DP Maths AI HL 2.2 Functions and Their Properties

DP Maths AI HL 2.2 Functions and Their Properties - Full notes

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DP Maths AI HL | 2.2 Functions and Their Properties

DP MATHEMATICS AI · TOPIC 2: FUNCTIONS

2.2

Functions and Their Properties

Full notes for Applications and Interpretation HL

Contents

1	Linear relationships and gradient
2	Equations of straight lines
3	The concept of a function
4	Domain and range
5	One-to-one and many-to-one functions
6	Key features of graphs
7	Drawing and sketching graphs with technology
8	Inverse functions
9	Piecewise functions
10	Direct and inverse variation
11	Linear models: cost, revenue and break-even
12	Quadratic models
13	Exponential models
14	Choosing a model and a sensible domain
15	Solving equations graphically
16	Interpreting graphs in context
17	Using technology efficiently
18	Mixed review: worked problems
19	Composite functions (HL)
20	Inverses, restricted domains and composites in models (HL)
21	Investigation: choosing a phone plan
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 practice exercise.

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 gradients and equations of lines and interpret them in context
use function notation and identify functions
state domains and ranges, including in context
identify key features of graphs with technology
find and interpret inverse functions
work with piecewise functions and direct and inverse variation
build linear, quadratic and exponential models and choose between them
solve equations graphically and interpret results

HL booklet: this booklet contains all the SL content plus sections on composite functions and inverse functions with restricted domains.

Guiding questions

How can functions describe relationships between real quantities? How do we choose the right function, and the right domain, for a situation?

How this topic is examined

Functions underpin the whole AI course. Paper 1 questions test notation, domain and range, gradients and simple inverses; Paper 2 questions build models from context, use the GDC to find key features and intersections, and ask for interpretations.

In context, always give units, state a sensible domain, and explain what intercepts, gradients and maxima mean.

Mathematics: Applications and Interpretation places strong emphasis on technology, modelling and interpretation. Most papers allow a GDC, so you are expected to use it efficiently, record the method you used, and interpret every answer in context.

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, including what you entered into a GDC
Show that	obtain the given result; every step must be shown and the result must not be assumed
Estimate	obtain an approximate value
Sketch	a diagram showing the general shape and key features, labelled
Interpret / Comment	explain the meaning of a result in context, or give a judgement
Justify / Explain	give valid reasons or evidence
Hence	use the previous result

How to use these notes

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or explanation of a key result
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
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), graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet

1 Linear relationships and gradient

Constant rate of change.

Think about it

A taxi charges a fixed $4 plus $2.20 per km. Sketch cost against distance. What do the 4 and the 2.20 mean on the graph?

The gradient m measures the rate of change of y with respect to x: the change in y for each unit increase in x. Positive gradients slope upwards, negative downwards, zero is horizontal; vertical lines have undefined gradient. In context, the units of the gradient are 'units of y per unit of x'.

Parallel lines have equal gradients; perpendicular lines have gradients with product −1 (m2 = −1/m1).

WORKED EXAMPLE 1 Gradient in context (basic)

A plant is 12 cm tall after 3 weeks and 30 cm after 9 weeks. Find the average growth rate.

Solution

m = (30 − 12)/(9 − 3)M1

= 3 cm per weekA1

Practice exercise 1

1. Find the gradient through (−2, 7) and (4, −5).

2. Find the gradient perpendicular to y = 0.4x − 3.

3. A car's fuel falls from 52 L to 31 L over 300 km. Interpret the gradient.

2 Equations of straight lines

Three useful forms.

Gradient-intercept form shows the gradient and y-intercept; point-gradient form is fastest when a point and gradient are known; general form (with integer a, b, d) is often requested. In context, c is usually the initial or fixed value.

WORKED EXAMPLE 2 Line through two points (standard)

Find the equation of the line through (2, 5) and (6, −3) in the form ax + by + d = 0.

Solution

m = −2A1

y − 5 = −2(x − 2)M1

2x + y − 9 = 0A1

WORKED EXAMPLE 3 Perpendicular line (standard)

Find the line through (3, 1) perpendicular to 3x − 4y = 12.

