DP Maths AA HL - Functions - Functions and Their Properties

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

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DP Maths AA HL | Functions and Their Properties

Functions and Their Properties

This topic contains 3 notes:

2.1 Lines and Functions
2.5 Composite and Inverse Functions
2.15 Inequalities and Modulus

DP MATHEMATICS AA · TOPIC 2: FUNCTIONS

2.1

Lines and Functions

Full notes for HL · Syllabus SL 2.1, 2.2, 2.3, 2.4

Contents

1	Gradient and equations of straight lines
2	Parallel and perpendicular lines
3	Distance, midpoint and intersections of lines
4	The concept of a function
5	Domain and range
6	Graphs of functions: drawing and sketching
7	Key features of graphs
8	Solving equations graphically
9	Forms of the equation of a line
10	Parallel and perpendicular lines in practice
11	Distances and midpoints in coordinate geometry
12	Perpendicular bisectors and circles
13	Linear models in context
14	Functions: notation, domain and range
15	Graphs and key features with technology
16	Solving equations graphically
17	Mixed review: worked problems
18	Additional exam-style practice
19	The angle a line makes with the axes
20	Exam-style questions with markscheme
21	Common misconceptions and links
22	Check your understanding
23	Summary and topic map
24	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
find the gradient of a line and interpret it as a rate of change
write the equation of a line in gradient-intercept, point-gradient and general forms
use gradients to identify parallel and perpendicular lines
find distances, midpoints and points of intersection
define a function, use function notation and apply the vertical line test
state the domain and range of a function using correct notation
draw and sketch graphs, identifying intercepts, turning points, asymptotes and symmetry
solve equations graphically, including with technology

This booklet covers SL 2.1, 2.2, 2.3, 2.4, which is common to SL and HL.

Guiding questions

How can relationships between quantities be described with functions? What features of a graph tell us most about a function?

How this topic is examined

Lines and basic function concepts are tested throughout both papers, often as the opening parts of longer questions. Domain and range notation must be precise, and sketches must show key features clearly.

When sketching, label intercepts, turning points and asymptotes, and show the correct shape and end behaviour. On Paper 2, give answers read from a GDC to 3 significant figures.

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 Gradient and equations of straight lines

Gradient and a point fix a line.

Think about it

A taxi charges a fixed $4 plus $2.50 per km. What do the 4 and the 2.50 represent on a graph of cost against distance? What would the graph look like if the fixed charge were removed?

The gradient of the line through (x1, y1) and (x2, y2) is the change in y divided by the change in x:

A positive gradient slopes upward from left to right; negative slopes downward; zero is horizontal; a vertical line has undefined gradient. In context, the gradient is a rate of change: in C = 2.5d + 4, the cost rises by $2.50 per km.

There are three standard forms of the equation of a line:

Gradient-intercept form y = mx + c shows the gradient m and the y-intercept c. Point-gradient form is quickest when you know a point and the gradient. General form ax + by + d = 0 (usually with integer a, b, d) is often requested in exam questions. Vertical lines have equations x = k; horizontal lines y = k.

WORKED EXAMPLE 1 Equation through two points (basic)

Find the equation of the line through (2, −1) and (6, 7), in the form ax + by + d = 0.

Solution

m = (7 − (−1)) ÷ (6 − 2) = 2A1

y + 1 = 2(x − 2)M1

2x − y − 5 = 0A1

WORKED EXAMPLE 2 Interpreting a linear model (standard)

The temperature T (°C) of an oven t minutes after switching on is modelled by T = 18t + 20 for 0 ≤ t ≤ 10. Interpret 18 and 20, and find when T = 155.

Solution

18: the temperature rises 18 °C per minute; 20: initial (room) temperatureA1 A1

18t + 20 = 155, t = 7.5 minutesA1

WORKED EXAMPLE 3 Gradient from general form (standard)

Find the gradient and intercepts of 3x + 4y = 24.

Solution

4y = −3x + 24, so y = −0.75x + 6: m = −3/4A1

y-intercept (0, 6); x-intercept (8, 0)A1

Practice exercise 1

1. Find the gradient of the line through (−3, 5) and (1, −3).

2. Find the equation of the line with gradient 2/3 through (3, −1), in the form ax + by + d = 0.

3. Find the x- and y-intercepts of 5x − 2y = 20.

4. A plumber charges $45 call-out plus $30 per hour. Write a formula and find the cost of 3.5 hours.

5. Find k if the line through (1, k) and (4, 10) has gradient 3.

2 Parallel and perpendicular lines

Equal gradients; negative reciprocal gradients.

Parallel lines have equal gradients: m1 = m2. Perpendicular lines have gradients whose product is −1: m1m2 = −1, so m2 = −1/m1 (the negative reciprocal). A horizontal line is perpendicular to a vertical line.

Why it works

Rotating a line through 90° turns a 'run' of 1 and 'rise' of m into a 'run' of −m and 'rise' of 1. The new gradient is 1/(−m) = −1/m, so the product of the gradients is −1.

WORKED EXAMPLE 4 Perpendicular bisector (standard)

Find the equation of the perpendicular bisector of A(1, 3) and B(5, 1).

Solution

Midpoint (3, 2); gradient AB = −2/4 = −½M1

Perpendicular gradient 2A1

y − 2 = 2(x − 3): y = 2x − 4A1

WORKED EXAMPLE 5 Showing a right angle (exam level)

Show that the triangle with vertices P(0, 0), Q(4, 2) and R(1, 8) has a right angle.

Solution

mPQ = 1/2; mQR = 6/(−3) = −2; mPR = 8M1

mPQ × mQR = −1A1

So PQ ⊥ QR: the right angle is at QR1

Practice exercise 2

1. Find the gradient of a line perpendicular to y = −3x + 1.

2. Are 2x + 4y = 7 and y = −½x + 3 parallel?

3. Find the equation of the line through (2, 5) perpendicular to 3x − y = 4.

4. For what k are the lines y = kx + 1 and 4x − 2y = 3 perpendicular?

3 Distance, midpoint and intersections of lines

Coordinate geometry tools.

The distance formula is Pythagoras' theorem applied to the horizontal and vertical changes. The midpoint is the average of the coordinates. The intersection of two lines is found by solving their equations simultaneously.

WORKED EXAMPLE 6 Intersection (standard)

Find where y = 2x − 1 meets 3x + 2y = 12.

