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FUNCTIONS II
Dr. Gabriel Obed Fosu
Department of Mathematics
Kwame Nkrumah University of Science and Technology
Google Scholar: https://scholar.google.com/citations?user=ZJfCMyQAAAAJ&hl=en&oi=ao
ResearchGate ID: https://www.researchgate.net/profile/Gabriel_Fosu2
Dr. Gabby (KNUST-Maths) Functions 1 / 41
Lecture Outline
1 Properties of Functions
Odd and Even Functions
Periodic Functions
Monotonic Functions
Bounded Functions
Maxima and Minima of Functions
2 Inverse Function
3 Sequence and Series
Dr. Gabby (KNUST-Maths) Functions 2 / 41
Properties of Functions Odd and Even Functions
Even Function
Let f be a function and Df its domain. We assume that if x ∈ Df then −x ∈ Df .
Definition (Even Function)
f is an even function if f (−x) = f (x).
Example
The functions f (x) = x2
, g(x) = −x4
+ 2x2
− 1, h(x) = cos(x) + x2
, i(x) = x sinx are even
functions since:
1 f (−x) = (−x)2
= x2
= f (x),
2 g(−x) = −(−x)4
+2(−x)2
−1 = g(x),
3 h(−x) = cos(−x)+(−x)2
= h(x), and
4 i(−x) = (−x)sin(−x) = −x(−sinx) = x sinx = i(x).
Dr. Gabby (KNUST-Maths) Functions 4 / 41
Properties of Functions Odd and Even Functions
Odd Function
Definition (Odd Function)
f is an odd function if f (−x) = −f (x).
Example
The functions f (x) = x, f (x) = −x3
+ 2x, f (x) = sin(x), and f (x) = csc(x), f (x) = tan(x) are
odd functions. Because
1 f (−x) = −x = −f (x)
2 f (−x) = −(−x)3
+2(−x) = x3
−2(x) = −f (x)
3 f (−x) = sin(−x) = −sin(x) = −f (x)
4 f (−x) =
1
sin(−x)
= −
1
sin(x)
= −f (x)
Dr. Gabby (KNUST-Maths) Functions 5 / 41
Properties of Functions Odd and Even Functions
Remarks
The graph of an even function is symmetric about the y-axis.
The graph of an odd function is symmetric about the origin.
Dr. Gabby (KNUST-Maths) Functions 6 / 41
Properties of Functions Periodic Functions
Periodic Functions
Definition
Let f be a function, and Df its domain, then f is a periodic function if there exists a
positive real number t such that f (x + t) = f (x) for all x ∈ Df .
The minimum of such t’s which is often denoted as T , is called the period of f .
Example
The trigonometric functions are periodic functions.
1 sin(x +2kπ) = sin(x +2π) = sin(x) for k ∈ Z, however, T = 2π.
2 cos(x +2kπ) = cos(x +2π) = cos(x),T = 2π.
3 tan(x +(2k +1)π) = tan(x +π) = tan(x),T = π.
Dr. Gabby (KNUST-Maths) Functions 7 / 41
Properties of Functions Monotonic Functions
Monotonic Functions
Let I be an open interval. x1 and x2 are two elements of I such that x1 < x2.
Definition
☛ f is an increasing function on I if f (x1)<f (x2).
☛ f is a decreasing function on I if f (x1)>f (x2).
Example
• The functions ex
, tan(x) and ax + b, where a > 0, are increasing on their respective
domains.
• The functions e−x
, cot(x) and ax + b, where a < 0, are decreasing on their respective
domains.