Solution

Given gradient 3/4, so perpendicular gradient −4/3A1

y − 1 = −(4/3)(x − 3): 4x + 3y − 15 = 0M1 A1

Practice exercise 2

1. Find the equation of the line with gradient −3 through (1, 4).

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

3. Find the intersection of y = 2x + 1 and x + y = 7.

3 The concept of a function

Each input has exactly one output.

A function assigns to each input exactly one output. We write f(x) = 3x + 2 or f: x ↦ 3x + 2, and f(4) = 14 is the output for input 4. Functions describe dependence: the output (dependent variable) depends on the input (independent variable).

A graph represents a function if every vertical line meets it at most once (the vertical line test). Circles and sideways parabolas fail this test.

WORKED EXAMPLE 4 Function notation (basic)

Given f(x) = x2 − 3x, find f(−2), f(a + 1) and solve f(x) = 10.

Solution

f(−2) = 10A1

f(a + 1) = a2 + 2a + 1 − 3a − 3 = a2 − a − 2A1

x2 − 3x − 10 = 0: x = 5 or −2A1

Practice exercise 3

1. Given g(t) = 50 − 4.9t2, find g(2) and interpret if g is a height in metres after t seconds.

2. Is the relation x = y2 a function of x?

4 Domain and range

Allowed inputs and resulting outputs.

The domain is the set of allowed inputs; the range is the set of outputs produced. In mathematics, domains are restricted by division by zero, square roots of negatives and logs of non-positive numbers. In context, domains are restricted by reality: time is non-negative, a number of items is a whole number, and a model may only be valid over a limited interval.

To find a range, use the graph: look for the lowest and highest values on the given domain, including endpoints and turning points.

Function	Domain	Range
f(x) = 2x − 1	x ∈ ℝ	f(x) ∈ ℝ
f(x) = x2 + 3	x ∈ ℝ	f(x) ≥ 3
f(x) = √(x − 4)	x ≥ 4	f(x) ≥ 0
f(x) = 1/x	x ≠ 0	f(x) ≠ 0
f(x) = 3(0.5)x	x ∈ ℝ	f(x) > 0

WORKED EXAMPLE 5 Range on a restricted domain (standard)

f(x) = x2 − 6x + 5 for 0 ≤ x ≤ 7. Find the range.

Solution

Vertex at x = 3: f(3) = −4M1

f(0) = 5, f(7) = 12A1

−4 ≤ f(x) ≤ 12A1

WORKED EXAMPLE 6 Domain in context (standard)

The height of a ball is h(t) = 1.5 + 12t − 4.9t2. Find a suitable domain.

Solution

h(t) = 0 when t = 2.57 (positive root)M1

0 ≤ t ≤ 2.57A1

Practice exercise 4

1. State the domain and range of f(x) = 5 − √(x + 2).

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

3. A company's profit is P(n) = 12n − 300 for n items. State a suitable domain.

5 One-to-one and many-to-one functions

The horizontal line test.

A function is one-to-one if different inputs always give different outputs (every horizontal line meets the graph at most once). It is many-to-one if some output comes from more than one input, like f(x) = x2 (f(2) = f(−2)). Only one-to-one functions have inverses.

Practice exercise 5

1. Classify f(x) = x3 and g(x) = x2 + 1.

2. Explain why f(x) = sin x is many-to-one.

6 Key features of graphs

What to identify and label.

Intercepts: y-intercept f(0); x-intercepts (zeros) where f(x) = 0. Maximum and minimum points (turning points). Asymptotes: lines the graph approaches. Symmetry and end behaviour. In AI these are usually found with the GDC, and interpreted in context: the y-intercept is often an initial value; a maximum may be the greatest profit or height; an asymptote is a long-term value.

Figure 1. Key features: intercepts and the maximum point.

WORKED EXAMPLE 7 Features with a GDC (standard)

For f(x) = x3 − 4x2 + x + 6, find the zeros and the turning points.

Solution

Zeros x = −1, 2, 3A1

Local max (0.131, 6.06); local min (2.54, −0.879)A1 A1

Practice exercise 6

1. Find the zeros and vertex of y = 2x2 − 8x + 3.

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

7 Drawing and sketching graphs with technology

Window, scale and labels.

To draw a graph, plot accurately with labelled axes and scales; to sketch, show the general shape and key features. When copying a GDC graph into your answer, choose a window that shows all key features, label axes, intercepts, turning points and asymptotes, and keep the shape accurate (curvature, end behaviour).