Solution

3x + 2(2x − 1) = 12, so 7x = 14M1

x = 2, y = 3: (2, 3)A1

Practice exercise 3

1. Find the distance between (−2, 3) and (4, −5).

2. Find the midpoint of (−1, 7) and (5, −3).

3. Find the intersection of x + y = 5 and 2x − y = 1.

4 The concept of a function

One input, exactly one output.

A function is a rule that assigns to each input exactly one output. We write f(x) = 2x + 3, or f: x ↦ 2x + 3. The value f(4) = 11 is the output when the input is 4; f(a + 1) = 2a + 5 means substituting a + 1 for x.

A mapping diagram shows inputs linked to outputs. A relation is a function if no input is linked to two outputs. On a graph, the vertical line test checks this: a graph represents a function if every vertical line meets it at most once. The circle x2 + y2 = 4 is not a function, because x = 0 gives y = ±2.

A function is one-to-one if each output comes from exactly one input (f(x) = 2x + 3), and many-to-one if some outputs come from several inputs (f(x) = x2, since f(2) = f(−2) = 4). This distinction matters for inverse functions (2.5).

WORKED EXAMPLE 7 Function notation (basic)

Given f(x) = 3x2 − 1, find (a) f(−2); (b) f(2a); (c) x such that f(x) = 26.

Solution

(a) 3(4) − 1 = 11A1

(b) 3(4a2) − 1 = 12a2 − 1A1

Practice exercise 4

1. If g(x) = 5 − 2x, find g(−3) and solve g(x) = −7.

2. If f(x) = x2 − 4x, find f(x + 1) and simplify.

3. Is y2 = x a function of x? Explain.

4. Is f(x) = x3 one-to-one or many-to-one?

5 Domain and range

Allowed inputs and possible outputs.

The domain is the set of input values for which the function is defined (or which are allowed in context). The range is the set of output values the function actually produces.

The natural domain is restricted by: division by zero (1/x needs x ≠ 0); square roots of negatives (√x needs x ≥ 0); logarithms of non-positive numbers (ln x needs x > 0); and context (time t ≥ 0, number of items must be a whole number).

To find the range, consider the smallest and largest outputs: use the vertex of a quadratic, the asymptote of a rational or exponential function, or a sketch.

Notation. Inequality: x ∈ ℝ, x ≥ 2. Interval: [2, ∞), where square brackets include the endpoint and round brackets exclude it. Set-builder: {x | x > 0}. Ranges are written in terms of the output: f(x) ≥ 3 or y ≥ 3.

Function	Domain	Range
f(x) = 2x + 1	x ∈ ℝ	f(x) ∈ ℝ
f(x) = x2	x ∈ ℝ	f(x) ≥ 0
f(x) = √(x − 3)	x ≥ 3	f(x) ≥ 0
f(x) = 1/x	x ≠ 0	f(x) ≠ 0
f(x) = ln x	x > 0	f(x) ∈ ℝ
f(x) = ex	x ∈ ℝ	f(x) > 0
f(x) = (x − 1)2 + 4	x ∈ ℝ	f(x) ≥ 4

WORKED EXAMPLE 8 Domain and range of a root function (standard)

Find the domain and range of f(x) = 3 + √(5 − x).

Solution

5 − x ≥ 0: domain x ≤ 5A1

√(5 − x) ≥ 0, so range f(x) ≥ 3A1

WORKED EXAMPLE 9 Restricted domain (exam level)

f(x) = x2 − 4x + 1 for 0 ≤ x ≤ 5. Find the range.

Solution

Vertex at x = 2: f(2) = −3 (minimum)M1

Endpoints: f(0) = 1, f(5) = 6A1

Range −3 ≤ f(x) ≤ 6A1

With a restricted domain, check the vertex and both endpoints.

Common mistake

Giving the range in terms of x. The range is a set of output values: write f(x) ≥ 3 or y ≥ 3, not x ≥ 3.

Practice exercise 5

1. State the domain of f(x) = 1/(x + 4).

2. State the domain and range of g(x) = √(2x − 6) + 1.

3. State the domain and range of h(x) = ln(x − 2).

4. Find the range of f(x) = 2x + 3 for −1 ≤ x < 4.

5. Find the range of f(x) = 5 − (x + 2)2.

6 Graphs of functions: drawing and sketching

Draw versus sketch.

To draw a graph means to plot it accurately, with labelled scales on both axes, either by plotting points or by copying a GDC display with a suitable window. To sketch means to show the general shape and the key features, labelled, without needing exact scale.

On Paper 2, use the GDC to graph functions and find features. On Paper 1 you must sketch using algebra: find intercepts by setting x = 0 and y = 0, find turning points by completing the square or calculus, and identify asymptotes from the equation.

Figure 1. A sketch shows intercepts and turning points.

GDC tip

Choose a window that shows all key features. If a graph looks like a straight line or disappears, zoom out; if turning points merge, zoom in. Use the 'zero', 'maximum', 'minimum' and 'intersect' functions rather than tracing by eye.

7 Key features of graphs

What to look for and label.

Intercepts: the y-intercept is f(0); the x-intercepts (zeros or roots) are the solutions of f(x) = 0.

Turning points: local maxima and minima, where the graph changes from increasing to decreasing or vice versa.

Asymptotes: lines that the graph approaches ever more closely. Vertical asymptotes occur where the function is undefined and the output grows without bound (x = 0 for 1/x). Horizontal asymptotes describe the value approached as x → ±∞ (y = 0 for e−x).

Symmetry: about the y-axis (even functions such as x2) or about the origin (odd functions such as x3). End behaviour: what happens as x → ±∞.

Figure 2. Vertical asymptote x = 2 and horizontal asymptote y = 1.

WORKED EXAMPLE 10 Identifying features (standard)

For f(x) = (x − 1)(x + 3), find the intercepts, the axis of symmetry and the vertex.

Solution

x-intercepts (1, 0) and (−3, 0); y-intercept (0, −3)A1

Axis x = (1 + (−3))/2 = −1A1

Vertex (−1, f(−1)) = (−1, −4)A1

Practice exercise 6

1. Write down the asymptotes of y = 3/(x + 1) − 2.

2. Find the intercepts of y = x2 − x − 6.

3. Use a GDC to find the coordinates of the turning points of y = x3 − 6x2 + 9x + 1.

8 Solving equations graphically

Where two graphs meet.

