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
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