GDC tip

Use zoom-fit or set the window from the context (for example 0 ≤ t ≤ 3 for a projectile). Use the 'zero', 'max', 'min' and 'intersect' tools rather than tracing, and record answers to 3 significant figures.

8 Inverse functions

Undoing a function.

The inverse f−1 reverses f: if f(a) = b then f−1(b) = a. Its graph is the reflection of y = f(x) in y = x. The domain of f−1 is the range of f. To find f−1(x) algebraically, write y = f(x), swap x and y, and rearrange.

In context, an inverse answers the reverse question: if C(n) gives the cost of n items, C−1(c) gives the number of items for cost c.

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

WORKED EXAMPLE 8 Inverse in context (standard)

Temperature in °F is F(C) = 1.8C + 32. Find F−1 and interpret F−1(98.6).

Solution

C = (F − 32)/1.8M1 A1

F−1(98.6) = 37: 98.6 °F is 37 °CA1

Practice exercise 7

1. Find f−1(x) for f(x) = (x + 4)/3.

2. If f(5) = 12, write down f−1(12).

3. Find the inverse of f(x) = 10x.

9 Piecewise functions

Different rules on different intervals.

A piecewise function uses different formulas on different parts of its domain. Tax brackets, postage costs, phone plans and speed-time graphs are naturally piecewise. When evaluating, choose the formula for the interval containing the input; when graphing, draw each piece on its own interval, showing included (●) and excluded (○) endpoints.

WORKED EXAMPLE 9 Evaluating a piecewise function (standard)

A data plan costs $30 for up to 5 GB, plus $8 per extra GB. Find the cost for 3 GB and 8.5 GB, and the data used for a bill of $70.

Solution

C(3) = $30; C(8.5) = 30 + 8(3.5) = $58A1 A1

30 + 8(x − 5) = 70: x = 10 GBM1 A1

Practice exercise 8

1. Income tax is 0% on the first $12 000 and 20% on income above. Write a piecewise function for the tax T(x) and find T(30 000).

2. Sketch the cost function for parking at $3 per hour or part of an hour, up to 4 hours.

10 Direct and inverse variation

Proportional relationships.

y varies directly as xn if y = kxn: doubling x multiplies y by 2n. y varies inversely as xn if y = k/xn: doubling x divides y by 2n. Find k from one pair of values. Examples: the area of a circle varies directly as r2; the time for a journey varies inversely as speed; light intensity varies inversely as the square of distance.

WORKED EXAMPLE 10 Inverse square (standard)

Light intensity I varies inversely as the square of distance d. I = 80 lux at d = 2 m. Find I at 5 m.

Solution

I = k/d2: k = 320M1 A1

I(5) = 320/25 = 12.8 luxA1

Practice exercise 9

1. y varies directly as √x and y = 12 when x = 9. Find y when x = 25.

2. The time t for a trip varies inversely as speed v; t = 3 h at 80 km/h. Find t at 96 km/h.

11 Linear models: cost, revenue and break-even

Business functions.

A cost function C(x) = fixed cost + variable cost × x; a revenue function R(x) = price × x; profit P(x) = R(x) − C(x). The break-even point is where R(x) = C(x), i.e. P(x) = 0.

WORKED EXAMPLE 11 Break-even (standard)

A company has fixed costs $2400 and variable costs $18 per item, selling each for $42. Find the break-even quantity and the profit on 300 items.

Solution

42x = 2400 + 18x: x = 100 itemsM1 A1

P(300) = 24(300) − 2400 = $4800A1

Figure 3. Cost and revenue lines meet at the break-even point.

Practice exercise 10

1. Fixed costs $500, variable cost $3.50, price $9. Find the break-even point.

2. Write the profit function and interpret its gradient.

12 Quadratic models

Curves with a maximum or minimum.

Quadratic models y = ax2 + bx + c describe projectiles, areas, and revenue when price affects demand. The vertex gives the maximum or minimum; the zeros give where the quantity is zero. A GDC quadratic regression fits data.

WORKED EXAMPLE 12 Projectile (standard)

A ball's height is h(t) = 1.8 + 14t − 4.9t2. Find the maximum height and when the ball lands.