The solutions of f(x) = g(x) are the x-coordinates of the points of intersection of y = f(x) and y = g(x). Equivalently, they are the zeros of y = f(x) − g(x). On Paper 2, equations that cannot be solved algebraically, such as ex = x + 2, are solved with the GDC's intersect function.

WORKED EXAMPLE 11 Solving with technology (standard)

Solve ex = x + 2.

Solution

Graph y = ex and y = x + 2; find intersectionsM1

x = −1.84 or x = 1.15 (3 s.f.)A1 A1

Figure 3. Two points of intersection.

WORKED EXAMPLE 12 Number of solutions (exam level)

By sketching, determine how many solutions x3 − 3x = k has for k = 1 and for k = 3.

Solution

y = x3 − 3x has max (−1, 2) and min (1, −2)M1

k = 1 lies between −2 and 2: three solutionsA1

k = 3 > 2: one solutionA1

Practice exercise 7

1. Solve ln x = 3 − x.

2. Solve x2 = 2x for x < 0.

3. For what values of k does x3 − 3x = k have exactly two solutions?

9 Forms of the equation of a line

Choosing the most useful form.

The gradient-intercept form y = mx + c shows the gradient and y-intercept; the point-gradient form y − y1 = m(x − x1) is quickest from a point and a gradient; the general form ax + by + d = 0 (integers a, b, d) is often required in answers. Converting between them is routine algebra.

WORKED EXAMPLE 13 Converting (basic)

Write y − 3 = −(2/5)(x + 1) in the form ax + by + d = 0.

Solution

5y − 15 = −2x − 2M1

2x + 5y − 13 = 0A1

WORKED EXAMPLE 14 Intercepts (standard)

Find the intercepts and gradient of 3x − 4y + 12 = 0.

Solution

x-intercept (−4, 0); y-intercept (0, 3)A1

Gradient 3/4A1

Practice exercise 8

1. Find the line with gradient −3 through (2, −1) in general form.

2. Find the gradient and y-intercept of 2x + 7y = 14.

10 Parallel and perpendicular lines in practice

Gradients tell the story.

Parallel lines have equal gradients; perpendicular lines have gradients whose product is −1. A horizontal line (gradient 0) is perpendicular to a vertical line (undefined gradient).

WORKED EXAMPLE 15 Perpendicular through a point (standard)

Find the line through (4, 1) perpendicular to 2x + 3y = 6.

Solution

Gradient of given line −2/3; perpendicular 3/2A1

y − 1 = (3/2)(x − 4): 3x − 2y − 10 = 0M1 A1

WORKED EXAMPLE 16 Unknown coefficient (standard)

Find k if kx + 2y = 5 is perpendicular to y = 4x − 1.

Solution

Gradient −k/2 = −1/4M1

k = 1/2A1

Practice exercise 9

1. Find the line through (−2, 5) parallel to y = −3x + 1.

2. Show that A(1, 1), B(4, 5), C(8, 2) form a right angle at B.

11 Distances and midpoints in coordinate geometry

Pythagoras on a grid.

WORKED EXAMPLE 17 Isosceles triangle (standard)

Show that P(1, 2), Q(5, 4), R(3, 8) form an isosceles triangle.

Solution

PQ = √20, QR = √20, PR = √40A1 A1

PQ = QR, so isoscelesR1

Practice exercise 10

1. Find the perimeter of the triangle with vertices (0, 0), (6, 0), (6, 8).

2. Find the point dividing (2, 3) to (8, 9) in the ratio 1 : 2.

12 Perpendicular bisectors and circles

Equidistant points.

The perpendicular bisector of AB passes through the midpoint and is perpendicular to AB; every point on it is equidistant from A and B. The perpendicular bisectors of the sides of a triangle meet at the centre of the circle through the three vertices.

WORKED EXAMPLE 18 Circumcentre (exam level)

Find the centre of the circle through (0, 0), (8, 0) and (0, 6).

Solution

Bisector of the first side: x = 4; of the second: y = 3M1 A1

Centre (4, 3), radius 5A1

Practice exercise 11

1. Find the perpendicular bisector of (−1, 4) and (5, 2).

2. Find the centre of the circle through (1, 1), (5, 1), (1, 7).

13 Linear models in context

Gradient and intercept interpreted.

A linear model y = mx + c describes a quantity that changes at a constant rate m from an initial value c. Interpret both in context, with units, and state a sensible domain.

WORKED EXAMPLE 19 Phone plan (standard)

A plan costs $15 per month plus $0.08 per minute. Write the monthly cost and interpret the gradient.

Solution

C = 15 + 0.08mA1

Each extra minute costs 8 centsR1

WORKED EXAMPLE 20 Temperature conversion (standard)

Water freezes at 0 °C = 32 °F and boils at 100 °C = 212 °F. Find F in terms of C and convert 37 °C.

Solution

Gradient 180/100 = 1.8: F = 1.8C + 32M1 A1

37 °C = 98.6 °FA1

Practice exercise 12

1. A candle is 24 cm long and burns 1.5 cm per hour. Write its length after t hours and the time it burns out.

2. Find where the plans C = 15 + 0.08m and C = 25 + 0.04m cost the same.

14 Functions: notation, domain and range

Inputs and outputs.

A function maps each input in its domain to exactly one output. State domains by considering division by zero, square roots of negative numbers and logarithms of non-positive numbers. Find ranges from the graph, including turning points and asymptotes.

Function	Domain	Range
√(4 − x)	x ≤ 4	y ≥ 0
1/(x + 2)	x ≠ −2	y ≠ 0
x2 − 6x + 10	x ∈ ℝ	y ≥ 1
ln(x − 1)	x > 1	y ∈ ℝ

WORKED EXAMPLE 21 Range on a domain (standard)

Find the range of f(x) = 2x2 − 8x + 3 for 0 ≤ x ≤ 5.

Solution

Vertex (2, −5)A1

f(0) = 3, f(5) = 13A1

−5 ≤ f(x) ≤ 13A1

Practice exercise 13

1. State the domain and range of f(x) = 3 + √(x + 1).

2. Find the range of g(x) = 1/x for x ≥ 2.

15 Graphs and key features with technology

Intercepts, extrema and asymptotes.

Use the GDC to find zeros, maxima, minima and intersections, and to sketch graphs. When you transfer a graph to paper, label the axes and the key points with coordinates and draw asymptotes as dashed lines.

