. Find (f ∘ g)(2) and (g ∘ f)(2).

Solution

g(2) = 6, f(6) = 17A1

f(2) = 5, g(5) = 27A1

WORKED EXAMPLE 13 Formula (standard)

Find (f ∘ g)(x) and (g ∘ f)(x).

Solution

f(g(x)) = 3(x2 + 2) − 1 = 3x2 + 5A1

g(f(x)) = (3x − 1)2 + 2 = 9x2 − 6x + 3A1

Practice exercise 8

1. f(x) = 2x + 5, g(x) = 1/x. Find (f ∘ g)(x) and (g ∘ f)(x).

2. f(x) = √x, g(x) = x − 3. Find (f ∘ g)(7).

3. f(x) = x2, find f(f(x)).

10 Domain and range of composite functions

The output of g must be acceptable to f.

The domain of f ∘ g consists of the x in the domain of g for which g(x) is in the domain of f. Find the range by considering the range of g fed into f.

WORKED EXAMPLE 14 Domain (exam level)

f(x) = √(x − 2), g(x) = x2 − 7. Find the domain of f ∘ g.

Solution

Need x2 − 7 ≥ 2, so x2 ≥ 9M1

x ≤ −3 or x ≥ 3A1

WORKED EXAMPLE 15 Range (standard)

f(x) = ex, g(x) = x2 + 1. Find the range of f ∘ g.

Solution

g(x) ≥ 1A1

f(g(x)) ≥ e: range [e, ∞)A1

Practice exercise 9

1. f(x) = ln x, g(x) = 4 − x2. Find the domain of f ∘ g.

2. f(x) = 1/x, g(x) = x − 5. Find the domain of f ∘ g.

11 Decomposing functions

Writing h as f ∘ g.

Many functions are composites of simpler ones: (3x + 1)5 = f(g(x)) with g(x) = 3x + 1 and f(x) = x5. Decomposition is essential for the chain rule and for transformations.

WORKED EXAMPLE 16 Decomposition (standard)

Write h(x) = √(x2 + 4) as f(g(x)).

Solution

g(x) = x2 + 4, f(x) = √xA1 A1

Practice exercise 10

1. Write e2x − 1 as a composite.

2. Write 1/(x − 3)2 as a composite of two functions.

12 Finding inverses algebraically

Swap and rearrange.

To find f−1(x): write y = f(x), swap x and y, and rearrange to make y the subject. The domain of f−1 is the range of f, and vice versa.

WORKED EXAMPLE 17 Rational function (exam level)

Find the inverse of f(x) = (2x + 1)/(x − 3), x ≠ 3.

Solution

x = (2y + 1)/(y − 3): xy − 3x = 2y + 1M1

y(x − 2) = 3x + 1A1

f−1(x) = (3x + 1)/(x − 2), x ≠ 2A1

WORKED EXAMPLE 18 Logarithmic (standard)

Find the inverse of f(x) = 3 + ln(2x), x > 0.

Solution

x − 3 = ln(2y), 2y = ex−3M1

f−1(x) = ½ex−3, x ∈ ℝA1

Practice exercise 11

1. Find the inverse of f(x) = (x − 4)/5.

2. Find the inverse of f(x) = 2ex + 1 and its domain.

3. Find the inverse of f(x) = (x + 1)/(x − 1).

13 Restricting domains to make inverses

One-to-one functions.

Only one-to-one functions have inverses. A many-to-one function such as a quadratic can be made one-to-one by restricting its domain, usually to one side of the vertex.

WORKED EXAMPLE 19 Quadratic (exam level)

f(x) = (x − 2)2 + 1, x ≥ 2. Find f−1(x) and its domain.

Solution

x − 1 = (y − 2)2, y ≥ 2M1

f−1(x) = 2 + √(x − 1), x ≥ 1A1 A1

Practice exercise 12

1. Find the largest domain x ≥ a for which f(x) = x2 + 6x is one-to-one.

2. Find the inverse of f(x) = x2 − 4, x ≤ 0.

14 Graphs of inverse functions

Reflection in y = x.

The graph of y = f−1(x) is the reflection of y = f(x) in the line y = x. Intercepts swap coordinates; vertical asymptotes become horizontal and vice versa.

Figure 2. A function and its inverse are reflections in y = x.

WORKED EXAMPLE 20 Asymptotes (standard)

f(x) = 3 + 1/(x − 2) has asymptotes x = 2 and y = 3. State the asymptotes of f−1.

Solution

Swap: x = 3 and y = 2A1

Practice exercise 13

1. f has range y > −1 and passes through (0, 2). Write down the domain of f−1 and one point on it.

2. Sketch y = x3 and its inverse.

15 Solving f(x) = f⁻¹(x)

Use the line y = x.

For an increasing function, the graphs of f and f−1 meet only on the line y = x. So solving f(x) = x is usually easier than solving f(x) = f−1(x).

WORKED EXAMPLE 21 Intersection (exam level)

f(x) = (x2 + 6)/5 for x ≥ 0. Solve f(x) = f−1(x).