Dr. Gabby (KNUST-Maths) Functions 8 / 41
Properties of Functions Monotonic Functions
Monotonic Decreasing Function
−5 −4 −3 −2 −1 1 2 3 4 5
−3
−2
−1
1
2
3
0
−3x +1 ↘
e−x
↘
Dr. Gabby (KNUST-Maths) Functions 9 / 41
Properties of Functions Monotonic Functions
Monotonic Increasing Function
−5 −4 −3 −2 −1 1 2 3 4 5
−3
−2
−1
1
2
3
0
2x −1 ↗
ex
↗
Dr. Gabby (KNUST-Maths) Functions 10 / 41
Properties of Functions Monotonic Functions
Monotonic Increasing and Decreasing
−5 −4 −3 −2 −1 1 2 3 4 5
−3
−2
−1
1
2
3
0
sinx ↗↘
Dr. Gabby (KNUST-Maths) Functions 11 / 41
Properties of Functions Monotonic Functions
Monotonic Functions
Example
Show that the function f (x) =
p
x −2 is an increasing function on its domain.
1 Df = [2,+∞).
2 For x1,x2 ∈ Df and x1 < x2,
3 2 < x1 < x2 =⇒ 0 < x1 −2 < x2 −2
4 =⇒ 0 <
p
x1 −2 <
p
x2 −2
5 =⇒ f (x1) < f (x2).
6 Thus, f is an increasing function on its domain.
Dr. Gabby (KNUST-Maths) Functions 12 / 41
Properties of Functions Monotonic Functions
Monotonic Functions
Example
Show that f (x) = (2− x)2
+1 decreases on (−∞,2] and increases on [2,+∞).
1 Df = R.
2 For x1,x2 ∈ (−∞,2],
3 x1 < x2 ≤ 2 =⇒ −x1 > −x2 > −2
4 =⇒ 2− x1 > 2− x2 > 0
5 =⇒ (2− x1)2
> (2− x2)2
> 0
6 =⇒ (2− x1)2
+1 > (2− x2)2
+1 > 1
7 =⇒ f (x1) > f (x2).
8 f is decreasing on (−∞,2].
1 For x1,x2 ∈ [2,+∞),
2 2 ≤ x1 < x2 =⇒ −2 > −x1 > −x2
3 =⇒ 0 > 2− x1 > 2− x2
4 =⇒ 0 < (2− x1)2
< (2− x2)2
5 =⇒ 1 < (2− x1)2
+1 < (2− x2)2
+1
6 =⇒ f (x1) < f (x2).
7 f is an increasing function on [2,+∞).
Dr. Gabby (KNUST-Maths) Functions 13 / 41
Properties of Functions Bounded Functions
Bounded Functions
Definition
A function is said to be bounded above if there is ū ∈ R such that f (x) ≤ ū for all x in the
domain of f .
Example
The function f (x) = x2
+ 1 defined on 0 ≤ x ≤ 1 is bounded above by 2 since f (x) ≤ 2 for
0 ≤ x ≤ 1.
Example
The function f (x) = 1/x defined on x ∈ N is bounded above by 1
Example
The function f (x) = sinx is bounded above by 1 for x ∈ R.
Dr. Gabby (KNUST-Maths) Functions 14 / 41
Properties of Functions Bounded Functions
Bounded Functions
Definition
A function, f , is said to be bounded below if there is ℓ ∈ R such that f (x) ≥ ℓ for all x in the
domain of f .
Example
The function f (x) = x−1 defined in [0,1] is bounded below by −1 since −1 ≤ f (x) for x ∈ [0,1].
Example
The function g(x) = |
p
x +1| is bounded below by 0 on the interval [0,4] since 0 ≤ g(x) for
x ∈ [0,4].
Dr. Gabby (KNUST-Maths) Functions 15 / 41
Properties of Functions Maxima and Minima of Functions
Maxima and Minima of Functions
Local(or relative) and Global(or absolute) Minimum
1 The function f is said to have a local minimum value at the point x0 if f (x0) ≤ f (x) for
all x in a neighbourhood of x0.
2 f is said to have a global minimum value at the point x0 if f (x0) ≤ f (x) for all x in the
domain of f .
3 In this case f is bounded below.
Local(or relative) and Global(or absolute) Maximum
1 If f (x) ≤ f (x0) for all x in a neighbourhood of x0, then f has a local maximum value at
the point x0.