Figure 4. A cubic with a local maximum and minimum.

WORKED EXAMPLE 22 GDC features (standard)

Find the zeros of y = x3 − 3x + 1.

Solution

x = −1.88, 0.347, 1.53A2

Practice exercise 14

1. Find the intersections of y = x2 and y = 2x.

2. Find the maximum of y = xe−x.

16 Solving equations graphically

Where graphs meet.

The solutions of f(x) = g(x) are the x-coordinates of the intersections of y = f(x) and y = g(x). This is the main method for equations that cannot be solved algebraically, such as ex = 3 − x or ln x = x − 2.

WORKED EXAMPLE 23 Graphical solution (standard)

Solve ln x = x − 2.

Solution

Graph both sidesM1

x = 0.159 and x = 3.15A1 A1

Practice exercise 15

1. Solve ex = 3 − x.

2. Solve x3 = 2x + 1.

17 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 24 Mixed problem (exam level)

The line L passes through A(2, 5) and B(6, −3). (a) Find its equation. (b) Find the line through C(0, −2) perpendicular to L. (c) Find the intersection.

Solution

(a) gradient −2: y = −2x + 9A1

(b) y = x/2 − 2A1

Practice exercise 16

1. Find the shortest distance from C(0, −2) to L in the example.

2. Find k if (k, 3) lies on the perpendicular bisector of (1, 1) and (5, 5).

18 Additional exam-style practice

Questions in the style of Papers 1 and 2.

WORKED EXAMPLE 25 Paper 1 question (exam level)

f(x) = (2x − 1)/(x + 3). (a) State the domain. (b) Find the intercepts. (c) Find the asymptotes.

Solution

(a) x ≠ −3A1

(b) (0.5, 0) and (0, −1/3)A1 A1

Practice exercise 17

1. Find the area of the triangle formed by 2x + y = 8 and the axes.

2. Find the angle between y = 2x and the x-axis.

19 The angle a line makes with the axes

Gradient as tan θ.

The gradient m of a line equals tan θ, where θ is the angle the line makes with the positive x-axis. The acute angle between two lines with gradients m1 and m2 follows from the difference of these angles.

WORKED EXAMPLE 26 Angle between lines (exam level)

Find the acute angle between y = 2x + 1 and y = −x + 4.

Solution

Angles with the x-axis: 63.4° and 135°M1 A1

Difference 71.6°: acute angle 71.6°A1

Practice exercise 18

1. Find the gradient of a line at 30° to the x-axis.

2. Find the acute angle between y = 3x and y = x/3.

20 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

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

Points A(−2, 5) and B(4, −1). The function f(x) = 2 + √(x + 4).

(a) Find the equation of line AB. [2]

(b) Find the equation of the line perpendicular to AB through B. [2]

(d) Find the x-coordinate where y = f(x) meets line AB. [4]

Markscheme

Part	Answer	Marks
(a)	m = −1	A1
	y = −x + 3	A1
(b)	gradient 1	A1
	y = x − 5	A1
(c)	x ≥ −4	A1
	f(x) ≥ 2	A1
(d)	2 + √(x + 4) = 3 − x	M1
	x + 4 = (1 − x)2: x2 − 3x − 3 = 0	A1
	x = (3 ± √21)/2	A1
	x = (3 − √21)/2 only, since 1 − x ≥ 0	R1

Examiner's comments. In (d), squaring introduces a spurious root that must be rejected with a reason.

Question 2 (Paper 2 style) (about 10 minutes)

Consider f(x) = x3 − 4x2 + x + 6 and g(x) = 2x.

(a) Find the zeros of f. [2]

(b) Find the coordinates of the local maximum of f. [2]

Markscheme

Part	Answer	Marks
(a)	x = −1, 2, 3	A2
(b)	(0.131, 6.06)	A2
(c)	intersections by GDC	M1
	x = −0.956, 1.35, 4.69	A2

Examiner's comments. Answers from a GDC should be given to 3 s.f.; listing all intersections in (c) was essential.

21 Common misconceptions and links

Students often think…	Actually…
perpendicular gradients are reciprocals	negative reciprocals
the range is given in terms of x	the range describes outputs
a sketch must be to scale	a sketch shows shape and key features
every graph is a function	use the vertical line test

How 2.1 connects to the rest of the course

Topic	Connection
2.5 Composite and inverse functions	domain and range
2.6 Quadratics	key features
3.1 Geometry	distance and midpoint in 3D
5.1 Differentiation	tangents and normals

22 Check your understanding

Section A: multiple choice

1. The gradient of 3x + 2y = 6 is

A. −3/2 B. 3/2 C. 3 D. −2/3

2. Perpendicular to gradient 4:

A. 1/4 B. −4 C. −1/4 D. 4

3. The domain of ln(x − 1) is

A. x ≥ 1 B. x > 1 C. x > 0 D. all reals

4. The range of x2 + 3 is

A. f(x) ≥ 0 B. all reals C. f(x) > 3 D. f(x) ≥ 3

5. A vertical line meets a function's graph

A. exactly twice B. at most once C. never D. infinitely often

6. The y-intercept of y = 2x − 7 is

A. −7 B. 2 C. 7 D. 3.5

Section B: structured

7. Show that the lines 2x − 3y = 1 and 3x + 2y = 4 are perpendicular. [2]

8. Find the domain and range of f(x) = 1/(x + 3) − 2. [2]

9. Sketch y = x(x − 3)2, labelling intercepts. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A

7. Gradients 2/3 and −3/2: product −1.

8. Domain x ≠ −3; range f(x) ≠ −2.

9. Intercepts (0, 0) crossing and (3, 0) touching; positive cubic shape.

23 Summary and topic map

Idea	Remember
Lines	y = mx + c; y − y1 = m(x − x1); ax + by + d = 0
Parallel / perpendicular	m1 = m2; m1m2 = −1
Distance, midpoint	Pythagoras; averages
Function	each input has exactly one output
Domain and range	inputs and outputs; restrictions
Key features	intercepts, turning points, asymptotes, symmetry
Intersections	solve f(x) = g(x)

Before the exam, make sure you…

•give ranges in terms of f(x)
•label key features on sketches
•use technology for intersections on Paper 2
•check for spurious roots

Lines and 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 5. Topic map for Lines and Functions.

