Chapter 1
Sets
Consider the set of faces on a die. Let us use numbers 1, 2, 3, 4, 5, 6 for the
number of dots on each face. Let S be the set of these possible outcomes.
We denote this as
S = {1, 2, 3, 4, 5, 6}.
Other sets are: people eye colour,
{blue, black, brown, green},
natural numbers,
{1, 2, 3, . . .},
a set of natural numbers next to each other
½
1.1 Subset
¾
{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6} .
⊆ or ⊂ (proper subset)
If E is sides of a die with an even number of dots then E = {2, 4, 6} and
E is a subset of S , denoted E ⊂ S (contained in S ). Let A be the set of
numbers of dots less than or equal to three, A = {1, 2, 3}. A is another
subset of S , A ⊆ S . Let B be the set of all odd numbers of dots on a
die. Then B = {1, 3, 5} and B ⊂ S but B is not a subset of A, denoted
B 6⊂ A.
1.2 Element of
∈
Each individual outcome is called an element of the set, denoted ∈.
4
1.3. Null Set
∅ or φ
5
3 is an element of S , 3 ∈ S .
3 is also an element of A, 3 ∈ A.
3 is not an element of E , 3 6∈ E .
{1, 2} is an element of the earlier set of natural numbers next to each
other,
½
¾
{1, 2} ∈ {1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6} .
1.3 Null Set
∅ or φ
The empty set, that is, the set that contains no elements, is called the
null set and is denoted ∅ or φ, or {}.
1.4 Union
S
The union of two sets, say A with B , is denoted A ∪ B and is the set
containing all the elements that are in A or B or both.
Examples
1. A = {1, 2, 3}, B = {1, 3, 5}
A ∪ B = {1, 2, 3, 5}
2. E = {2, 4, 6}
E ∪ B = {1, 2, 3, 4, 5, 6}.
Note: Duplicates of elements need not be shown in sets that describe
something and it is best to leave them out. For example,
A ∪ B = {1, 1, 2, 3, 5} = {1, 2, 3, 5};
the duplicate 1's are superuous. For sets, all we need to know is whether
an element is a member of the set or not.
6
Chapter 1. Sets
1.5 Intersection
T
The intersection of two sets, say A and B , is denoted A ∩ B , and is the
set containing those elements that are in both A and B .
Examples
1. A = {1, 2, 3}, B = {1, 3, 5}, E = {2, 4, 6}.
A ∩ B = {1, 3},
E ∩ A = {2},
E ∩ B = φ.
1.6 Complement
Suppose A ⊂ S .
The complement of A is all the elements that are not in A.
It is denoted Ā.
Note: A must be a subset of the other set.
Examples
1. A = {1, 2, 3}, E = {2, 4, 6}, S = {1, 2, 3, 4, 5, 6}, so A ⊂ S
and E ⊂ S .
Ā = {4, 5, 6},
Ē = {1, 3, 5}.
Note:
1. A ∪ Ā = S,
A ∩ Ā = φ.
2. If A 6⊂ B then the complement of A with respect to B doesn't make
sense. Also A ∪ Ā 6= B .
1.7. Mutually exclusive or disjoint (No symbol)
7
1.7 Mutually exclusive or disjoint (No symbol)
Two sets are said to be mutually exclusive or disjoint if they have no
elements in common, that is, A ∩ B = φ. For example, if B = {1, 3, 5}
and E = {2, 4, 6}, then B ∩ E = φ, and B and E are said to be mutually
exclusive or disjoint.
A = {1, 2, 3} is not mutually exclusive to B or E as A ∩ B = {1, 3}
and A ∩ E = {2}.
1.8 Cardinality
We will dene cardinality for nite sets only; innite sets (for example,
the natural numbers) need a more complicated denition.
The cardinality of a set A, denoted |A|, is the number of elements of
A.
Note, since we are using sets that describe something, elements appearing more than once are counted once only. That is, any duplicates
are omitted from our counting.