2 The maximum is global if f (x) ≤ f (x0) for all x in the domain of f .
3 In this case f is bounded above.
Dr. Gabby (KNUST-Maths) Functions 16 / 41
Properties of Functions Maxima and Minima of Functions
Maxima and Minima of Functions
−5 −4 −2 2 4
−3
−1
1
2
3
0
global min
f (x) = 1+(x +1)2
,
Dr. Gabby (KNUST-Maths) Functions 17 / 41
Properties of Functions Maxima and Minima of Functions
Maxima and minima
−5 −4 −2 2 4
−3
−1
1
2
3
0
local max
local min
g(x) = 1−2x −3x2
+2x3
Dr. Gabby (KNUST-Maths) Functions 18 / 41
Properties of Functions Maxima and Minima of Functions
Maxima and minima
−5 −4 −2 2 4
−3
−1
1
2
3
0
global min
local max
local min
h(x) = (x −1)(−x +3)2
(x),
Dr. Gabby (KNUST-Maths) Functions 19 / 41
Inverse Function
Inverse Functions
1 An inverse function is a function that undoes the action of the another function.
2 A function g is the inverse of a function f if whenever y = f (x) then x = g(y)
3 In other words, applying f and then g is the same thing as doing nothing. We can
write this in terms of the composition of f and g as
g(f (x)) = x
4 A function f has an inverse function only if for every y in its range there is only one
value of x in its domain for which f (x) = y
5 This inverse function is unique and is frequently denoted by f −1
and called f inverse.
Dr. Gabby (KNUST-Maths) Functions 21 / 41
Inverse Function
Inverse Functions
Given the function f (x) we want to find the inverse function, f −1
(x)
1. First, replace f (x) with y.
2. Replace every x with a y and replace every y with an x.
3. Solve the equation from Step 2 for y. This is the step where mistakes are most often
made so be careful with this step.
4. Replace y with f −1
(x). In other words, we’ve managed to find the inverse at this point!
5. Verify your work by checking that (f ◦ f −1
)(x) = x and (f −1
◦ f )(x) = x are both true.
Example
Given f (x) = 3x −2 find f −1
(x)
Dr. Gabby (KNUST-Maths) Functions 22 / 41
Inverse Function
1 Given the function f
f (x) = 3x −2
2 First, replace f (x) with y.
y = 3x −2
3 Replace every x with a y and replace every y with an x.
x = 3y −2
4 Solve the equation from Step 2 for y.
y =
1
3
(x +2)
5 Replace y with f −1
(x).
f −1
(x) =
x
3
+
2
3
Dr. Gabby (KNUST-Maths) Functions 23 / 41
Inverse Function
We can verify the results, we check that (f ◦ f −1
)(x) = x
(f ◦ f −1
)(x) = f [f −1
(x)]
= f
hx
3
+
2
3
i
= 3
hx
3
+
2
3
i
−2
= x +2−2
= x
Dr. Gabby (KNUST-Maths) Functions 24 / 41
Inverse Function
Graph of a Function and Its Inverse
There is an interesting relationship between the graph of a function and the graph of its
inverse. Here is the graph of the function and inverse from the first two examples.
In both cases the graph of the inverse is a reflection of the actual function about the line
y = x . This will always be the case with the graphs of a function and its inverse.
Dr. Gabby (KNUST-Maths) Functions 25 / 41
Sequence and Series
Sequence and Series
Definition
A sequence is an ordered set of numbers that most often follows some rule (or pattern) to
determine the next term in the order.
Example
x,x2
,x3
,x4
, ... is a sequence of numbers, where each successive term is multiplied by x.
Definition
A series is a summation of the terms of a sequence. The greek letter sigma Σ is used to
represent the summation of terms of a sequence of numbers.
Dr. Gabby (KNUST-Maths) Functions 27 / 41
Sequence and Series
Sequence and Series
Series are typically written in the following form:
n
X
i=1
ai = a1 + a2 + a3 ···+ an
where the index of summation, i takes consecutive integer values from the lower limit, 1
to the upper limit, n. The term ai is known as the general term.