24 Answers to practice exercises

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

Practice exercise 1

1. −2

2. 2x − 3y − 9 = 0

3. (4, 0) and (0, −10)

4. C = 30h + 45; $150

5. k = 1

Practice exercise 2

1. 1/3

2. yes: both have gradient −½

3. y = −x/3 + 17/3

4. k = −½

Practice exercise 3

1. 10

2. (2, 2)

3. (2, 3)

Practice exercise 4

1. 11; x = 6

2. x2 − 2x − 3

3. no: x = 4 gives y = ±2

4. one-to-one

Practice exercise 5

1. x ≠ −4

2. x ≥ 3; g(x) ≥ 1

3. x > 2; h(x) ∈ ℝ

4. 1 ≤ f(x) < 11

5. f(x) ≤ 5

Practice exercise 6

1. x = −1, y = −2

2. (−2, 0), (3, 0), (0, −6)

3. max (1, 5), min (3, 1)

Practice exercise 7

1. x = 2.21

2. x = −0.767

3. k = ±2

Practice exercise 8

1. 3x + y − 5 = 0

2. −2/7; 2

Practice exercise 9

1. y = −3x − 1

2. gradients 4/3 and −3/4: product −1

Practice exercise 10

1. 24

2. (4, 5)

Practice exercise 11

1. y = 3x − 3

2. (3, 4)

Practice exercise 12

1. L = 24 − 1.5t; 16 hours

2. m = 250 minutes

Practice exercise 13

1. x ≥ −1; f(x) ≥ 3

2. 0 < g(x) ≤ ½

Practice exercise 14

1. x = −0.767, 2, 4

2. (1, 0.368)

Practice exercise 15

1. 0.792

2. x = −1, −0.618, 1.62

Practice exercise 16

1. distance from C to (4.4, 0.2) = √24.2 = 4.92

2. x + y = 6: k = 3

Practice exercise 17

1. 16

2. 63.4°

Practice exercise 18

1. √3/3

2. 53.1°

DP MATHEMATICS AA · TOPIC 2: FUNCTIONS

2.5

Composite and Inverse Functions

Full notes for HL · 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. HL students extend this topic in booklet 2.13.

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

DP MATHEMATICS AA · TOPIC 2: FUNCTIONS

2.15

Inequalities and Modulus Functions

Full notes for HL · Syllabus AHL 2.15, 2.16

Contents

1	Solving polynomial inequalities
2	Solving inequalities graphically
3	Inequalities with rational expressions
4	The modulus function
5	Graphs of y = \|f(x)\|
6	Graphs of y = f(\|x\|)
7	Modulus equations
8	Modulus inequalities
9	The modulus function
10	Graphs of y = \|f(x)\|
11	Graphs of y = f(\|x\|)
12	Modulus equations
13	Modulus inequalities
14	Polynomial inequalities with sign diagrams
15	Rational inequalities
16	Inequalities with technology
17	Mixed review: worked problems
18	Additional exam-style practice
19	Modulus in context
20	Exam-style questions with markscheme
21	Common misconceptions and links
22	Check your understanding
23	Summary and topic map
24	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
solve polynomial inequalities using critical values and sign analysis
solve inequalities graphically, including with technology
solve rational inequalities without multiplying by expressions of unknown sign
understand the modulus function as distance and sketch y = \|x\|
sketch y = \|f(x)\| and y = f(\|x\|) from y = f(x)
solve modulus equations by considering cases or squaring
solve modulus inequalities algebraically and graphically

This booklet covers Additional Higher Level (AHL) content (AHL 2.15, 2.16). It is studied by HL students only.

Guiding questions

How do we solve inequalities when the sign of an expression changes? How does taking the modulus transform a graph?

How this topic is examined

Inequalities and modulus functions are tested on Paper 1 (algebraic methods and sketches) and Paper 2 (graphical solutions with technology). They often appear as the final part of a function question.

Never multiply both sides of an inequality by an expression unless you know its sign. Check modulus solutions in the original equation.

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 Solving polynomial inequalities

Find critical values, then test signs.

Think about it

Why is it wrong to solve 1/x < 2 by multiplying both sides by x to get 1 < 2x? Try x = −1.

For a polynomial inequality: bring everything to one side so the other side is zero; factorize; find the critical values (the roots); then determine the sign of the expression in each interval between critical values, using a sign table or a sketch.

A factor (x − a) changes sign at x = a. A squared factor (x − a)2 does not change sign: it is never negative. Include critical values in the answer for ≤ or ≥, exclude them for < or >.

WORKED EXAMPLE 1 A cubic inequality (standard)

Solve (x + 2)(x − 1)(x − 4) > 0.

Solution

Critical values −2, 1, 4M1

Signs: x < −2 negative; −2 < x < 1 positive; 1 < x < 4 negative; x > 4 positiveM1

−2 < x < 1 or x > 4A1

WORKED EXAMPLE 2 Rearranging first (exam level)

Solve x3 ≥ 4x.

Solution

x3 − 4x ≥ 0: x(x − 2)(x + 2) ≥ 0M1

Critical values −2, 0, 2; positive cubicA1

−2 ≤ x ≤ 0 or x ≥ 2A1

Common mistake

Dividing both sides of x3 ≥ 4x by x. This loses solutions and may reverse the inequality when x < 0.

Practice exercise 1

1. Solve (x − 3)(x + 1)(x − 5) ≤ 0.

2. Solve x3 < x.

3. Solve (x − 2)2(x + 3) > 0.

4. Solve x4 − 5x2 + 4 < 0.

2 Solving inequalities graphically

Where one graph lies above another.

To solve f(x) > g(x), find where the graph of y = f(x) lies above y = g(x). The boundaries of the solution set are the intersection points, found algebraically or with technology. Equivalently, find where y = f(x) − g(x) is above the x-axis.

WORKED EXAMPLE 3 With technology (standard)

Solve ex > 3 − x2.

Solution

Intersections at x = −1.68 and x = 0.834M1 A1

ex is above outside these: x < −1.68 or x > 0.834A1

Practice exercise 2

1. Solve ln x < 2 − x.

2. Solve x2 > 2x for x ∈ ℝ.

3 Inequalities with rational expressions

Avoid multiplying by unknown signs.

For inequalities such as (x + 1)/(x − 2) ≥ 3, there are two safe methods. Method 1: subtract to get zero on one side, combine into a single fraction, find critical values from the numerator and denominator, and analyse signs. Method 2: multiply both sides by the square of the denominator, which is always positive (for x ≠ 2), and solve the resulting polynomial inequality.