Examples
1. A = {1, 2, 3}, |A| = 3.
2. |{1, 2, 3, 5}| = 4.
3. |φ| = 0.
4. |{1, 1, 2, 3}| = 3 (duplicate 1 not counted again).
8
Chapter 1. Sets
1.9 Some relationships among sets
1.9.1 Associative laws
(A ∪ B) ∪ C = A ∪ (B ∪ C)
(A ∩ B) ∩ C = A ∩ (B ∩ C)
1.9.2 Distributive laws
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
1.9.3 De Morgan's laws
A ∪ B = Ā ∩ B̄
A ∩ B = Ā ∪ B̄
1.10. Venn Diagrams
9
1.10 Venn Diagrams
Venn Diagrams are a graphical aid in set theory.
Examples
1. S = {1, 2, 3, 4, 5, 6}, A = {1, 2, 3}, B = {1, 3, 5}, E =
{2, 4, 6}.
Ā
A
1, 2, 3
4, 5, 6
S
A ∪ Ā = S − union of complements.
A
E
1, 3,
2
5
4, 6
S
A ∪ E = {1, 3, 2, 4, 6} −union
A ∩ E = {2}
−intersection
10
Chapter 1. Sets
Examples ctd. S = {1, 2, 3, 4, 5, 6}, A = {1, 2, 3}, B = {1, 3, 5},
E = {2, 4, 6}.
φ
B
E
1, 3, 5
2, 4, 6
S
B ∩ E = φ − mutually exclusive or disjoint.
1.11 Inclusion - exclusion principle
|A ∪ B| = |A| + |B| − |A ∩ B|
Note: |A ∪ B| is not simply |A| + |B| because we are ignoring duplicate
elements.
Outline proof: Assume A ∩ B 6= φ.
A
B
|A| − |A ∩ B|
|B| − |A ∩ B|
|A ∩ B|
1.12. Example
11
Then
|A ∪ B| = (|A| − |A ∩ B|) + |A ∩ B|
+(|B| − |A ∩ B|)
= |A| + |B| − |A ∩ B|
If we have 3 sets then
|A ∪ B ∪ C| = |A| + |B| + |C|
−|A ∩ B| − |A ∩ C| − |B ∩ C|
+|A ∩ B ∩ C|.
1.12 Example
A batch of 75 electric motors are inspected for faults. 13 have scratched
paintwork, 6 have a loose cord, 10 have a missing pulley, 3 have scratched
paintwork and a loose cord, 6 have scratched paintwork and a missing
pulley, 4 have a loose cord and a missing pulley and 2 have all three
faults.
(a) How many of the motors have no faults?
(b) How many of the motors have scratched paintwork but do not have
a loose cord or a missing pulley?
(c) How many have scratched paintwork and a loose cord but do not
have a missing pulley?
Let S be the set of motors with scratched paintwork.
Let L be motors with a loose cord.
Let M be motors with a missing pulley.
12
Chapter 1. Sets
1.12.1 Solution by use of a Venn diagram
57
S
L
1
6
1
2
2
4
2
M
|S| = 75
|S ∩ L ∩ M | = 2 (all three faults)
|S ∩ L| = 3 (scratched and loose cord)
3 − 2 = 1 (scratched and loose cord and not a missing pully)
|S ∩ M | = 6 (scratched and missing pully)
6−2=4
|L ∩ M | = 4 (loose cord and missing pully)
4−2=2
|S| = 13 (scratched paintwork)
13 − 1 − 2 − 4 = 6
|L| = 6 (loose cord)
6−1−2−2=1
|M | = 10 (missing pully)
(a) 10 − 4 − 2 − 2 = 2
Therefore total motors with defects (|S ∪ L ∪ M |) is
6 + 1 + 1 + 4 + 2 + 2 + 2 = 18.
Therefore number having no faults is 75 − 18 = 57.
