Download Chapter 2 Sets and Functions

Sets and Functions
Set Notation
A set is a collection of objects (usually numbers)
Notation
Examples
 means is a member of
4  {1,2,3,4}
 means is not a member of
5 {1,2,3,4}
set brackets
{ }..empty set
 means Subset which is
part of a set
Set of prime numbers between 8
and 10 is { }
{1,2,3,4}  {0,1,2,3,4,5}
Standard sets
The set of NATURAL NUMBERS
The set of WHOLE NUMBERS
The set of INTEGERS
N  {1,2,3,4.....}
W  {0,1,2,3,4.....}
Z  {..., 2,1,0,1,2,3,4.....}
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
N
5,6
0
W Z
Q
-1,-2 0.5
2, 
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
Domain
B
•3
•7
•9
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.
Graph Notation.
Not a function
A Function
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
Range is {-1, 0, 3 , 8}
Formula
8
A
•-2
•-1
•0
•1
•2
B
•8
•3
•0
•-1
domain
range
Arrow diagram
-2
2
Graph
Composition of Functions
h(x) = 4x - 3, can be thought of as a composition of two functions:
Multiply by 4 and then
x
•-1
•0
•1
f(x) = 4x
4x
•-4
•0
•4
subtract 3
g(x) = x - 3
4x - 3
•-7
•-3
•1
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))
g(0) = 0 + 3
b) f(g(-5))
c) g(f(2))
d) g(f(-1))
g(-5)
= -5 + 3
f(2) = 2(2)
f(-1) = 2(-1)
=3
f(g(0)) = f(3)
= -2
f(g(-5)) = f(-2)
= 2(3)
= 2(-2)
=4+3
= -2 + 3
=6
= -4
=7
=1
=4
g(f(2)) = g(4)
= -2
g(f(-1)) = g(-2)
2. If f(x) = 2x, g(x) = x + 3, Find
(a) h(x) = g(f(x))
(b) k(x) = f(g(x))
(a) h(x) = g(f(x))
(b) k(x) = f(g(x))
f(x) = 2x
g(x) = x + 3
g(f(x)) = g(2x)
f(g(x)) = f(x + 3)
= 2(x + 3)
= 2x + 3
k(x) = 2x + 6
h(x) = 2x + 3
In general f(g(x))

g(f(x))
1
( x  0), find
x
(b) k ( x)  g ( f ( x))
3. If f ( x)  x 2  2 and g ( x) 
(a) h( x)  f ( g ( x))
1
(a) g ( x) 
x
(b) f ( x)  x 2  2
g ( f ( x))  g ( x 2  2)
1
f ( g ( x))  f  
x
1
 2
x 2
2
1
  2
x

1
2
2
x
x0
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
A
f
1
f
B
Is used to denote this
function
f 1
The sets must be in 1 - 1 correspondence for this function to exist.
f maps Set A to set B and f
1
maps Set B to set A.
f -1(f(x)) = f ( f -1 (x)) = x
Graphs of Inverse Functions
2– 2
0.5
1
1.5
0.5
1
1.5
0.5
1
1.5
2
– 2
0.5
1
1.5
The graph of an inverse function is found by reflecting the graph
in the line y = x.
y
y  f ( x)
2
1.5
1
0.5
– 2
– 1.5
– 1
– 0.5
0.5
– 0.5
– 1
– 1.5
– 2
1
1.5
2
x
Graphs of Inverse Functions
2– 2
0.5
1
1.5
0.5
1
1.5
0.5
1
1.5
2
– 2
0.5
1
1.5
The graph of an inverse function is found by reflecting the graph
in the line y = x.
y
y  f ( x)
2
1.5
1
0.5
– 2
– 1.5
– 1
– 0.5
0.5
– 0.5
– 1
– 1.5
– 2
1
1.5
2
x
Graphs of Inverse Functions
2– 2
0.5
1
1.5
0.5
1
1.5
0.5
1
1.5
2
– 2
0.5
1
1.5
The graph of an inverse function is found by reflecting the graph
in the line y = x.
y
y  f ( x)
2
1.5
1
y  f 1 ( x)
0.5
– 2
– 1.5
– 1
– 0.5
0.5
– 0.5
– 1
– 1.5
– 2
1
1.5
2
x
10
2
4
6
8
– 10
2
4
6
8
2– 10
4
6
8
10
2
4
6
8
This works for all functions that have an inverse.
y
10
y  g ( x)
8
6
4
2
– 10
– 8
– 6
– 4
– 2
2
– 2
– 4
– 6
– 8
– 10
4
6
8
10
x
10
2
4
6
8
– 10
2
4
6
8
2– 10
4
6
8
10
2
4
6
8
This works for all functions that have an inverse.
y
10
y  g ( x)
8
6
4
2
– 10
– 8
– 6
– 4
– 2
2
– 2
– 4
– 6
– 8
– 10
4
6
8
10
x
10
2
4
6
8
– 10
2
4
6
8
2– 10
4
6
8
10
2
4
6
8
This works for all functions that have an inverse.
y  g 1 ( x)
y
10
y  g ( x)
8
6
4
2
– 10
– 8
– 6
– 4
– 2
2
– 2
– 4
– 6
– 8
– 10
4
6
8
10
x
Exponential and Logarithmic Functions
f (x) = ax , x ϵ R
is called an exponential function to base a, a ϵ R, a ≠ 0
It 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”
if
Hence, if
f ( x)  a
x
then f
1
( x)  log a x
1
f ( x)  log a x then f ( x)  a
x
Exp and Log Graphs
f(x) = ax
passes through (0,1) and (1,a)
f-1(x) = logax passes through (1,0) and (a,1)
Determine the equation of the graphs shown below.
(i)
(i)
y
y
(2,25)
(1,4)
1
1
x
y  ax
using (1,4)
x
y  ax
4  a1
25  a 2
a4
a5
 y  4x
 y  5x
using (2,25)
Determine the inverse of the following functions.
(a)
y  2x
(a) y  log 2 x
(b)
y  6x
(b) y  log 6 x
Graphs of Functions
Standard Graphs you should know.
y
y
x
x
y  mx  c
y
y  ax 2  bx  c
x
y  ax3  bx 2  cx  d
y
x
1
y
x
y
y
x
x
y  cos x
y  sin x
y
x
y  tan x
Let us consider:
y  f ( x)  a and y  f ( x)  a
all points moved up by ‘a’ units if a is positive and down if a is
negative.
y  f ( x)  a
y  f ( x)
a
a
a
a
a
1. Describe the transformation of the following graphs.
y
2
2
x and y  x  5
5
5
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.
A and C are the roots of the quadratic.
y
y
They occur when y = 0.
C
A
B
x
x
x 2  3x  0
x( x  3)  0
x  0 or 3
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 - 31.5 = -2.25
C(1.5, -2.25)
(ii)
y  f ( x)
y
y  f ( x)  2
y
y
C`
A`
C
A
x
B
(iii) A` (0,2)
x
x
C` (3,2) B` (1.5, -0.25)
B`
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.
y
(Right if a is negative)
y = f (x)
a
x
y = f (x - a)
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.
y
(Left if a is positive)
y = f (x)
a
y = f (x + a)
x
Now Consider y = -f (x)
y
y = f (x)
x
y = - f (x)
Page 40 Exercise 3G
The y coordinates become negative.
The graph is reflected in the X axis
Now Consider y = f (-x)
y
y = f (x)
x
y = f (-x)
The x coordinates become negative.
The graph is reflected in the Y axis
Now Consider y = k f (x)
y
2
1
x
– 1
y  sin x
– 2
let us consider y  k sin x where k  1
Now Consider y = k f (x)
y
y  k sin x, k  1
2
1
x
– 1
– 2
let us consider y  k sin x where k  1
let us consider y  k sin x where k  1
Now Consider y = k f (x)
y
y  k sin x, k  1
2
y  k sin x, k  1
1
x
– 1
– 2
let us consider y  k sin x where k  1
let us consider y  k sin x where k  1
To obtain the graph of y  kf ( x)
The graph is stretched vertically by a factor k when k  1
The graph is compressed vertically by a factor k when k  1
Now Consider y = f (kx)
y
2
1
x
– 1
y  sin x
– 2
let us consider y  sin(kx) where k  1
Now Consider y = f (kx)
y
2
y  sin kx, k  1
1
x
– 1
– 2
let us consider y  sin(kx) where k  1
let us consider y  sin(kx) where k  1
Now Consider y = f (kx)
y
2
1
y  sin kx, k  1
x
– 1
– 2
let us consider y  sin(kx) where k  1
let us consider y  sin(kx) where k  1
To obtain the graph of y  f (kx)
The graph is compressed horizontally by a factor k when k  1
The graph is stretched horizontally by a factor k when k  1
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
y
(3,9)
2
x
x
y
y
Using the point (0,2)
y  ax  b
2  a0  b
2  1 b
b 1
(3,9)
2
y  ax 1
x
x
Using the point (3,9)
y  ax 1
9  a3  1
a3  8
a2
y  2x  1
Graphs of Logarithmic Functions
The exponential function y  3x passes through the points (0,1) and (1,3).
The inverse function y  log3 x is a reflection of y  3x in the y  x.
Hence (0,1) and (1,3) are reflected to the points (1,0) and (3,1).
y
y  ax
(1,a)
(0,1)
(a,1)
x
(1,0)
y  log a x
In general, since a0  1 and a1  a then:
log a 1  0 and log a a  1
1. Part of the graph of y  log a x is shown. Find the value of a.
y  log a x
y
y
Using the point (5,1)
(5,1)
x
x
1
1  log a 5
a5