Example
5
X
i=1
i = 1+2+3+4+5 = 15 (1)
6
X
k=3
2k
= 23
+24
+25
+26
= 8+16+32+64 = 120 (2)
4
X
k=1
kk
= 11
+22
+33
+44
= 1+4+27+256 = 288 (3)
Dr. Gabby (KNUST-Maths) Functions 28 / 41
Sequence and Series
Finite Series
Definition (Finite and Infinite Series)
A finite series is a summation of a finite number of terms. An infinite series has an infinite
number of terms and an upper limit of infinity.
Properties of Finite Series
The following are the properties for addition/subtraction and scalar multiplication of series.
For some sequence ai and bi and a scalar k then:
n
X
i=1
kai ±bi = k
n
X
i=1
ai ±
n
X
i=1
bi
Dr. Gabby (KNUST-Maths) Functions 29 / 41
Sequence and Series
Theorems of Finite Series
1 The following theorems give formulas to calculate series with common general terms.
These formulas, along with the properties listed above, make it possible to solve any
series with a polynomial general term, as long as each individual term has a degree
of 3 or less.
n
X
i=1
1 = n (4)
n
X
i=1
c = nc (5)
n
X
i=1
i =
n(n +1)
2
(6)
n
X
i=1
i2
=
n(n +1)(2n +1)
6
(7)
n
X
i=1
i3
=
µ
n(n +1)
2
¶2
(8)
(9)
Dr. Gabby (KNUST-Maths) Functions 30 / 41
Sequence and Series
Types of Sequences
There are two main types of sequences.
1 An arithmetic sequence is one in which successive terms differ by the same amount.
For example, {3, 6, 9, 12, ···}. Note each term is obtained by adding 3 to the previous
term. This is called the common difference denoted as d = 6−3
2 A geometric sequence is one in which the quotient of any two successive terms is a
constant. For example, {3, 9, 27, 81, ···}. Note each term is obtained by multiplying
the previous term by 3. This is called the common ratio denoted by r = 9
3 .
Similarly, there are also arithmetic series and geometric series, which are simply
summations of arithmetic and geometric sequences, respectively.
Dr. Gabby (KNUST-Maths) Functions 31 / 41
Sequence and Series
Theorems for Arithmetic and Geometric Series
Arithmetic Series
Suppose we have the following arithmetic series,
{a +(a +d)+(a +2d)+···+(a +(n −1)d}
Then,
n−1
X
k=0
a +kd =
n
2
(2a +(n −1)d) (10)
Geometric Series
Suppose we have the following geometric series,
{a + ar + ar2
+...+ ar(n−1)
}
Then,
n−1
X
k=0
ark
= a
µ
rn
−1
r −1
¶
(11)
Dr. Gabby (KNUST-Maths) Functions 32 / 41
Sequence and Series
Example
Solve the series
20
X
i=1
2i2
+7i
20
X
i=1
2i2
+7i =
20
X
i=1
2i2
+
20
X
i=1
7i
= 2
µ
20(20+1)(2(20)+1)
6
¶
+7
µ
20(20+1)
2
¶
= 2
µ
17220
6
¶
+7(210)
= 5740+1470
= 7210
Dr. Gabby (KNUST-Maths) Functions 33 / 41
Sequence and Series
Example
Find the sum of the geometric series: 2+8+32+128+...+8192
1 We know that
8
2
= 4 and that
32
8
= 4, so r = 4 for this geometric series.
2 The initial value represents our a value, so a = 2.
3 Before we can write the series in summation notation, we must determine the upper
limit of the summation.
8192 = arn−1
= (2)(4n−1
)
4096 = 4n−1
46
= 4n−1
6 = n −1
n = 7
Dr. Gabby (KNUST-Maths) Functions 34 / 41
Sequence and Series
So, we can rewrite the series as
P6
k=0
2(4k
). From the formula for the sum of a geometric
series,
6
X
k=0
2(4k
) = 2
µ
(4)7
−1
4−1
¶
= 2(5461)
= 10922
Therefore, the sum of the series, 2+8+32+128+...+8192 is 10922
Dr. Gabby (KNUST-Maths) Functions 35 / 41
Sequence and Series
Binomial Series
In this final section of this chapter we are going to look at another series representation
for a function. Before we do this let’s first recall the following theorem.