Solution

Vertex at t = 1.43 s: h = 11.8 mA1 A1

h(t) = 0: t = 2.98 sA1

WORKED EXAMPLE 13 Revenue model (exam level)

At a price of $p, a shop sells 600 − 20p items. Find the price that maximizes revenue.

Solution

R = p(600 − 20p)M1

Vertex p = 15A1

$15, revenue $4500A1

Practice exercise 11

1. A rectangle has perimeter 36 m. Find the dimensions for maximum area.

2. Use quadratic regression on (0, 2), (1, 5), (2, 10), (3, 17).

13 Exponential models

Constant percentage change.

Exponential models y = kax + c describe growth and decay (see the exponential functions booklet). Interpret k + c as the initial value, a as the growth factor and c as the asymptote (long-term value).

WORKED EXAMPLE 14 Interpreting an exponential model (standard)

V(t) = 18 000(0.85)t + 2000 gives a car's value after t years. Interpret each parameter.

Solution

Initial value 20 000; loses 15% of the depreciating part each yearA1

Long-term value approaches $2000 (scrap value)A1

Practice exercise 12

1. For V(t) above, find when the value falls below $8000.

14 Choosing a model and a sensible domain

Fit the shape and the story.

Choose a model by the pattern: constant differences (linear), a single maximum or minimum (quadratic), constant ratios or an asymptote (exponential), or a power relationship (variation). Then restrict the domain to values where the model makes sense, and comment on extrapolation.

Feature in context	Suggested model
equal increases per unit time	linear
rises then falls symmetrically	quadratic
percentage growth or decay	exponential
levels off to a limit	exponential with asymptote (or logistic, HL)
y proportional to a power of x	power (variation)

Practice exercise 13

1. Suggest a model for the height of water in a draining bath that decreases quickly at first and then more slowly.

2. Why should a linear model for a child's height not be used for adults?

15 Solving equations graphically

Intersections.

The solutions of f(x) = g(x) are the x-coordinates of the intersections of y = f(x) and y = g(x). On Paper 2, this is often the fastest method, especially for equations mixing different types of function.

WORKED EXAMPLE 15 Graphical solution (standard)

Solve 2x = 3x + 1.

Solution

Graph both sides; intersectionsM1

x = 0 and x = 3.54A1 A1

Practice exercise 14

1. Solve x2 = 5 − x.

2. Solve 0.5x3 = 4 − x.

16 Interpreting graphs in context

What does the graph tell us?

In context, describe features in words with units: where the graph is increasing (the quantity is growing), steep (changing quickly), maximum (the greatest value and when it occurs), intercepts (starting value or when the quantity becomes zero), and asymptotes (long-term behaviour).

WORKED EXAMPLE 16 Interpretation (exam level)

The concentration of a drug is C(t) = 20te−0.5t mg/L. Find the maximum concentration and when it occurs, and describe the long-term behaviour.

Solution

Maximum at t = 2 h: C = 14.7 mg/LA1 A1

As t → ∞, C → 0: the drug is eliminatedR1

Practice exercise 15

1. For C(t) above, find the times when C > 10 mg/L.

17 Using technology efficiently

A checklist.

Before solving: set an appropriate window. While solving: use built-in tools (zero, max, min, intersect, table). After solving: write the method in words ('using the GDC, the intersection of ... is ...'), give answers to 3 s.f., and sketch the graph if asked.

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 17 Mixed problem 1 (exam level)

f(x) = 3x − 2 and g(x) = x2. Find the values of x for which f(x) > g(x).

Solution

x2 − 3x + 2 < 0M1

(x − 1)(x − 2) < 0: 1 < x < 2A1

WORKED EXAMPLE 18 Mixed problem 2 (exam level)

A gym charges $50 joining plus $35 per month; another charges $60 per month with no joining fee. After how many months is the first gym cheaper overall?

Solution

50 + 35m < 60mM1

m > 2: from month 3A1

Practice exercise 16

1. Find the inverse of f(x) = 5 − 2x and solve f(x) = f−1(x).

2. The range of f(x) = ax + b for 0 ≤ x ≤ 4 is 3 ≤ f(x) ≤ 11 (a > 0). Find a and b.

19 Composite functions (HL)

One function after another.

The composite (f ∘ g)(x) = f(g(x)) applies g first, then f. In context, composites chain processes: if C(n) is the cost of n items and n(t) the number made after t hours, C(n(t)) is the cost after t hours. Order matters: f(g(x)) ≠ g(f(x)) in general.

WORKED EXAMPLE 19 Composite (standard)

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

Solution

f(g(x)) = 2x2 + 3A1

g(f(x)) = (2x + 3)2A1

(f ∘ g)(−2) = 11A1

WORKED EXAMPLE 20 Composite in context (exam level)

A factory makes n(t) = 40t items in t hours, and the cost of n items is C(n) = 500 + 6n. Find C(n(t)) and the cost of an 8-hour shift.

Solution

C(n(t)) = 500 + 240tA1

C(n(8)) = $2420A1

Practice exercise 17

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

2. Show that f(x) = (x + 2)/(x − 1) satisfies f(f(x)) = x.

20 Inverses, restricted domains and composites in models (HL)

Making functions invertible.

A many-to-one function can be made invertible by restricting its domain, for example f(x) = x2 for x ≥ 0. In models, the natural domain often does this automatically: the area A = πr2 for r ≥ 0 has inverse r = √(A/π). f(f−1(x)) = x and f−1(f(x)) = x on the appropriate domains.

WORKED EXAMPLE 21 Restricted inverse (exam level)

f(x) = (x − 3)2 + 1 for x ≥ 3. Find f−1(x) and its domain.

Solution

x − 1 = (y − 3)2, y ≥ 3M1

f−1(x) = 3 + √(x − 1), x ≥ 1A1 A1

Practice exercise 18

1. The volume of a cube is V(s) = s3. Find s(V) and interpret.

2. Find the largest domain x ≥ a for which f(x) = x2 − 6x has an inverse.

21 Investigation: choosing a phone plan

A Paper 3 style exploration.

Plan A: $25 per month including 10 GB, then $4 per extra GB. Plan B: $40 per month including 20 GB, then $2 per extra GB. Plan C: $2.50 per GB with no monthly fee. 1. Write piecewise functions A(g), B(g), C(g) for the monthly cost when g GB are used. 2. Draw all three on the same axes for 0 ≤ g ≤ 40. 3. Find the ranges of g for which each plan is cheapest. 4. A family's usage varies between 15 and 30 GB per month. Which plan would you recommend, and why? 5. Suggest one other factor, not captured by the model, that might affect the decision.

Answers. 1. A = 25 for g ≤ 10, 25 + 4(g − 10) above; B = 40 for g ≤ 20, 40 + 2(g − 20) above; C = 2.5g. 3. C is cheapest for g < 16 (equal to A at g = 10); B is cheapest for g > 16; A is never strictly cheapest. 4. Plan B: for 16 to 30 GB it is cheapest, and at 15 GB it costs only $2.50 more than C. 5. Network coverage, contract length, roaming charges, speed limits.

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 10 minutes)

f(x) = 4 − 2x and g(x) = x2 + 1.

(a) Find f(−3). [1]

(b) Find f−1(x). [2]

(d) Solve f(x) = g(x). [3]

Markscheme

Part	Answer	Marks
(a)	10	A1
(b)	x = 4 − 2y	M1
	f−1(x) = (4 − x)/2	A1
(c)	g(x) ≥ 1	A1
(d)	x2 + 2x − 3 = 0	M1
	(x + 3)(x − 1) = 0	A1
	x = −3 or 1	A1

Examiner's comments. The range must be in terms of g(x) or y.

Question 2 (Paper 2 style) (about 15 minutes)

A company sells x units per week. Its weekly cost is C(x) = 1200 + 15x dollars and its revenue is R(x) = 60x − 0.05x2 dollars, for 0 ≤ x ≤ 1000.

(a) Write down the fixed cost. [1]

(b) Find the profit function. [2]

(d) Find the maximum profit and the number of units at which it occurs. [3]

Markscheme

Part	Answer	Marks
(a)	$1200	A1
(b)	P(x) = 45x − 0.05x2 − 1200	A2
(c)	P(x) = 0	M1
	x = 27.5	A1
	x = 872	A1
(d)	vertex x = 450	M1 A1
	$8925	A1

Examiner's comments. Interpret break-even points in units per week.

