Sets and Functions

Sets and Functions

Set Notation

Notation Examples

ofmember a ismeans

ofmember a is notmeans

}4,3,2,1{4

}4,3,2,1{5

{ }..empty set

set bracketsSet of prime numbers between 8 and 10 is { }

means Subset which is part of a set

}5,4,3,2,1,0{}4,3,2,1{

A set is a collection of objects (usually numbers)

Standard sets

.....}4,3,2,1{NThe set of NATURAL NUMBERS

The set of WHOLE NUMBERS .....}4,3,2,1,0{W

.....}4,3,2,1,0,1,2{..., ZThe set of INTEGERS

The set of RATIONAL NUMBERS Q...the set of all numbers which can be written as fractions, i.e. all of the above plus ½, -0.9….etc

The set of REAL NUMBERS ...the set of all rational numbers plus irrationals e.g 2, pi etc.

R

Q

0.5

2,

Z

-1,-2

W

0

N5,6

Functions and mappings.A function or mapping from a set A to a set B is a rule that relates each element in set A to ONE and only ONE element in set B.

The set of elements in set A is called the DOMAIN

The set of elements in set B is called the RANGE

A•1•2•3

B•3•7•9

Domain Range

This is known as an arrow diagram.

A•1•2•3

B•3•7•9

This is not a function or mapping as 2 is mapped to 7 and 9.

A•1•2•3

B•3•7•9

This is a function as each element in set A is mapped to one and only one element in set B.

Not a function A Function

Graph Notation.

8

-2 2

1. The function f, defined by f(x) = x² - 2x, has domain

{-2, -1, 0, 1, 2}. Find the range.

f(x) = x² - 2x, f(-2) = (-2)² - 2(-2) = 8 f(-1) = (-1)² - 2(-1) = 3 f(0) = 0² - 2(0) = 0 f(1) = 1² - 2(1) = -1

f(2) = 2² - 2(2) = 0

domain

A•-2•-1•0•1•2

B•8•3•0•-1

range

Formula Arrow diagram Graph

Range is {-1, 0, 3 , 8}

Composition of Functionsh(x) = 4x - 3, can be thought of as a composition of two functions:

Multiply by 4 and then subtract 3

x•-1•0•1

4x•-4•0•4

4x - 3•-7•-3•1

f(x) = 4x g(x) = x - 3

h(x) = 4x - 3,

h(x) is f applied first then g applied to the result. h is a function of a function and written h(x) = g(f(x)) and read as g of f of x

1. f(x) = 2x, g(x) = x + 3, evaluate:

a) f(g(0)) b) f(g(-5)) c) g(f(2)) d) g(f(-1))

a) f(g(0)) b) f(g(-5)) c) g(f(2))

d) g(f(-1)) g(0) = 0 + 3

= 3 f(g(0)) = f(3)

= 2(3)

= 6

g(-5) = -5 + 3

= -2f(g(-5)) = f(-2)

f(2) = 2(2)

= 4

= 4 + 3

= 7

f(-1) = 2(-1)

= -2

= -2 + 3

= 1

g(f(2)) = g(4)

g(f(-1)) = g(-2)

= 2(-2)

= -4

2. If f(x) = 2x, g(x) = x + 3, Find

(a) h(x) = g(f(x))

(b) k(x) = f(g(x))

f(x) = 2x

g(f(x)) = g(2x)

= 2x + 3

g(x) = x + 3

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

= 2(x + 3)

(a) h(x) = g(f(x))

h(x) = 2x + 3

(b) k(x) = f(g(x))

In general f(g(x)) g(f(x))

k(x) = 2x + 6

2 13. If ( ) 2 and ( ) ( 0), find

(a) ( ) ( ( )) ( ) ( ) ( ( ))

f x x g x xx

h x f g x b k x g f x

1( ) ( )a g xx

1( ( ))f g x fx

21 2x

2

1 2x

2( ) ( ) 2b f x x

2( ( )) ( 2)g f x g x

2

12x

0x

The denominator of a function can never be zero as it will be undefined.

Inverse Functions.The inverse function is a function which reverses or ‘undoes’

a function

1f Is used to denote this function

A Bf

1f

The sets must be in 1 - 1 correspondence for this function to exist.

f -1(f(x)) = f ( f -1 (x)) = x

f maps Set A to set B and maps Set B to set A.1f

Graphs of Inverse Functions

The graph of an inverse function is found by reflecting the graph in the line y = x.

y

x

0.5

0.5

1

1

1.5

1.5

2

2

– 0.5

– 0.5

– 1

– 1

– 1.5

– 1.5

– 2

– 2

0.5

0.5

1

1

1.5

1.5

2

2

– 0.5

– 0.5

– 1

– 1

– 1.5

– 1.5

– 2

– 2

( )y f x

Graphs of Inverse Functions

The graph of an inverse function is found by reflecting the graph in the line y = x.

y

x

0.5

0.5

1

1

1.5

1.5

2

2

– 0.5

– 0.5

– 1

– 1

– 1.5

– 1.5

– 2

– 2

0.5

0.5

1

1

1.5

1.5

2

2

– 0.5

– 0.5

– 1

– 1

– 1.5

– 1.5

– 2

– 2

( )y f x

Graphs of Inverse Functions

The graph of an inverse function is found by reflecting the graph in the line y = x.

y

x

0.5

0.5

1

1

1.5

1.5

2

2

– 0.5

– 0.5

– 1

– 1

– 1.5

– 1.5

– 2

– 2

0.5

0.5

1

1

1.5

1.5

2

2

– 0.5

– 0.5

– 1

– 1

– 1.5

– 1.5

– 2

– 2

( )y f x

1( )y f x

This works for all functions that have an inverse.