(b) 6
(c) 1
1.12. Example
13
1.12.2 Solution by formula
(a)
|S ∪ L ∪ M | = |S| + |L| + |M | − |S ∩ L| − |S ∩ M |
−|L ∩ M | + |S ∩ L ∩ M |
= 13 + 6 + 10 − 3 − 6 − 4 + 2
= 18.
Therefore 75 − 18 = 57 have no faults.
(b) number of motors with scratched paint but no loose cord and no
missing pulley is
|S| − |S ∩ (L ∪ M )|
(L ∪ M represents loose cord or missing pulley)
Now
|S ∩ (L ∪ M )| = |S| + |L ∪ M | − |S ∪ (L ∪ M )|
= 13 + |L ∪ M | − 18.
|L ∪ M | = |L| + |M | − |L ∩ M |
= 6 + 10 − 4 = 12.
Therefore
|S ∩ (L ∪ M )| = 13 + 12 − 18 = 7.
Therefore
|S| − |S ∩ (L ∪ M )| = 13 − 7 = 6.
(c) number scratched and loose cord and not a missing pulley is the
number scratched and a loose cord, |S ∪ L|, less the number that are
scratched and have a loose cord and have a missing pulley, |S ∩ L ∩ M |.
Therefore
|S ∩ L| − |S ∩ L ∩ M | = 3 − 2 = 1.
Please do Exercises 1A and 1B.
14
Chapter 1. Sets
1.13 Sets in Calculus
The Natural numbers, denoted by N, is the set of all the counting
numbers starting from 1 (not 0; computer scientists and French-educated
people start at 0).
N = {1, 2, . . .}
The Integers, denoted Z (from German zahl for number), is the set of
all counting numbers, negative counting numbers and 0.
Z = {. . . , −2, −1, 0, 1, 2, . . .}
The Rational numbers, denoted Q, is the set of all quotients of integers
with denominator 6= 0.
a
Q = {x : x = , where a, b ∈ Z, b 6= 0}.
b
eg.
1
,
2
2
2= ,
1
0
,
6
27
,
4
27
,
3
3.56,
3.56̇.
The Irrational numbers, √
with√no special symbol, have no recurring
decimal, for example surds 2, 3.1, . . . , and transcendental numbers
e ' 2.71828, π ' 3.1415 . . .. A number is irrational if it is not rational.
It turns out that irrational numbers are either transcendental, or not
transcendental (called algebraic). A number x is transcendental if taking
any powers of x and and multiples of the powers and adding them together we can never get zero, unless we always take zero multiples. Note
that if x is rational, say x = a/b, then of course bx
√− a = 0, so a/b is not
transcendental. √
It is algebraic. Now what about
√ 3? Although it is not
2
rational, if x = 3, we have that x − 3 = 0 so 3 is not transcendental.
However, for π , for example, no matter how you take multiples of
1, π, π 2, π 3, . . .
and add them together, you can never get zero, unless you do the trivial thing of taking zero times each one. So π is transcendental (not
1.13. Sets in Calculus
15
√
√
√
algebraic). What
2 + √3? We saw that 3 is not transcenden√ about
2
tal, and
since
(
2)
− 2 = 0, 2 isn't transcendental either. Well, if
√
√
x = 2 + 3, note that
√
√
x = 2 + 2 6 + 3 = 5 + 2 6,
2
so that
√
x − 5 = 2 6,
2
and then
(x2 − 5)2 = 24,
hence
(x2 − 5)2 − 24 = 0.
Multiplying this out we have
x4 − 10x2 + 1 = 0
√ √
can
if x = 2 + 3, so it is not transcendental. So, an irrational number
√
be transcendental (such as π ) or not transcendental (such as 2).
The Real numbers, denoted R, is the union of Q and irrational numbers. eg.
1,
√
3,
e,
π,
6
,
5
3.145̇,
√
2 6.
Note: N ⊆ Z ⊆ Q ⊆ R.
1.13.1 Some set notation
We can list the elements of sets, eg.
{−1, 0, 1}, {0, e}, N = {1, 2, 3, . . .},
or specify a rule, ie.