Question 3 (Paper 2 style) (about 10 minutes)

The time T minutes to fill a tank varies inversely as the flow rate r litres per minute. T = 45 when r = 12.

(a) Find T in terms of r. [2]

(b) Find T when r = 20. [1]

Markscheme

Part	Answer	Marks
(a)	T = k/r	M1
	T = 540/r	A1
(b)	27 minutes	A1
(c)	decreasing curve from (5, 108) to (30, 18)	A2

Examiner's comments. Label endpoints on the sketch.

23 Common misconceptions and links

Students often think…	Actually…
f−1(x) = 1/f(x)	f−1 reverses f
the range is described with x	use f(x) or y
any model works for any domain	restrict the domain to the context
a vertical line can meet a function twice	only once

How 2.2 connects to the rest of the course

Topic	Connection
2.5 Polynomials and rational functions	further models
2.9 Transformations	changing graphs
1.5 Exponential functions	growth models
5.3 Derivatives	rates of change

24 Check your understanding

Section A: multiple choice

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

A. −1.5 B. 1.5 C. 3 D. −2/3

2. The range of x2 − 4 is

A. x ≥ −4 B. f(x) ≥ 0 C. f(x) ≥ −4 D. ℝ

3. Break-even occurs when

A. profit is maximum B. revenue equals cost C. cost is zero D. price equals cost

4. y varies inversely as x means

A. y = kx B. y = x − k C. y = k − x D. y = k/x

5. The graph of f−1 is the reflection of f in

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

Section B: structured

6. Find the domain and range of f(x) = 2 + √(6 − x). [2]

7. Find the line perpendicular to y = 4x − 1 through (8, 3). [2]

Answers

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

6. x ≤ 6; f(x) ≥ 2

7. y = −x/4 + 5

25 Summary and topic map

Idea	Remember
Lines	gradient = rate of change; y = mx + c
Function	one output per input
Domain and range	inputs and outputs; context restrictions
Inverse	reverse process; reflect in y = x
Piecewise	different rules on intervals
Variation	y = kxn; y = k/xn
Models	linear, quadratic, exponential; choose and restrict

Before the exam, make sure you…

•interpret in context with units
•state a sensible domain
•use GDC tools and record the method
•label sketches

Functions and Their Properties 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 4. Topic map for Functions and Their Properties.

26 Answers to practice exercises

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

Practice exercise 1

1. −2

2. −2.5

3. −0.07 L per km: it uses 7 L per 100 km

Practice exercise 2

1. y = −3x + 7

2. (4, 0) and (0, −10)

3. (2, 5)

Practice exercise 3

1. 30.4: the height after 2 s is 30.4 m

2. no: x = 4 gives y = ±2

Practice exercise 4

1. x ≥ −2; f(x) ≤ 5

2. 0.5 ≤ f(x) ≤ 8

3. n ∈ ℕ, up to production capacity

Practice exercise 5

1. f one-to-one, g many-to-one

2. many inputs give the same output, e.g. sin 0 = sin π = 0

Practice exercise 6

1. x = 0.419, 3.58; vertex (2, −5)

2. x = 1 and y = 3

Practice exercise 7

1. 3x − 4

2. 5

3. log x

Practice exercise 8

1. T = 0 for x ≤ 12 000; 0.2(x − 12 000) above: $3600

2. step function: 3, 6, 9, 12

Practice exercise 9

1. 20

2. 2.5 h

Practice exercise 10

1. 91 items (90.9)

2. P = 5.5x − 500: $5.50 profit per extra item

Practice exercise 11

1. 9 m by 9 m

2. y = x2 + 2x + 2

Practice exercise 12

1. 18 000(0.85)t < 6000: t > 6.76, so after 7 years

Practice exercise 13

1. exponential decay

2. growth stops; extrapolation invalid

Practice exercise 14

1. x = −2.79, 1.79

2. x = 1.67

Practice exercise 15

1. 0.715 < t < 4.31 hours

Practice exercise 16

1. f−1(x) = (5 − x)/2; x = 5/3

2. a = 2, b = 3

Practice exercise 17

1. x ≥ 4

2. simplify f(f(x)) to x

Practice exercise 18

1. s = ∛V: side length from volume

2. x ≥ 3

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