Values that make the denominator zero are never included in the solution.

WORKED EXAMPLE 4 Method 1 (standard)

Solve (x + 1)/(x − 2) ≥ 3.

Solution

(x + 1)/(x − 2) − 3 ≥ 0, so (−2x + 7)/(x − 2) ≥ 0M1

Critical values 2 (excluded) and 3.5A1

Sign analysis: 2 < x ≤ 3.5A1

WORKED EXAMPLE 5 Method 2 (exam level)

Solve 2x/(x + 1) < 1.

Solution

Multiply by (x + 1)2: 2x(x + 1) < (x + 1)2M1

(x + 1)(2x − x − 1) < 0, so (x + 1)(x − 1) < 0A1

−1 < x < 1A1

Common mistake

Multiplying by (x − 2) without considering that it may be negative, which reverses the inequality and gives a wrong or incomplete answer.

Practice exercise 3

1. Solve 1/x < 2.

2. Solve (x − 3)/(x + 2) ≤ 0.

3. Solve 3/(x − 1) > x + 1.

4. Solve (2x + 1)/(x − 3) ≥ 1.

4 The modulus function

Distance from zero.

The modulus (absolute value) |x| is the distance of x from zero on the number line, so it is never negative:

|a − b| is the distance between a and b on the number line. Key results (for a > 0): |x| < a ⇔ −a < x < a; |x| > a ⇔ x < −a or x > a; |ab| = |a||b|; |a/b| = |a|/|b|; |x|2 = x2; √(x2) = |x|.

The graph of y = |x| is a V shape with its vertex at the origin, gradient −1 for x < 0 and 1 for x > 0. y = |ax + b| is a V with vertex at x = −b/a.

Practice exercise 4

1. Evaluate |−7| − |3 − 8|.

2. Write |x − 2| < 5 without modulus signs.

3. Explain why √(x2) = |x| rather than x.

5 Graphs of y = |f(x)|

Reflect the negative part.

To sketch y = |f(x)| from y = f(x): keep the parts of the graph on or above the x-axis; reflect the parts below the x-axis in the x-axis. The result is never below the x-axis. Roots become points where the graph touches the axis (often with a sharp corner, a cusp).

Figure 1. y = |f(x)| reflects the negative part upward.

Practice exercise 5

1. Sketch y = |2x − 4|, labelling the vertex and intercepts.

2. Sketch y = |ln x|, labelling the key point.

3. State the range of y = |sin x|.

6 Graphs of y = f(|x|)

Mirror the right half.

To sketch y = f(|x|) from y = f(x): keep the part of the graph for x ≥ 0; discard the part for x < 0; reflect the kept part in the y-axis. The result is always an even function (symmetric about the y-axis), because f(|−x|) = f(|x|).

Figure 2. y = f(|x|) mirrors the right half in the y-axis.

Transformation	Rule	Result
y = \|f(x)\|	reflect parts below the x-axis upward	never negative
y = f(\|x\|)	keep x ≥ 0 and reflect in the y-axis	always even

Practice exercise 6

1. Sketch y = |x|2 − 4|x| for −5 ≤ x ≤ 5.

2. Sketch y = e|x|.

3. Sketch y = |f(|x|)| for f(x) = x − 2.

7 Modulus equations

Consider cases or square both sides.

|f(x)| = k (k ≥ 0) means f(x) = k or f(x) = −k. |f(x)| = g(x): solve f(x) = g(x) and f(x) = −g(x), then check each solution, since g(x) must be non-negative. |f(x)| = |g(x)|: square both sides, f2 = g2, or solve f = ±g. A sketch shows how many solutions to expect.

WORKED EXAMPLE 6 Modulus equation (standard)

Solve |2x − 3| = x + 1.

Solution

2x − 3 = x + 1 gives x = 4A1

2x − 3 = −(x + 1) gives x = 2/3A1

Both satisfy x + 1 ≥ 0: x = 4 or x = 2/3R1

WORKED EXAMPLE 7 Rejecting a solution (exam level)

Solve |x + 4| = 2x − 1.

Solution

x + 4 = 2x − 1 gives x = 5A1

x + 4 = −(2x − 1) gives x = −1A1

At x = −1, 2x − 1 = −3 < 0: reject. x = 5R1

Practice exercise 7

1. Solve |3x − 1| = 5.

2. Solve |x − 2| = |2x + 1|.

3. Solve |x2 − 4| = 3.

4. Solve |x| = x2 − 6.

8 Modulus inequalities

Use the definitions or square.

|f(x)| < k ⇔ −k < f(x) < k. |f(x)| > k ⇔ f(x) < −k or f(x) > k. |f(x)| > |g(x)| ⇔ f2 > g2, which can be solved as a polynomial inequality. For |f(x)| compared with a non-modulus expression, use a sketch and the intersection points.

WORKED EXAMPLE 8 Squaring (standard)

Solve |x − 2| > |2x + 1|.

Solution

Square: (x − 2)2 > (2x + 1)2M1

3x2 + 8x − 3 < 0, (3x − 1)(x + 3) < 0A1

−3 < x < 1/3A1

WORKED EXAMPLE 9 Graphical modulus inequality (exam level)

Solve |x2 − 4| < 3.

Solution

−3 < x2 − 4 < 3, so 1 < x2 < 7M1

x2 > 1: |x| > 1; x2 < 7: |x| < √7A1

−√7 < x < −1 or 1 < x < √7A1

Practice exercise 8

1. Solve |3x − 1| < 5.

2. Solve |x + 3| ≤ 2|x|.

3. Solve |2x − 1| > x + 2.

9 The modulus function

Distance from zero.

|x| = x if x ≥ 0 and −x if x < 0. Geometrically, |x − a| is the distance between x and a on the number line. The graph of y = |x| is a V shape with vertex at the origin.

WORKED EXAMPLE 10 Evaluating (basic)

Evaluate |3 − 7| + |−2| − |5|.

Solution

4 + 2 − 5M1

= 1A1

WORKED EXAMPLE 11 Distance interpretation (standard)

Describe the set of x with |x − 3| < 2.

Solution

Points within 2 of 3R1

1 < x < 5A1

Practice exercise 9

1. Solve |x + 1| = 4.

2. Describe |x + 2| ≥ 5 as an inequality without modulus.

10 Graphs of y = |f(x)|

Reflect the negative parts.