Binomial Theorem
If n is any positive integer then,
(a +b)n
=
n
X
i=0
Ã
n
i
!
an−i
bi
= an
+nan−1
b +
n (n −1)
2!
an−2
b2
+···+nabn−1
+bn
where, Ã
n
i
!
=
n (n −1)(n −2)···(n −i +1)
i!
; i = 1,2,3,...n
Ã
n
0
!
= 1
Dr. Gabby (KNUST-Maths) Functions 36 / 41
Sequence and Series
Example
(2x −3)4
=
4
X
i=0
Ã
4
i
!
(2x)4−i
(−3)i
=
Ã
4
0
!
(2x)4
+
Ã
4
1
!
(2x)3
(−3)+
Ã
4
2
!
(2x)2
(−3)2
+
Ã
4
3
!
(2x)(−3)3
+
Ã
4
4
!
(−3)4
= (2x)4
+4(2x)3
(−3)+
3
2
(2x)2
(−3)2
+4(2x)(−3)3
+(−3)4
= 16x4
−96x3
+216x2
−216x +81
Dr. Gabby (KNUST-Maths) Functions 37 / 41
Sequence and Series
Binomial Series
If k is any number and |x| < 1 then,
(1+ x)k
=
∞
X
n=0
Ã
k
n
!
xn
= 1+kx +
k (k −1)
2!
x2
+
k (k −1)(k −2)
3!
x3
+···
where, Ã
k
n
!
=
k (k −1)(k −2)···(k −n +1)
n!
; n = 1,2,3,...
Ã
k
0
!
= 1
Example
Write down the first four terms in the binomial series for
p
9− x
Dr. Gabby (KNUST-Maths) Functions 38 / 41
Sequence and Series
We have to rewrite the terms into the form (1+ xk)k
required, that is
p
9− x = 3
³
1−
x
9
´1
2
= 3
³
1+
³
−
x
9
´´1
2
So, k = 1
2 and xk = −(x/9)
Then the binomial series is given,
p
9− x = 3
³
1+
³
−
x
9
´´1
2
= 3
∞
X
n=0
Ã
1
2
n
!
³
−
x
9
´n
= 3
"
1+(
1
2
)(−
x
9
)+
1
2 (−1
2 )
2
(−
x
9
)
2
+
1
2 (−1
2 )(−3
2 )
6
(−
x
9
)
3
+···
#
So the first four terms are
= 3−
x
6
−
x2
216
−
x3
3888
−···
Dr. Gabby (KNUST-Maths) Functions 39 / 41
Sequence and Series
Exercise
1 Find the period of the following functions
1) f (x) = sin(2x), 2) f (x) = cos(−2x +π/3), 3) f (x) = x −sin(x).
2 Find the domain of:
1)f (x) =
1
2x −6
2)f (x) =
1− x
1+ x
3)f (x) =
x3
−2x
x(−x −6)
4)f (x) = 3x −1−
1
2x −6
5)f (x) =
x
1−2x + x2
6)f (x) =
x2
−2x
(x −3)(1− x2)
3 Determine whether the functions below are even, odd or neither.
1) f (x) = ex2
−1
+ln(|x|+1), 2) f (x) = x2
−2
x(1−x2)
, 3) f (x) = x2
sin(x)
4) f (x) = x
p
|x|−1, 5) f (x) = ln
¡
tanx −e|x|
¢
, 6) f (x) = x −1.