{x : −2 < x ≤ 1, x ∈ Z}, {y : y + ey = x, x ≤ 1}.
16
Chapter 1. Sets
: means such that
∈ means is an element of, or belongs to, or is a member of.
For example, x ∈ R means x is a real number.
Please do Exercise 1C.
Chapter 2
Indices and Logarithms
2.1 Indices
Denition 2.1.1 If a ∈ R and n ∈ N then
n factors
z
}|
{
an = a × a × a × · · · × a .
n is called the exponent or power or index to which a is raised.
Rules
1. Let a ∈ R and a 6= 0. Then
a
−n
1
= n
a
and
³
0
a =1
For a > 0, dene
1
an =
´
a
1
−1
0
1= =a ·a =a .
a
√
n
a,
the positive number x for which xn = a.
2. Let a ∈ R and b ∈ R. If p, q ∈ Q, let p =
provided all the terms exist,
√
(a) a = a = ( a) = n am
p
m
n
√
n
m
(b) ap aq = ap+q
(c) (ab)p = ap bp
(d) (ap )q = apq
17
m
n
and q = dc . Then,
18
Chapter 2. Indices and Logarithms
Note:
•
√
1
n
a = a n means the nth root of a.
√
√
√
• 1 = 1, 4 = 2, 9 = 3, . . . so by convention we take the positive
value.
Examples
1.
√
9=3
2. (−27)1/2 is not real
3. (−27)1/3 = −3
4. (−8)2/3 = ((−8)1/3 )2 = (−2)2 = 4
5. (−8)
− 32
¡
¢
2/3 −1
= (−8)
= 4−1 =
1
4
6. (2a2 )3 = 23 (a2 )3 = 8a6
7.
¡
a+a
¢
−1 −1
µ
¶−1
1
= a+
a
µ 2
¶−1
a +1
=
a
1
= ³ 2 ´
a +1
a
=
8.
√
√
27 + 3 =
=
=
=
a
a2 + 1
√
√
9×3+ 3
√ √
√
9 3+ 3
√
√
3 3+ 3
√
4 3
2.1. Indices
19
Examples ctd
9.
√
10.
√
√125
5
11.
√1
5
12.
2
√
3 2
√
5√ 5
5
=
=
√1
5
2
3
×
= ×
8×
√
√ √
2 = 2 2 2
= 2×2
= 4
=5
√
√5
5
√1
2
=
×
√
5
5
√
√2
2
2
3
= ×
13.
√
2
√ =
5− 3
=
=
=
√
2
2
=
√
2
3
√
√
2
( 5 + 3)
√
√ · √
√
( 5 − 3) ( 5 + 3)
√
√
2( 5 + 3)
√
√
2
( 5) − ( 3)2
√
√
2( 5 + 3)
√ 5 −√3
5+ 3
20
Chapter 2. Indices and Logarithms
2.2 Logarithms
Denition 2.2.1 Let b be a positive real number. We will dene for each
positive real number N a number denoted logb N , called the logarithm base
b of N , by the rule
x = logb N
if and only if
bx = N.
Hence logb N is an exponent (a power ); it is the power b must be raised
to to give N . Of course, b is the base.
Rules for logarithms
Let x and y be positive real numbers and let p be a real number.
(i) logb xy = logb x + logb y
(ii) logb xp = p logb x
(iii) logb x =
logc x
(change of base)
logc b
Trivial example: 3 = log4 64 =
log2 64
log2 4
=
6
2
= 3.
Proof of (i). We have that x = blogb x and y = blogb y , so starting with
the left hand side of (i) we have
¡
¢
logb xy = logb blogb x × blogb y
¡ log x+log y ¢
b
= logb b b
, by indices rule 2(b),
= logb x + logb y.
2
You see, therefore, that rule (i) is really just another way to say that
bpbq = bp+q . Can you see that rule (ii) is really just another way to say
(ac)d = acd (indices rule 2(d))?
2.2. Logarithms
21
Special Cases
logb b = 1
logb 1 = 0
logb bp = p logb b = p (surely obvious without using rule (ii)!)