The graph of y = |f(x)| keeps the parts of y = f(x) on or above the x-axis and reflects the parts below the axis in the x-axis. Zeros of f become points where the graph touches or bounces off the axis.

Figure 3. The part below the axis is reflected.

WORKED EXAMPLE 12 Sketch (standard)

Describe y = |x2 − 2x − 3|.

Solution

Roots −1 and 3; between them the curve is reflectedA1

Local maximum (1, 4)A1

Practice exercise 10

1. Sketch y = |2x − 4|.

2. Sketch y = |sin x| for 0 ≤ x ≤ 2π.

11 Graphs of y = f(|x|)

Mirror the right-hand side.

y = f(|x|) uses only the values of f for x ≥ 0, reflected in the y-axis. The graph is always symmetric about the y-axis (an even function).

Figure 4. f(|x|) mirrors the right half.

WORKED EXAMPLE 13 Sketch (standard)

Describe y = (|x| − 1)2 − 2.

Solution

Right half of the parabola with vertex (1, −2), mirroredA1

Minima at (±1, −2), local maximum (0, −1)A1

Practice exercise 11

1. Sketch y = sin|x|.

2. Sketch y = e|x|.

12 Modulus equations

Square, or split into cases.

To solve |f(x)| = |g(x)|, either square both sides (f2 = g2) or solve f = g and f = −g. For |f(x)| = g(x), split into cases and check each solution satisfies g(x) ≥ 0.

WORKED EXAMPLE 14 Two moduli (standard)

Solve |2x − 1| = |x + 4|.

Solution

2x − 1 = x + 4 gives x = 5A1

2x − 1 = −x − 4 gives x = −1A1

WORKED EXAMPLE 15 Checking solutions (exam level)

Solve |x − 3| = 2x.

Solution

x − 3 = 2x gives x = −3 (reject: 2x < 0)M1 R1

−x + 3 = 2x gives x = 1 ✓A1

Practice exercise 12

1. Solve |3x + 2| = 7.

2. Solve |x2 − 5| = 4.

13 Modulus inequalities

Distances and squaring.

|f(x)| < a means −a < f(x) < a; |f(x)| > a means f(x) < −a or f(x) > a. For |f(x)| < |g(x)|, square both sides (both sides are non-negative) and solve the resulting inequality.

WORKED EXAMPLE 16 Two moduli (exam level)

Solve |x + 2| > |2x − 1|.

Solution

Square: x2 + 4x + 4 > 4x2 − 4x + 1M1

3x2 − 8x − 3 < 0: (3x + 1)(x − 3) < 0A1

−1/3 < x < 3A1

Practice exercise 13

1. Solve |2x − 5| ≤ 3.

2. Solve |x − 1| < |x + 3|.

14 Polynomial inequalities with sign diagrams

Mark the roots.

Factorize, mark the zeros on a number line, and find the sign of the expression in each interval. The sign changes at simple roots and stays the same at double roots.

WORKED EXAMPLE 17 Quartic (exam level)

Solve (x − 1)2(x + 2)(x − 4) ≥ 0.

Solution

Sign diagram: positive for x < −2 and x > 4; zero at −2, 1, 4M1 A1

x ≤ −2 or x = 1 or x ≥ 4A1

Practice exercise 14

1. Solve x3 − x2 − 6x > 0.

2. Solve (x + 1)2(3 − x) < 0.

15 Rational inequalities

Never multiply by an unknown sign.

Two safe methods: (1) move everything to one side, combine into a single fraction and use a sign diagram of numerator and denominator; (2) multiply both sides by the square of the denominator, which is positive. Exclude values that make the denominator zero.

WORKED EXAMPLE 18 Method 1 (exam level)

Solve (x + 1)/(x − 2) ≤ 3.

Solution

(x + 1 − 3x + 6)/(x − 2) = (7 − 2x)/(x − 2) ≤ 0M1 A1

Sign diagram: x < 2 or x ≥ 3.5A1

Practice exercise 15

1. Solve 1/(x − 3) > 2.

2. Solve x/(x + 1) ≥ x.

16 Inequalities with technology

Graphs make it visible.

To solve f(x) > g(x) graphically, find the intersections of y = f(x) and y = g(x) and read off where the graph of f is above the graph of g. This is the standard method for inequalities involving exponentials, logarithms or trigonometric functions.

WORKED EXAMPLE 19 Graphical inequality (standard)

Solve ex > x + 2.

Solution

Intersections x = −1.84 and 1.15M1 A1

x < −1.84 or x > 1.15A1

Practice exercise 16

1. Solve ln x < x − 2.

2. Solve |x2 − 4| < 3x with technology.

17 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 20 Mixed problem (exam level)

f(x) = |x − 2| − 3. (a) Sketch f. (b) Solve f(x) = ½x.

Solution

(a) V with vertex (2, −3), intercepts (−1, 0) and (5, 0)A1

(b) x ≥ 2: x − 5 = x/2, x = 10A1

x < 2: −x − 1 = x/2, x = −2/3; x = 10 or −2/3A1

Practice exercise 17

1. Find the number of solutions of |x2 − 4| = k for k = 2, 4 and 5.

2. Solve |x| + |x − 2| = 4.

18 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 21 Paper 1 question (exam level)

Solve |x2 − 3x| < 2.

Solution

−2 < x2 − 3x < 2M1

x2 − 3x + 2 > 0: x < 1 or x > 2; x2 − 3x − 2 < 0: (3 − √17)/2 < x < (3 + √17)/2A1 A1

(3 − √17)/2 < x < 1 or 2 < x < (3 + √17)/2A1

Practice exercise 18

1. Solve (x − 2)/(x + 1) < (x + 1)/(x − 2).

19 Modulus in context

Tolerances and errors.

Modulus inequalities describe tolerances: a part must have length L with |L − 50| ≤ 0.2 mm, so 49.8 ≤ L ≤ 50.2. Absolute error is |measured − true|; percentage error divides by the true value.

WORKED EXAMPLE 22 Tolerance (standard)

A bolt must be 12 mm ± 0.05 mm. Write this with a modulus and find the acceptable range.