Dr. Gabby (KNUST-Maths) Functions 40 / 41
END OF LECTURE
THANK YOU
Dr. Gabby (KNUST-Maths) Functions 41 / 41

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3. Functions II.pdf

  • 1. FUNCTIONS II Dr. Gabriel Obed Fosu Department of Mathematics Kwame Nkrumah University of Science and Technology Google Scholar: https://scholar.google.com/citations?user=ZJfCMyQAAAAJ&hl=en&oi=ao ResearchGate ID: https://www.researchgate.net/profile/Gabriel_Fosu2 Dr. Gabby (KNUST-Maths) Functions 1 / 41
  • 2. Lecture Outline 1 Properties of Functions Odd and Even Functions Periodic Functions Monotonic Functions Bounded Functions Maxima and Minima of Functions 2 Inverse Function 3 Sequence and Series Dr. Gabby (KNUST-Maths) Functions 2 / 41
  • 3. Properties of Functions Odd and Even Functions Even Function Let f be a function and Df its domain. We assume that if x ∈ Df then −x ∈ Df . Definition (Even Function) f is an even function if f (−x) = f (x). Example The functions f (x) = x2 , g(x) = −x4 + 2x2 − 1, h(x) = cos(x) + x2 , i(x) = x sinx are even functions since: 1 f (−x) = (−x)2 = x2 = f (x), 2 g(−x) = −(−x)4 +2(−x)2 −1 = g(x), 3 h(−x) = cos(−x)+(−x)2 = h(x), and 4 i(−x) = (−x)sin(−x) = −x(−sinx) = x sinx = i(x). Dr. Gabby (KNUST-Maths) Functions 4 / 41
  • 4. Properties of Functions Odd and Even Functions Odd Function Definition (Odd Function) f is an odd function if f (−x) = −f (x). Example The functions f (x) = x, f (x) = −x3 + 2x, f (x) = sin(x), and f (x) = csc(x), f (x) = tan(x) are odd functions. Because 1 f (−x) = −x = −f (x) 2 f (−x) = −(−x)3 +2(−x) = x3 −2(x) = −f (x) 3 f (−x) = sin(−x) = −sin(x) = −f (x) 4 f (−x) = 1 sin(−x) = − 1 sin(x) = −f (x) Dr. Gabby (KNUST-Maths) Functions 5 / 41
  • 5. Properties of Functions Odd and Even Functions Remarks The graph of an even function is symmetric about the y-axis. The graph of an odd function is symmetric about the origin. Dr. Gabby (KNUST-Maths) Functions 6 / 41
  • 6. Properties of Functions Periodic Functions Periodic Functions Definition Let f be a function, and Df its domain, then f is a periodic function if there exists a positive real number t such that f (x + t) = f (x) for all x ∈ Df . The minimum of such t’s which is often denoted as T , is called the period of f . Example The trigonometric functions are periodic functions. 1 sin(x +2kπ) = sin(x +2π) = sin(x) for k ∈ Z, however, T = 2π. 2 cos(x +2kπ) = cos(x +2π) = cos(x),T = 2π. 3 tan(x +(2k +1)π) = tan(x +π) = tan(x),T = π. Dr. Gabby (KNUST-Maths) Functions 7 / 41
  • 7. Properties of Functions Monotonic Functions Monotonic Functions Let I be an open interval. x1 and x2 are two elements of I such that x1 < x2. Definition ☛ f is an increasing function on I if f (x1)<f (x2). ☛ f is a decreasing function on I if f (x1)>f (x2). Example • The functions ex , tan(x) and ax + b, where a > 0, are increasing on their respective domains. • The functions e−x , cot(x) and ax + b, where a < 0, are decreasing on their respective domains. Dr. Gabby (KNUST-Maths) Functions 8 / 41
  • 8. Properties of Functions Monotonic Functions Monotonic Decreasing Function −5 −4 −3 −2 −1 1 2 3 4 5 −3 −2 −1 1 2 3 0 −3x +1 ↘ e−x ↘ Dr. Gabby (KNUST-Maths) Functions 9 / 41