√
1
logb n x = logb x1/n = logb x
n
x
logb = logb xy −1 = logb x + logb y −1 = logb x − logb y
y
There are no rules to simplify
log(x + y),
log(x − y).
How are you going to remember this?
We saw that
logb x + logb y = logb xy,
so how can
logb x + logb y
equal
logb(x + y)?
I think this mistake is made every year by several people. So,
log(x + y) 6= log x + log y,
and
log(x − y) 6= log x − log y.
2.2.1 Natural logarithms
Natural logarithms are logarithms to the base e and are denoted ln x
or sometimes loge x. See footnote1
e is a transcendental number that satises
e ' 2.718281828
1 What's
d x
dx e
= ex .
the big deal about e? Well, it turns out that if you want a function f (x) whose derivative is itself, setting
f (x) = ex does the trick. This is the most important function in mathematics.[WRU]
22
Chapter 2. Indices and Logarithms
Examples
If
ln 2 ' 0.6931
ln 3 ' 1.0986
ln 5 ' 1.6094 as approximations,
and
then
1. ln 8 = ln 23 = 3 ln 2 ' 3 × 0.6931.
2. ln 6 = ln(2 × 3) = ln 2 + ln 3 ' 0.6931 + 1.09864.
3. ln(2 + 3) = ln 5 ' 1.6094 (6= ln 2 + ln 3).
4. ln 21 = − ln 2 ' −0.6931.
5. (ln 2)(ln 3) ' 0.6931 × 1.0986 (6= ln 6).
6. ln(2 − 3) = ln(−1) is not real.
7. ln
√
2 = ln 21/2 = 12 ln 2 ' 12 × 0.6931.
8. ln e3 = 3 ln e = 3.
9. log2 e =
loge e
loge 2
=
1
ln 2
'
1
0.6931 .
2.2. Logarithms
23
Examples
1. If ln i = − R
L t − ln I , solve for i.
Solution: Exponentiate both sides:
R
eln i = e− L t−ln I ,
hence
and so
R
1
i = e− L teln I ,
1 R
i = e− L t .
I
2. If ln j = RL
t + ln(j − j0 ), solve for j . This is an exercise for you.
As above, exponentiate both sides.
Ans:
j0eRL/t
.
j = RL/t
e
−1
Please do Exercises 1D. (They are coming...)
Chapter 3
Factorization
3.1 Binomial Expansion
(a + b)0
(a + b)1
(a + b)2
(a + b)3
(a + b)4
(a + b)5
=
1
=
a+b
=
a2 + 2ab + b2
=
a3 + 3a2b + 3ab2 + b3
=
a4 + 4a3b + 6a2b2 + 4ab3 + b4
= a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
···
Alternatively, the binomial expansion may be calculated by:
µ
(a + b)n =
n
0
µ
+
¶
¶
n
an +
an−1b +
an−2b2 + · · ·
2
¶
µ
¶
µ ¶
n
n
n
a2bn−2 +
abn−1 +
bn,
n−2
n−1
n
where
µ
µ
n
r
n
1
¶
µ
=
n!
(n − r)!r!
¶
are the binomial coecients. Note that
n! = 1 × 2 × 3 × 4 × 5 × · · · × (n − 1) × n (n factors),
and 0! = 1.
eg. (a + b)2 =
µ
2
0
¶
µ
a2 +
2
1
¶
µ
ab +
24
2
2
¶
b2 = a2 + 2ab + b2.
3.2. Factorization
25
3.2 Factorization
Factorization of an algebraic expression consists of rewriting the given
algebraic expression as a product. The terms in the product are called
the factors. Factorization is an application of the distributive law:
a(b + c) = ab + ac.
Examples
1. ac + d − c + ad = c(a − 1) − d(a − 1) = (c − d)(a − 1).
2.
x2 y 2 +5xy+6
xy+2
=
(xy+2)(xy+3)
xy+2
= xy + 3, provided xy 6= −2.