y

x

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

( )y g x

This works for all functions that have an inverse.

y

x

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

( )y g x

This works for all functions that have an inverse.

y

x

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

( )y g x

1( )y g x

Exponential and Logarithmic Functionsf (x) = ax , x ϵ R

is called an exponential function to base a, a ϵ R, a ≠ 0It is read “a to the x”

loga x is the logarithmic function and is the inverse of the exponential function. It is read as “log to the base a of x”

Hence, if 1( ) log then ( ) xaf x x f x a

( ) xf x a 1( ) logaf x x if then

f(x) = ax passes through (0,1) and (1,a)

f-1(x) = logax passes through (1,0) and (a,1)

Exp and Log Graphs

Determine the equation of the graphs shown below.

y

x

1(1,4)

(i)y

x

1

(2,25)

(i)

xy a using (1,4)14 a

4a 4xy

xy a using (2,25)225 a

5a 5xy

Determine the inverse of the following functions.

( ) 2 ( ) 6x xa y b y

2( ) loga y x 6( ) logb y x

Graphs of Functions

Standard Graphs you should know.

y

x

y mx c

y

x

2y ax bx c

y

x

3 2y ax bx cx d

y

x

1yx

y

x

siny x

y

x

cosy x

y

x

tany x

Let us consider: ( ) and ( )y f x a y f x a

all points moved up by ‘a’ units if a is positive and down if a is negative.

a

aa a

a

( )y f x

( )y f x a

1. Describe the transformation of the following graphs.2 2 and 55 5

y x y x

The graph has been moved down vertically 5 units.

2. Part of the graph of f (x) = x2 – 3x is shown below.(i) Determine the values of A, B and C.(ii) Sketch the graph of y = f (x) +2(iii) State the coordinates of the images of A, B and C.

y

x

A

B

C x

y A and C are the roots of the quadratic.They occur when y = 0.

2 3 0x x ( 3) 0x x

0 or 3x

A(0,0) C(3,0)

The turning point B is mid way between the roots. x = 1.5

When x = 1.5, y = 1.52 - 31.5 = -2.25 C(1.5, -2.25)

y

x

A

B

C x

y( )y f x

y

x

( ) 2y f x

A` C`

B`

(ii)

(iii) A` (0,2) C` (3,2) B` (1.5, -0.25)

y = f (x)

y = f (x - a)

Now Consider y = f (x + a) and y = f (x - a)

a

(Right if a is negative)

All points are moved left by ‘a’ units if a is positive and right if a is negative.

x

y

y = f (x)

Now Consider y = f (x + a) and y = f (x - a)

All points are moved left by ‘a’ units if a is positive and right if a is negative.

x

y

a

(Left if a is positive)

y = f (x + a)

y = f (x)

y = - f (x)

The y coordinates become negative.

Now Consider y = -f (x)

x

y

The graph is reflected in the X axis

Page 40 Exercise 3G

y = f (x)

The x coordinates become negative.

Now Consider y = f (-x)

x

y

The graph is reflected in the Y axis

y = f (-x)

Now Consider y = k f (x)

y

x

1

1

2

2

– 1

– 1

– 2

– 2

siny x

let us consider sin where 1y k x k

Now Consider y = k f (x)

y

x

1

1

2

2

– 1

– 1

– 2

– 2

let us consider sin where 1y k x k

sin , 1y k x k

let us consider sin where 1y k x k

Now Consider y = k f (x)

y

x

1

1

2

2

– 1

– 1

– 2

– 2

let us consider sin where 1y k x k

sin , 1y k x k

let us consider sin where 1y k x k

sin , 1y k x k

To obtain the graph of ( )y kf x

The graph is stretched vertically by a factor when 1k k

The graph is compressed vertically by a factor when 1k k

Now Consider y = f (kx)

y

x

1

1

2

2

– 1

– 1

– 2

– 2

siny x

let us consider sin( ) where 1y kx k

Now Consider y = f (kx)

y

x

1

1

2

2

– 1

– 1

– 2

– 2

sin , 1y kx k

let us consider sin( ) where 1y kx k let us consider sin( ) where 1y kx k

Now Consider y = f (kx)

y

x

1

1

2

2

– 1

– 1

– 2

– 2

sin , 1y kx k

let us consider sin( ) where 1y kx k let us consider sin( ) where 1y kx k

To obtain the graph of ( )y f kx

The graph is compressed horizontally by a factor when 1k k

The graph is stretched horizontally by a factor when 1k k

Graphs of related exponential functions

As we saw previously, the equation of an exponential function can be determined from 2 points on the graph.

1. Part of the graph y = a x + b is shown below. Find the values of a and b and hence the equation of the curve.

y

x

2

(3,9)

x

y

y

x

2

(3,9)

x

y Using the point (0,2)xy a b

02 a b 2 1 b

1b

1xy a

Using the point (3,9)1xy a

39 1a 3 8a

2a 2 1xy

Graphs of Logarithmic FunctionsThe exponential function 3 passes through the points (0,1) and (1,3).xy

3The inverse function log is a reflection of 3 in the .xy x y y x

Hence (0,1) and (1,3) are reflected to the points (1,0) and (3,1).

y

x

(0,1)

(1,0)

(1,a)

(a,1)

xy a

logay x

0 1In general, since 1 and then:a a a

log 1 0 and log 1a a a

1. Part of the graph of log is shown. Find the value of .ay x a

1

(5,1)

x

yy

x

logay x

Using the point (5,1)1 log 5a

5a