  • 9. Properties of Functions Monotonic Functions Monotonic Increasing Function −5 −4 −3 −2 −1 1 2 3 4 5 −3 −2 −1 1 2 3 0 2x −1 ↗ ex ↗ Dr. Gabby (KNUST-Maths) Functions 10 / 41
  • 10. Properties of Functions Monotonic Functions Monotonic Increasing and Decreasing −5 −4 −3 −2 −1 1 2 3 4 5 −3 −2 −1 1 2 3 0 sinx ↗↘ Dr. Gabby (KNUST-Maths) Functions 11 / 41
  • 11. Properties of Functions Monotonic Functions Monotonic Functions Example Show that the function f (x) = p x −2 is an increasing function on its domain. 1 Df = [2,+∞). 2 For x1,x2 ∈ Df and x1 < x2, 3 2 < x1 < x2 =⇒ 0 < x1 −2 < x2 −2 4 =⇒ 0 < p x1 −2 < p x2 −2 5 =⇒ f (x1) < f (x2). 6 Thus, f is an increasing function on its domain. Dr. Gabby (KNUST-Maths) Functions 12 / 41
  • 12. Properties of Functions Monotonic Functions Monotonic Functions Example Show that f (x) = (2− x)2 +1 decreases on (−∞,2] and increases on [2,+∞). 1 Df = R. 2 For x1,x2 ∈ (−∞,2], 3 x1 < x2 ≤ 2 =⇒ −x1 > −x2 > −2 4 =⇒ 2− x1 > 2− x2 > 0 5 =⇒ (2− x1)2 > (2− x2)2 > 0 6 =⇒ (2− x1)2 +1 > (2− x2)2 +1 > 1 7 =⇒ f (x1) > f (x2). 8 f is decreasing on (−∞,2]. 1 For x1,x2 ∈ [2,+∞), 2 2 ≤ x1 < x2 =⇒ −2 > −x1 > −x2 3 =⇒ 0 > 2− x1 > 2− x2 4 =⇒ 0 < (2− x1)2 < (2− x2)2 5 =⇒ 1 < (2− x1)2 +1 < (2− x2)2 +1 6 =⇒ f (x1) < f (x2). 7 f is an increasing function on [2,+∞). Dr. Gabby (KNUST-Maths) Functions 13 / 41
  • 13. Properties of Functions Bounded Functions Bounded Functions Definition A function is said to be bounded above if there is ū ∈ R such that f (x) ≤ ū for all x in the domain of f . Example The function f (x) = x2 + 1 defined on 0 ≤ x ≤ 1 is bounded above by 2 since f (x) ≤ 2 for 0 ≤ x ≤ 1. Example The function f (x) = 1/x defined on x ∈ N is bounded above by 1 Example The function f (x) = sinx is bounded above by 1 for x ∈ R. Dr. Gabby (KNUST-Maths) Functions 14 / 41
  • 14. Properties of Functions Bounded Functions Bounded Functions Definition A function, f , is said to be bounded below if there is ℓ ∈ R such that f (x) ≥ ℓ for all x in the domain of f . Example The function f (x) = x−1 defined in [0,1] is bounded below by −1 since −1 ≤ f (x) for x ∈ [0,1]. Example The function g(x) = | p x +1| is bounded below by 0 on the interval [0,4] since 0 ≤ g(x) for x ∈ [0,4]. Dr. Gabby (KNUST-Maths) Functions 15 / 41
  • 15. Properties of Functions Maxima and Minima of Functions Maxima and Minima of Functions Local(or relative) and Global(or absolute) Minimum 1 The function f is said to have a local minimum value at the point x0 if f (x0) ≤ f (x) for all x in a neighbourhood of x0. 2 f is said to have a global minimum value at the point x0 if f (x0) ≤ f (x) for all x in the domain of f . 3 In this case f is bounded below. Local(or relative) and Global(or absolute) Maximum 1 If f (x) ≤ f (x0) for all x in a neighbourhood of x0, then f has a local maximum value at the point x0. 2 The maximum is global if f (x) ≤ f (x0) for all x in the domain of f . 3 In this case f is bounded above. Dr. Gabby (KNUST-Maths) Functions 16 / 41
  • 16. Properties of Functions Maxima and Minima of Functions Maxima and Minima of Functions −5 −4 −2 2 4 −3 −1 1 2 3 0 global min f (x) = 1+(x +1)2 , Dr. Gabby (KNUST-Maths) Functions 17 / 41
  • 17. Properties of Functions Maxima and Minima of Functions Maxima and minima −5 −4 −2 2 4 −3 −1 1 2 3 0 local max local min g(x) = 1−2x −3x2 +2x3 Dr. