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FUNCTIONS
Reference: Croft & Davision, Chapter 6 p.125
http://www.math.utep.edu/sosmath
Basic Concepts of Functions
A function is a rule which operates on an input and produces a
single output from that input.
Consider the function given by the rule: 'double the input'.
Functions
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Page 1
e.g.1 Given f (x) = 2x + 1 find:
(a) f (3)
(b) f (0)
(d) f (a)
(e) f (2a)
(g) f ( t + 1 )
a) 2(3)+1=7
d) 2(a)+1=2a+1
g) 2(t+1)+1=2t+3
(c) f (–1 )
(f ) f (t)
b) 2(0)+1=1
e) 2(2a)+1=4a+1
c) 2(-1) +1=-1
f) 2(t)+1=2t+1
End of Block Exercise
p.129
Functions
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Page 2
.
The Graph of a Function
A function may be represented
in graphical form.
The function f (x) = 2x is shown in
the figure.
We can write: y  f ( x )  2 x
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Page 3
In the function y = f (x), x is the independent variable
and y is the dependent variable.
The set of x values used as input to the function is
called the domain of the function
The set of values that y takes is called the range of the
function.
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Page 4
f (t)
e.g.2 The figure shows
the graph of the function
f (t) given by
f (t )  t 2 ,  3  t  3
(a) State the domain of
the function.
[-3, 3]
(b) State the range of
the function by
[0, 9]
inspecting the graph.
f (t )  t 2
9-
-3 -2 -1 0
t
1
2
3
e.g.3 Explain why the value t = 0 must be excluded
from the domain of the function f (t) = 1/t.
∵ 1/0 is undefined
End of Block Exercise
p.135
Functions
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Page 5
Determine the domain of each of the following functions:
(a)
f x   x  5
All real number
(b)
1
g t  
t
All real number except 0
2
hs   s  2
(c)
(d)
r x  
x3
2x  9x  5
2
(,  )
(, )
x0
For S≧2
[ 2,)
x3
2 x  1x  5
All real number except 5 & -0.5
(,  )
x  5 or  0.5
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Page 6
Composition of Functions
Reference URL:
http://archives.math.utk.edu/visual.calculus/0/compositions.5/
When the output from one function is used as the input
to another function - Composite Function
Consider
f (t )  2t  1, g (t )  t 2
f ( g (t ))  f (t 2 )  2t 2  1
2
(
2
t

1
)
g ( f (t ))  g ( 2t  1) 
Note : f ( g (t ))  g ( f (t ))
End of Block Exercise
p.141
Functions
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Page 7
One-to-many rules
f
x

only one y
Note:
x   x is not a function
One-to-many rule is not a function.
e.g.
But functions can be one-to-one or many-to-one.
f (x) = 5x +1 is an example of one-to-one function.
f (t )  t
Functions
2
is an example of many-to-one function.
V-01
Page 8
Inverse of a Function
x
 f ( x )  x
f
f
1
f 1 ( x ) is the notation used to denote the inverse
function of f (x). The inverse function, if exists, reverse
the process in f (x).
e.g.4 Find the inverse function of g ( x )  4 x  3
End of Block Exercise
p.148
Functions
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Page 9
Solution of e.g.4 :
The inverse function, g-1, must take an input 4x – 3 and
give an output x. That is,
g-1(4x-3) = x
Let Z = 4x-3, and transpose this to give x = (Z+3)/4
Then,
g-1(Z) = (Z+3)/4
Writing with x as its argument instead of Z gives
g-1(x) = (x+3)/4
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Page 10
f-1(3x-8)=x --------------------------------- Step 1
Class Exercises
Find the inverse functions for
the following functions.
1.
f(x) = 3x – 8
Let Z=3x-8, then x=(Z+8)/3 ------------Step 2
And then, f-1(Z) =(Z+8)/3 ----------------Step 3
Writing with x instead of Z, then,
f-1(x)=(x+8)/3 ----------------------------Step 4
Let Z=8-7x, then x=(8-Z)/7
2.
g(x) = 8 – 7x
Writing with x instead of Z, then,
g-1(x)=(8-x)/7
3.
f(x) = (3x – 2)/x
Let Z=(3x-2)/x, then x=-2/(Z-3)
Writing with x instead of Z, then,
f-1(x)=-2/(x-3)
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Page 11
TRIGONOMETRIC FUNCTIONS
Reference: Croft & Davision, Chapter 9
http://www.math.utep.edu/sosmath
L
Angles
r
Two main units of angle measures:
degree
90o, 180o
radian
 , 1/2 
Unit Conversion
 radian = 180o
e.g.1. Convert 127o in radians.

180
127

 x 
 0.706 rad
x 127
180
Trigonometric
Functions
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 r
Circle Angle 2

Arc Angle

Circumference 2r

Arc Length
L
L  r
  1 rad  Circle Angle  2  360
Page 12
Trigonometric functions
y
y
;
r
x
cos   :
r
y
tan   ;
x
sin  
P(x,y)
r
y
x

x
1
sin 
1
sec  
cos 
1
cot  
tan 
csc  
Reference URL: http://home.netvigator.com/~leeleung/sinBox.html
Trigonometric
Functions
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Page 13
The sign of a trigonometric ratio depends on the
quadrants in which  lies.
The sign chart will help you to remember this.
y
sin’+’ve
S
T
tan’+’ve
Trigonometric
Functions
All ‘+’ve
A
C
x
cos ‘+’ve
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Page 14
Reference Angle: 
y
II
y
I


O

x
O
  1800  
III
y

y
x
IV

O
    180o
Trigonometric
Functions
x


O

x

  3600  
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Page 15
Reduction Principle
sin    sin 
Where the sign depends on
S
cos    cos 
T
tan    tan 
A
C
2
1
=sin (1800-300 )
30
○
(b) cos 210 (c) tan 315
2
3
=cos(1800+300)
=tan(3600-450)
=sin300
=-cos300
=-tan450
=1/2 or 0.5
=
=-1
Trigonometric
Functions
○
1
e.g.2 Without using a calculator, find
(a) sin 150
45
 3/2
V-01
60
1
Page 16
○
Negative Angles
Negative angles are angles generated by
clockwise rotations.

x
sin(  )   sin 
Therefore
cos(  )  cos 
tan(  )   tan 
e.g.3
Find (a) sin(-30o)
=-sin 30
(b) cos (-300o)
o
o
o
=cos(360 - 60 )
o
=cos 60
=-1/2
=1/2
Trigonometric
Functions
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Page 17
Trigonometric graphs
Consider the function y = A sin x, where A is a positive
constant. The number A is called the amplitude.
Trigonometric
Functions
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Page 18
Example
State the amplitude of each of the following functions:
1.
2.
3.
4.
-2≦y ≦2
-4.7≦y ≦4.7
-2/3≦y ≦2/3
-0.8≦y≦0.8
y = 2 sin x
y = 4.7cos x
y = (2 sin x) / 3
y = 0.8cos x
Trigonometric
Functions
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Page 19
Simple trigonometric equations
Notation :
If sin  = k then  = sin-1k ( sin-1 is written as inv sin or
arcsin). Similar scheme is applied to cos and tan.
e.g.4 Without using a calculator, solve sin  =  0.5,
where 0o    360o
o
o
sin-1 0.5 = 210 , 330
e.g.5 Solve cos 2 =  0.4 , where 0    2
cos-1(-0.4)= 2
o
o
113.58 = 2 or 246.4 = 2
o
= 56.8
Trigonometric
Functions
o
or =123.2
V-01
<--- WRONG UNIT
Page 20
e.g.5 Solve cos2 =  0.4 , where 0    2
cos-1(-0.4) = 2 = 113.580 = 0.631 rad
Thus: 2 =  - 1.16;  + 1.16; 3 - 1.16; 3 + 1.16, …..
= 1.98, 4.3, 8.26, 10.58, ……
Thus:  = 0.99, 2.15, 4.13 or 5.29 rad
S
 /2, …
A
2 =113.580
, 3, …
1.16 rad
1.16 rad
0, 2 …
1.16 rad
T
Trigonometric
Functions
V-01
1.16 rad
3 /2, …
C
Page 21
TRIGONOMTRIC EQUATIONS
Reference: Croft & Davision, Chapter 9, Blocks 5, 6, 7
Some Common Trigonometric Identities
A trigonometric identity is an equality which contains one or
more trigonometric functions and is valid for all values of the
angles involved.
e.g.
sin 2   cos2   1
(1)
tan 2  
1  sec 2 
(2)
cot  
1  csc 
(3)
2
Trigonometric
Identities
2
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Page 22
Exercise: Derive (2) and (3) from (1)
sin   cos   1
2
2
sin   cos 
1

2
cos 
cos 2 
2
2
sin  cos 
2


sec

2
cos  cos 2 
tan 2   1  sec 2 
2
Trigonometric
Identities
2
sin 2   cos 2   1
sin 2   cos 2 
sin 2 
sin 2  cos 2 

2
sin  sin 2 
1
1
tan 2 
cot 2   1
V-01

1
sin 2 
 csc 2 
 csc 2 
 csc 2 
Page 23
e.g.1 (a) Solve
2x  x 1  0
2
2x
-1
x
1
(2x -1) (x +1) = 0
 x = ½ or x = -1
2x2 + (x)(-1) + 2x + 1(-1)
= 2x2 + x - 1
(b) Using (a), or otherwise, solve
2cos2 1 sin , 0   2
2(1-sin2θ) - sinθ-1 = 0
2 - 2sin2θ- sinθ-1 = 0
2sin2θ+ sinθ-1 = 0
sinθ= 0.5
or
sinθ= -1
θ= 30°, 150°, 270°
= π/6, 5π/6, 3π/2
Trigonometric
Identities
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Page 24
e.g.2 Solve
4 sec 2   3 tan   5, 0    2
4(tan 2   1)  3 tan   5
4tanθ
4 tan 2   4  3 tan   5  0
4 tan   3 tan   1  0
2
or
-1
4tan2θ – 4tanθ+ tanθ - 1
= 4tan2θ – 3tanθ - 1
(4tanθ+ 1) (tanθ- 1) = 0
tanθ= 1
tanθ
1
tanθ= - 0.25
θ= 45°, 225°, 166° or 346°
= π/4, 5π/4, 0.92π or 1.92π
End of Block Exercise: p.336
Trigonometric
Identities
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Page 25
Solving equations with given identities
e.g.3 Using the compound angle formula
sin( A  B)  sin A cos B  cos A sin B
find the acute angle  such that
2 sin 23 cos  2 cos 23 sin   1
2
1
45
○
1
2(sin 23 cos   cos 23 sin  )  1
30
○
2 sin( 23   )  1
sin( 23   )  1
2

23    30

3
2
60

1
  7 
Trigonometric
Identities
○
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Page 26
e.g.4 Using the double-angle formula
sin 2 A  2 sin A cos A
solve sin2 = sin  , where 0º <360 º
2 sin  cos   sin 
2 sin  cos   sin   0
sin  (2 cos   1)  0
To make the answer to be 0,
either sinθ=0
or
2cosθ-1=0
cosθ=0.5
θ= 0° or 180°
θ= 60° or 300°
Trigonometric
Identities
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Page 27
e.g.5 Using the double-angle formula
cos 2 A  cos 2 A  sin 2 A  2 cos 2 A  1  1  2 sin 2 A
solve cos2 = sin  , where 0   2.
1  2 sin 2   sin 
2 sin 2   sin   1  0
(2 sin   1) (sin   1)  0
sinθ= 0.5
or
2sinθ
-1
sinθ
1
2sin2θ – sinθ+ 2sinθ - 1
= 2sin2θ + sinθ - 1
sinθ= -1
θ= 30°, 150° or 270°
= π/6, 5π/6 or 3 π/2
Trigonometric
Identities
V-01
Page 28
Engineering waves
Reference: Croft &Davison , pp 348
Often voltages and currents vary with time and may be
represented in the form
A sin(  t   ) or A cos( t   )
where A : Amplitude of the combined wave
 : Angular frequency (rad/sec) of the combined wave
(Affect wave width)
 : Phase angle (left and right movement)
t : time in second
Example
State (i) the amplitude and (ii) the angular frequency of the
following waves:
i) 2
ii) 5 phase angle=0
(a)
y = 2 sin 5t
(b)
y = sin (t/2)
i)
Trigonometric
Identities
1
ii) ½
V-01
phase angle =0
Page 29
Trigonometric
Identities
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Page 30
The period, T, of both y = A sin ω t and y = A cos ω t is
given by T = (2π)/ω
Example
State the period of each of the following functions:
1.
y = 3 sin 6t
2π/6
=π/3
2.
y = 5.6 cosπ t
2π/π = 2
The frequency, f, of a wave is the number of cycles
completed in 1 second. It is measured in hertz (Hz).
T=1/f
Example
State the period and frequency of the following waves:
1.
y = 3 sin 6 t
T=π/3,
2.
y = 5.6cosπ t
T=2,
Trigonometric
Identities
f = 3/π
f=½
V-01
Page 31
E.g. 6 (a) Find the maximum and minimum value of
5 sin ( t + 0.93 )
∵ max sin θ=1 and min sin θ=-1
∴ max = 5
and
min = -5
(b) Solve 5 sin ( t + 0.93) = 3.8, where 0  t  2
3.8
 0.76
5
t  0.93  sin 1 (0.76)  49.46
sin( t  0.93) 
t  0.93 
49.46