Solution

Solve f(x) = x: x2 − 5x + 6 = 0M1

x = 2 or x = 3A1 A1

Practice exercise 14

1. f(x) = 2x − 3. Solve f(x) = f−1(x).

2. f(x) = ex − 2. Find the intersections of f and f−1 using technology.

16 Inverse of a composite

Undo in reverse order.

(f ∘ g)−1 = g−1 ∘ f−1: to undo 'g then f', undo f first, then g (like taking off shoes and socks).

WORKED EXAMPLE 22 Composite inverse (standard)

f(x) = 2x, g(x) = x + 3. Verify (f ∘ g)−1 = g−1 ∘ f−1.

Solution

f(g(x)) = 2x + 6, inverse (x − 6)/2A1

g−1(f−1(x)) = x/2 − 3 = (x − 6)/2 ✓A1

Practice exercise 15

1. f(x) = x3, g(x) = 2x − 1. Find (f ∘ g)−1(x).

17 Self-inverse functions

f(f(x)) = x.

A function is self-inverse if f−1 = f, so f(f(x)) = x. Its graph is symmetric about y = x. Examples: f(x) = 1/x, f(x) = a − x, f(x) = (ax + b)/(cx − a).

WORKED EXAMPLE 23 Proving self-inverse (exam level)

Show that f(x) = (3x + 2)/(x − 3) is self-inverse.

Solution

f(f(x)) = (3(3x + 2)/(x − 3) + 2)/((3x + 2)/(x − 3) − 3)M1

= (9x + 6 + 2x − 6)/(3x + 2 − 3x + 9) = 11x/11 = xA1 AG

Practice exercise 16

1. Show that f(x) = 5 − x is self-inverse.

2. Find a for which f(x) = (2x + 1)/(x + a) is self-inverse.

18 Composite and inverse functions in context

Chaining processes.

In context, a composite chains two processes and an inverse answers the reverse question. Example: if d(t) is the distance travelled after t hours and c(d) the fuel cost for distance d, then c(d(t)) is the cost after t hours, and d−1 tells us the time for a given distance.

WORKED EXAMPLE 24 Currency conversion (standard)

1 GBP = 1.25 USD and a $3 fee is charged: U(p) = 1.25p − 3. Find U−1 and interpret U−1(97).

Solution

p = (U + 3)/1.25M1 A1

U−1(97) = 80: £80 is needed to receive $97A1

Practice exercise 17

1. A balloon's radius is r(t) = 2t + 1 cm and its volume V(r) = (4/3)πr3. Find V(r(t)) and V when t = 2.

19 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 25 Mixed problem (exam level)

f(x) = 2x − 1, g(x) = (x + 3)/2. (a) Find (f ∘ g)(x). (b) Hence explain the relationship between f and g.

Solution

(a) f(g(x)) = x + 3 − 1 = x + 2A1

(b) Not inverses: (f ∘ g)(x) ≠ x. The inverse of f is (x + 1)/2R1 A1

Practice exercise 18

1. f(x) = x2 − 1 and g(x) = √(x + 3). Find the domain and range of f ∘ g.

2. Find x if f(g(x)) = 8 where f(x) = 3x + 2 and g(x) = x2.

20 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 26 Paper 1 question (exam level)

f(x) = e2x and g(x) = ln(x + 1), x > −1. (a) Find (f ∘ g)(x). (b) Find f−1(x). (c) Solve (f ∘ g)(x) = 9.

Solution

(a) e2 ln(x + 1) = (x + 1)2M1 A1

(b) ½ ln xA1

Practice exercise 19

1. f(x) = 4 − x2, x ≥ 0. Find f−1(x) and state its domain and range.

21 Graphical composition

Reading composites from graphs.

Given graphs or tables of f and g, evaluate f(g(a)) by reading g(a) from the graph of g and then f of that value from the graph of f. Tables make this straightforward.

x	0	1	2	3	4
f(x)	3	5	2	0	1
g(x)	2	4	3	1	0

WORKED EXAMPLE 27 From a table (standard)

Use the table to find f(g(1)), g(f(2)) and f−1(0).

Solution

f(g(1)) = f(4) = 1A1

g(f(2)) = g(2) = 3; f−1(0) = 3A1 A1

Practice exercise 20

1. Find (g ∘ g)(0).

2. Find x such that f(g(x)) = 3.

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

f(x) = 3x − 2 and g(x) = 2/(x + 1), x ≠ −1.

(a) Find (g ∘ f)(x). [2]

(b) Find g−1(x). [3]

(d) State the range of g. [1]

Markscheme

Part	Answer	Marks
(a)	g(3x − 2) = 2/(3x − 1)	M1 A1
(b)	x = 2/(y + 1)	M1
	y + 1 = 2/x	A1
	g−1(x) = 2/x − 1, x ≠ 0	A1
(c)	6/(x + 1) − 2 = 1	M1
	x + 1 = 2	A1
	x = 1	A1
(d)	g(x) ≠ 0	A1

Examiner's comments. In (b), the domain of the inverse is often omitted.

Question 2 (Paper 1 style) (about 12 minutes)

f(x) = x2 − 4x + 7, x ≥ 2.