3. a3 + a2 b + ab2 + b3 = a2 (a + b) + b2 (a + b) = (a2 + b2 )(a + b).
4. x2 −z 2 −3x−3z = (x+z)(x−z)−3(x+z) = (x+z)(x−z −3).
3.3 Standard Factorization
1. Dierence of powers
an − b n
(n is an integer)
is always factorizable and (a − b) is one of the factors.
2. Sum or powers
Case 1
a2n+1 + b2n+1
(2n + 1 is odd)
is always factorizable and (a + b) is one factor.
Case 2
a2n + b2n
always factorizable.
is not
26
Chapter 3. Factorization
Examples
1. a2 − b2 = (a − b)(a + b).
2. a3 − b3 = (a − b)(a2 + ab + b2 ).
3. a3 + b3 = (a + b)(a2 − ab + b2 ).
4. a6 + b6 = (a2 )3 + (b2 )3 = (a2 + b2 )(a4 − a2 b2 + b4 ).
Note: Once one factor is known, the other factor may be found by
long division. For example, factorize x3 + x2 − 5x − 2:
x2 +
¯ 3
x − 2 ¯ x + x2 −
x3 − 2x2
0 + 3x2 −
3x2 −
Therefore
3x + 1
5x − 2
5x
6x
x − 2
x − 2
0 + 0
x3 + x2 − 5x − 2 = (x − 2)(x2 + 3x + 1).
Another example: factorize a4 − b4 :
a3
¯ 4
a − b¯ a + 0
a4 − a3 b
a3 b
a3 b
+ a2b + ab2 + b3
+ 0 + · · · − b4
+ 0
− a2 b 2
a2 b 2 + 0
a2b2 − ab3
ab3 − b4
ab3 − b4
0 − 0
Also, a3 + a2 b + ab2 + b3 is further factorizable: note that if we set
3.4. Simplication of Algebraic Fractions
27
a = −b we get −b3 + b2b − bb2 + b3 = 0. It follows that1 (a + b) is a
factor of a3 + a2 b + ab2 + b3 . By long division (try it) the other factor is
(a2 + b2). Therefore
a4 − b4 = (a − b)(a + b)(a2 + b2).
Of course we could have seen this more easily just using dierence of
squares:
a4 − b4 = (a2)2 − (b2)2 = (a2 − b2)(a2 + b2) = (a − b)(a + b)(a2 + b2).
Further Examples
1.
x4
x3
3x −
= x(3 − )
9
9
x
= (27 − x3)
9
x
= (3 − x)(9 + 3x + x2).
9
2.
a6 − 26a3 − 27 = (a3)2 − 26a3 − 27
= (a3 − 27)(a3 + 1)
and each of these factors can be further factorized (do it!).
3.4 Simplication of Algebraic Fractions
There are no set methods to simplify algebraic fractions. The guidelines
are:
1. Try combining fractions over a common denominator.
1 Let's
say f (x) = (x − 1)(x2 + 3x + 1). It's clear because of the factor (x − 1) that f (x) = 0 when x = 1. From a
factor we get a root. Well, for polynomials it works the other way around. From a root we get a factor. For example,
if f (x) = x3 + 2x2 − 2x − 1, and we guess that f (x) = 0 when x = 1 (which you can check), then we know that (x − 1)
must be a factor of x3 + 2x2 − 2x − 1, which you can also check by long division. To relate this to the above, let
f (a) = a3 + a2 b + ab2 + b3 . Then f (a) = 0 when a = −b means (a + b) must be a factor of a3 + a2 b + ab2 + b3 .
28
Chapter 3. Factorization
2. Factorize numerator and/or denominator.
Examples
1.
1 1 1×b+1×a a+b
+ =
=
.
a b
ab
ab
2.
y 2 − xy − 2x2
(y − 2x)(y + x)
=
12x2 − 3y 2
3(4x2 − y 2)
(y − 2x)(y + x)
=
3(2x + y)(2x − y)
(y + x)
= −
.
3(2x + y)
3.
1 + a1
1 − a12
=
=
=
=
¡
¢ 2
1
1+ a a
¡
¢
1 − a12 a2
a2 + a
a2 − 1
a(a + 1)
(a + 1)(a − 1)
a
.
a−1
4.
1
1
x + 1 + (x − 1)
+
=
x−1 x+1
(x − 1)(x + 1)
2x
=
.
(x − 1)(x + 1)
Please do Exercise set 1E and 1F (when you get them!).
Chapter 4
Functions
4.1 Function Notation
A function is a mathematical statement of a rule which associates to each
of a given set of objects precisely one object. When you ask, What is the
distance from BSB to KB? you would not expect two dierent answers.
This is the idea of a function. So in this example you show that you
believe in functions, or at least one function, in this case the function
that takes any pair of cities and associates the distance between them.
A function is an answer-giver. The answer, in this case, depends upon
what pair of cities is taken.
An example of the notation used for functions is:
f (x) = ax2 + bx + c,
where a, b, and c are some xed numbers that we do not know, for the
present.
f is the function name or symbol. The round braces (parentheses) on
the left indicate what variable(s) the function depends upon, in this case
just x. f depends upon x because, as you can see, if we change x, we get
a dierent answer on the right hand side, which is the rule for nding
the answer.
Note: The variable name x is irrelevant. f could also be written as
f (t) = at2 + bt + c.
The variable is treated as a place holder. Here, we have substituted t for
x.
The idea, of course, is to substitute numbers, for example
f (2) = a × 22 + b × 2 + c = 4a + 2b + c.
29
30
Chapter 4. Functions
We would say that when x = 2 we have f (x) = 2. You can do silly
things, too, like
f (elephant) = a × elephant2 + b × elephant + c.
We can prohibit such silliness by insisting that f only applies to numbers.
Not so silly is to substitute another function for x, for example
f (z 2 + 6) = a(z 2 + 6)2 + b(z 2 + 6) + c,
or
µ ¶2
1
b
f (1/x) = a
+ + c.
x
x
4.1.1 Implicitly-dened relations
Functions, or more correctly relations1 may be implicitly dened, for
example
f (x) + xf 2(x) = 6a.
[f 2 (x) means (f (x))2 ]
Here, f is the function (relation) name and it is dependent upon x. See
example 2 below.
1 because
they break the requirement that a function only ever give one answer
4.1. Function Notation
31
Examples
1. Suppose g(x) =
g(0) =
1−x
1+x .
Then
1−0
1−1
= 1, g(1) =
= 0, g(−1) is not dened,
1+0
1+1
and
1 − t2
g(t ) =
,
1 + t2
2
g(1 + t) =
1 − (1 + t)
−t
=
.
1 + (1 + t) 2 + t
2. Suppose y(x) + x y 2 (x) = 1.
Then to nd y(1), we have y(1) + 1 × y 2 (1) = 1, hence y(1)2 +
y(1) = 1 and so
y(1) =
−1 ±
√
√
1 + 4 × 1 × 1 −1 ± 5
=
.
2
2
Note here that y(t2 ) must satisfy y(t2 ) + t2 y(t2 ) = 1.
4.1.2 Zeros or Roots of a function
If f is a function, then a real number a such that f (a) = 0 is called a
zero of f or a root of the equation f (x) = 0.
Examples
1.
f (x) = x2 + x
The roots are given by x2 + x = 0. Factorizing: x2 + x = 0 =⇒
x(x + 1) = 0. Hence the roots are x = 0 and x = −1.
32
Chapter 4. Functions
4.2 Two special functions
4.2.1 Linear functions
A function of the form
(a, b are constants)
f (x) = ax + b
is called a linear function. It's graph is a straight line. It has one zero
− ab (for a 6= 0). For example, for f (x) = 2x + 1, clearly f (x) = 0 when
x = − 12 .
4.2.2 Quadratic functions
A function of the form
f (x) = ax2 + bx + c
(a, b, c constants, a 6= 0)
is called a quadratic function.