180
t  0.93  0.86rad or 2.28rad
t  0.07rad or 1.35rad
t  6.21rad or 1.35rad
What ? –ve rad ???
End of Chapter Exercise: p.360
Trigonometric
Identities
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Page 32
Formula for Reference (Given in the Exam)
sin  A  B   sin A cos B  cos A sin B sin A  sin B  2 sin  A  B  cos A  B 

 

2
2

 

cos A  B   cos A cos B  sin A sin B
tan A  tan B
tan  A  B  
1  tan A tan B
 A B   A B 
sin A  sin B  2 cos
 sin 

2
2

 

 A B   A B 
cos A  cos B  2 cos
 cos

 2   2 
 A B   A B 
cos A  cos B  2 sin 
 sin 

2
2

 

2 sin A cos B  sin  A  B   sin  A  B 
2 cos A cos B  cos A  B   cos A  B 
2 sin A sin B  cos A  B   cos A  B 
Trigonometric
Identities
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Page 33
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Reference: Croft & Davision, Chapter 8 p.253
http://www.math.utep.edu/sosmath
x
The exponential function is y  e
where e = 2.71828182…..
y  ex
y
30
25
Properties
(1) e x e y  e x  y
(2)
ex
e
y
e
x y
(3) ( e x ) r  e rx
20
(5) e x   as x  
10
x
(6) e  0 as x  
5
1
(7) e x  0 for all x
(4) e 0  1
Page 34
15
V-01
-3
-2
-1 0
1
2
3
x
e.g. 1 Simplify
(a)
e2 x e x
e 3 x
b e 4 x  (e 2 x  1) 2
e
2 x  x  ( 3 x )
 e 4 x  [( e 2 x ) 2  2e 2 x (1)  1]
e
6x
 e  (e  2e  1)
4x
4x
 e 4 x  e 4 x  2e 2 x  1
 2e 2 x  1
Exercise: p.259
Page 35
2x
V-01
Applications : Laws of growth and decay
(A)Growth curve
y
y  Ae kx ,
k 0
A
x
0
e.g. Change of electrical resistance (R) with temp. 
R  R0e
Page 36
V-01
(B) Decay Curve
y
A
y  Ae  kx ,
k 0
x
0
e.g. Discharge of a capacitor q (t )  Qe
Exercise: p.259
Page 37
V-01

t
CR
Class Exercise
Q=50, C=0.25 and R=2
a) When t=1, q(t) = ?
q(1) = 6.77
b) When R is double, q(1) = ?
q(1) = 18.39
Page 38
V-01
Logarithmic Functions
If N  a x then x  loga N , where a  1.
The number a is called the base of the logarithm.
e.g. 16  2 ,
4
 log 2 16  4
125  53 ,
 log 5 125  3
 0.01  10 2 ,
 log 10 0.01  2
In particular,
If N  10 then x  log N (common logarithm)
x
If N  e x then x  nN
(natural logarithm )
Exercise: p.271
Page 39
V-01
Properties of
nx
(1) nA  nB  n( AB)
(5) nx   as x  
 A
(2) nA  nB  n 
 B
(6) nx   as x  0
(3) nAn  nnA
(7) domain  (0, )
(4) n1  0
(8) range  (, )
y  nx
y
1
1
2
3
4
0
-1
5
x
-2
-3
Exercise: p.275
V-01
Page 40
(a) 5  3e 2 x (b) 36 
t
72(1  e 3
Solving equations
5
2x
e.g.2
 e Solve
3
5
ln  2 x
3
1 5
x
ln
2 3
x  0.255
Page 41
0.5  1  e
0.5  e
 4.92 
) (c) 2.58  n

x


t
3
2.58  ln 4.92  ln x
t
3
ln x  ln 4.92  2.58
ln x  0.987
t
ln 0.5 
3
t  2.079
x  0.373
V-01
e.g.3 The decay of current in an inductive circuit is given by
i  50e 0.1t
t 0
Find (a) the current when t=0;
(b) the value of the current when t=3;
(c) the time when the value of the current is 15.
i  50e
0.1t
for t  (0)
i  50e
 0.1( 0 )
i  50e
0.1( t )
for i  15
for t  3
i  50e
i  50e 0.1t
 0.1( 3)
i  50  1
i  50 x0.741
i  50
i  37.04
15  50e 0.1t
15
 e 0.1t
50
ln 0.3  0.1t
t  12.04
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