(a) Write f(x) in the form (x − h)2 + k. [2]

(b) Find f−1(x), stating its domain. [4]

Markscheme

Part	Answer	Marks
(a)	(x − 2)2 + 3	A2
(b)	x = (y − 2)2 + 3	M1
	y = 2 ± √(x − 3)	A1
	f−1(x) = 2 + √(x − 3)	R1
	x ≥ 3	A1
(c)	reflection in y = x; endpoints (2, 3) and (3, 2)	A2

Examiner's comments. Choosing the positive root must be justified by the restricted domain.

23 Common misconceptions and links

Students often think…	Actually…
f−1(x) = 1/f(x)	f−1 undoes f
(f ∘ g)(x) means f first	g first
f2(x) = [f(x)]2	f2(x) = f(f(x))
keeping both ± roots for an inverse	select the root matching the restricted domain

How 2.5 connects to the rest of the course

Topic	Connection
2.1 Functions	domain and range
1.5 Logarithms	inverse of the exponential
2.13 Further properties (HL)	self-inverse functions
5.6 Chain rule	differentiating composites

24 Check your understanding

Section A: multiple choice

1. If f(x) = x + 2 and g(x) = 3x, (f ∘ g)(1) =

A. 5 B. 9 C. 6 D. 3

2. The inverse of f(x) = 2x is

A. 1/(2x) B. 2/x C. x/2 D. −2x

3. The graph of f−1 is a reflection in

A. the x-axis B. y = x C. the y-axis D. y = −x

4. A function has an inverse if it is

A. many-to-one B. even C. quadratic D. one-to-one

5. If the range of f is y > 0, the domain of f−1 is

A. all reals B. x > 0 C. x ≥ 0 D. x < 0

Section B: structured

6. Find f−1(x) for f(x) = √(2x − 3) and state its domain. [3]

7. Show that f(x) = (x + 1)/(x − 1) is self-inverse. [3]

8. Explain why f(x) = x2, x ∈ ℝ has no inverse. [2]

Answers

1. A 2. C 3. B 4. D 5. B

6. f−1(x) = (x2 + 3)/2, x ≥ 0.

7. f(f(x)) = x after simplifying.

8. Many-to-one: f(2) = f(−2); fails the horizontal line test.

25 Summary and topic map

Idea	Remember
Composite	(f ∘ g)(x) = f(g(x)); g first
Composite domain	g(x) must lie in the domain of f
Inverse	f−1(f(x)) = x; one-to-one only
Finding f−1	swap x and y, rearrange, state domain
Graph	reflect in y = x
Restricting	make one-to-one; choose the correct root

Before the exam, make sure you…

•apply the inner function first
•state domains of inverses
•check one-to-one
•verify f(f−1(x)) = x

Composite and Inverse Functions on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 3. Topic map for Composite and Inverse Functions.

26 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. 3x + 4; 3x + 12

2. 9; 23

3. 8x − 7

4. x3

5. x

Practice exercise 2

1. x ≥ 4

2. x ≠ ±3

3. x ≠ 0; range ℝ

Practice exercise 3

1. g(x) = x2 + 4, f(x) = √x

2. g(x) = 2x + 1, f(x) = ex

3. g(x) = x − 3, f(x) = 1/x2

Practice exercise 4

1. 3x + 5

2. ∛(x − 2)

3. 3/x − 1, x ≠ 0

4. (ex + 4)/2, x ∈ ℝ

5. 3

Practice exercise 5

1. (7, 2)

2. y = 1 and x = 3

3. y = 2x through (0, 1); y = log2 x through (1, 0)

Practice exercise 6

1. √x − 1, x ≥ 0

2. −√(4 − x), x ≤ 4

3. x ≥ −2

Practice exercise 7

1. x = 2 or x = −8

2. 5

3. f(g(x)) = x + 3 − 3 = x

Practice exercise 8

1. 2/x + 5; 1/(2x + 5)

2. 2

3. x4

Practice exercise 9

1. −2 < x < 2

2. x ≠ 5

Practice exercise 10

1. g(x) = 2x − 1, f(x) = ex

2. g(x) = x − 3, f(x) = 1/x2

Practice exercise 11

1. 5x + 4

2. ln((x − 1)/2), x > 1

3. f is self-inverse: (x + 1)/(x − 1)

Practice exercise 12

1. x ≥ −3

2. −√(x + 4), x ≥ −4

Practice exercise 13

1. x > −1; (2, 0)

2. y = ∛x, reflection in y = x

Practice exercise 14

1. x = 3

2. f(x) = x: x = −1.84 and 1.15

Practice exercise 15

1. (∛x + 1)/2

Practice exercise 16

1. f(f(x)) = 5 − (5 − x) = x

2. a = −2

Practice exercise 17

1. (4/3)π(2t + 1)3; 523.6 cm3

Practice exercise 18

1. x ≥ −3; range y ≥ −1 (f(g(x)) = x + 2 for x ≥ −3)

2. x = ±√2

Practice exercise 19

1. √(4 − x); x ≤ 4; f−1(x) ≥ 0

Practice exercise 20

1. g(2) = 3

2. g(x) = 0: x = 4

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