Solution

|d − 12| ≤ 0.05A1

11.95 ≤ d ≤ 12.05A1

Practice exercise 19

1. A thermostat keeps T within 1.5 °C of 21 °C. Write this as a modulus inequality.

20 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

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

(a) Solve 2x/(x + 1) < 1. [4]

(b) Sketch y = |x2 − 4| and hence solve |x2 − 4| = 3. [4]

Markscheme

Part	Answer	Marks
(a)	2x/(x + 1) − 1 < 0, (x − 1)/(x + 1) < 0	M1 A1
	critical values −1, 1	A1
	−1 < x < 1	A1
(b)	W-shaped graph, cusps at (±2, 0), local maximum (0, 4)	A1 A1
	x2 − 4 = ±3	M1
	x = ±√7, ±1	A1
(c)	(x + 3)2 ≤ 4x2: 3x2 − 6x − 9 ≥ 0	M1
	(x − 3)(x + 1) ≥ 0	A1
	x ≤ −1 or x ≥ 3	A1

Examiner's comments. In (a), multiplying by (x + 1) without care led many candidates to an incomplete answer.

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

Let f(x) = 2x − x2.

(a) Sketch y = f(|x|), labelling intercepts and turning points. [3]

(b) Solve f(|x|) = −3. [3]

Markscheme

Part	Answer	Marks
(a)	even graph; zeros at 0, ±2	A1
	maxima at (±1, 1)	A1
	correct shape (cusp at origin)	A1
(b)	2\|x\| − \|x\|2 = −3: \|x\|2 − 2\|x\| − 3 = 0	M1
	\|x\| = 3 (reject −1)	A1
	x = ±3	A1
(c)	−3 ≤ 2x − x2 ≤ 3: right inequality always true	M1
	x2 − 2x − 3 ≤ 0	A1
	−1 ≤ x ≤ 3	A1

Examiner's comments. In (b), treating |x| as a single variable gives a neat quadratic.

21 Common misconceptions and links

Students often think…	Actually…
multiplying inequalities by expressions of unknown sign	use sign analysis or the squared denominator
\|x\| < 3 means x < 3	−3 < x < 3
f(\|x\|) reflects negative y-values	that is \|f(x)\|
all solutions of \|f(x)\| = g(x) are valid	check g(x) ≥ 0

How 2.15 connects to the rest of the course

Topic	Connection
2.6 Quadratic inequalities	sign analysis
2.11 Transformations	reflections
2.12 Polynomials	roots and signs
1.12 Complex numbers	modulus of a complex number

22 Check your understanding

Section A: multiple choice

1. |x| < 4 means

A. −4 < x < 4 B. x < 4 C. x > −4 D. x < −4 or x > 4

2. |−7| =

A. −7 B. 0 C. 7 D. 49

3. y = |f(x)| is never

A. above the x-axis B. below the x-axis C. symmetric about y = x D. periodic

4. y = f(|x|) is always

A. odd B. one-to-one C. increasing D. even

5. x(x − 3) < 0 for

A. x < 0 B. 0 < x < 3 C. x > 3 D. x < 0 or x > 3

Section B: structured

6. Solve x3 ≥ 4x. [3]

7. Solve |3x − 1| < 5. [2]

8. Explain the difference between y = |f(x)| and y = f(|x|). [2]

Answers

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

6. x(x − 2)(x + 2) ≥ 0: −2 ≤ x ≤ 0 or x ≥ 2.

7. −5 < 3x − 1 < 5: −4/3 < x < 2.

8. |f(x)| reflects negative parts upward; f(|x|) mirrors the right half across the y-axis.

23 Summary and topic map

Idea	Remember
Polynomial inequalities	critical values and sign analysis
Graphical	where one graph is above another
Rational	one side zero, or multiply by the squared denominator
Modulus	distance from zero; \|x\| < a ⇔ −a < x < a
\|f(x)\|	reflect negative parts
f(\|x\|)	mirror the right half
Equations	cases or squaring; check

Before the exam, make sure you…

•never multiply by unknown signs
•check modulus solutions
•square both sides for |f| vs |g|
•sketch to confirm the number of solutions

Inequalities and Modulus 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 5. Topic map for Inequalities and Modulus.

24 Answers to practice exercises

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

Practice exercise 1

1. x ≤ −1 or 3 ≤ x ≤ 5

2. x < −1 or 0 < x < 1

3. x > −3, x ≠ 2

4. −2 < x < −1 or 1 < x < 2

Practice exercise 2

1. 0 < x < 1.56

2. x < −0.767 or 2 < x < 4

Practice exercise 3

1. x < 0 or x > ½

2. −2 < x ≤ 3

3. x < −2 or 1 < x < 2

4. x ≤ −4 or x > 3

Practice exercise 4

1. 2

2. −3 < x < 7

3. for x < 0, √(x2) is positive, equal to −x

Practice exercise 5

1. V with vertex (2, 0); y-intercept (0, 4)

2. touches the axis at (1, 0); asymptote x = 0

3. 0 ≤ y ≤ 1

Practice exercise 6

1. W shape: roots at 0, ±4; minima at (±2, −4)

2. symmetric, minimum (0, 1)

3. W shape with zeros at ±2, value 2 at x = 0

Practice exercise 7

1. x = 2 or x = −4/3

2. x = −3 or x = 1/3

3. x = ±√7, ±1

4. x = ±3

Practice exercise 8

1. −4/3 < x < 2

2. x ≤ −1 or x ≥ 3

3. x < −1/3 or x > 3

Practice exercise 9

1. x = 3 or −5

2. x ≤ −7 or x ≥ 3

Practice exercise 10

1. V with vertex (2, 0)

2. two positive arches

Practice exercise 11

1. even; equals sin x for x ≥ 0

2. V-like curve with minimum (0, 1)

Practice exercise 12

1. 5/3, −3

2. x = ±3, ±1

Practice exercise 13

1. 1 ≤ x ≤ 4

2. x > −1

Practice exercise 14

1. −2 < x < 0 or x > 3

2. x > 3

Practice exercise 15

1. 3 < x < 3.5

2. x < −1 or x = 0

Practice exercise 16

1. 0 < x < 0.159 or x > 3.15

2. 1 < x < 4

Practice exercise 17

1. 4, 3, 2

2. x = −1 or 3

Practice exercise 18

1. (6x − 3)/((x + 1)(x − 2)) > 0: −1 < x < 1/2 or x > 2

Practice exercise 19

1. |T − 21| ≤ 1.5

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