The constant a is called the coecient of x2 .
The constant b is called the coecient of x.
The constant c is called the constant term (or coecient of x0 ).
Its graph is a parabola. It has up to two real zeros:
√
−b ± b2 − 4ac
.
x=
2a
The term b2 − 4ac is called the discriminant.
If
b2 − 4ac = 0 roots real and equal
b2 − 4ac > 0 distinct roots, both real
b2 − 4ac < 0 distinct roots, both complex.
The sum of the roots is − ab . (can you see that from the formula?)
The product of the roots is ac . (can you show this 2 )
2
which you can check is c/a.
Ã
−b +
√
b2 − 4ac
2a
!Ã
−b −
√
b2 − 4ac
2a
!
=
b2 − (b2 − 4ac)
,
4a2
4.2. Two special functions
33
You don't need the fancy formula to see that the sum of the roots is
− ab , or that the product of the roots is ac . Let us call the roots α (alpha)
and β (beta). So
√
−b + b2 − 4ac
α=
,
2a
β=
−b −
√
b2 − 4ac
.
2a
These are where f (x) = ax2 + bx + c is zero. Getting α and β really
means we are writing
µ
b
c
ax2 + bx + c = a x2 + x +
a
a
¶
= a(x − α)(x − β).
Multiplying that last expression out gives us
ax2 − aαx − aβx + aαβ = ax2 − a(α + β)x + aαβ,
(check this)
and if this is to still be ax2 + bx + c, we must have
b = −a(α + β),
and c = aαβ,
giving the sum of the roots
b
a
to be minus the coecient of x divided by a, and the product of the roots
c
αβ =
a
to be the constant term divided by a. See the following example.
α+β =−
34
Chapter 4. Functions
Examples
1.
f (x) = x2 − 8x + 4
roots are
x =
−(−8) ±
p
√
= 4 ± 2 3.
(−8)2 − 4 × 1 × 4
2×1
The roots are distinct and real.
(−8)
The sum of the roots is − 1 = 8. Note that this is minus the
coecient of x in the quadratic. (a = 1)
The product of the roots is 41 = 4. Note that this is the constant
in the quadratic. (still, a = 1)
4.2. Two special functions
35
4.2.2.1 Roots by completing the square
We attempt to put ax2 + bx + c into the form (A + B) + C , knowing that
(A + B)2 = A2 + 2AB + B 2:
µ
ax2 + bx + c =
=
=
=
=
=
¶
b
c
a x2 + x +
a
a
Ã
!
µ ¶2 µ ¶2
b
b
b
c
a x2 + 2 · x ·
+
−
+
2a
2a
2a
a
"
µ ¶2 #
µ ¶2
b
b
b
+
a x2 + 2 · x ·
−a
+c
2a
2a
2a
¤
£
a A2 + 2AB + B 2 + C,
b
where A = x, B = ,
µ 2a¶2
4ac − b2
b
=
C =c−a
2a
4a
a (A + B)2 + C
µ
¶2
b
4ac − b2
a x+
+
.
2a
4a
We have completed the square. To get the roots we solve
µ
b
a x+
2a
¶2
4ac − b2
+
= 0.
4a
36
Chapter 4. Functions
Examples
1. Find the roots of 2x2 − 5x − 3 = 0.
Solution:
Ã
!
µ ¶2 µ ¶2
5
5
3
5
2x2 − 5x − 3 = 2 x2 − 2 · x +
−
−
4
4
4
2
"µ
#
¶2
5
25 + 24
= 2 x−
−
4
16
"µ
#
¶2
5
49
= 2 x−
−
.
4
16
h¡
i
¢
5 2
49
Solve 2 x − 4 − 16 = 0. Therefore
µ
5
x−
4
so
x−
therefore x =
7+5 −7+5
4 ,
4 .
Please do Exercises 1G, 1H.
¶2
=
49
,
16
7
5
=± ,
4
4
The roots are x = 3 or x = − 12 .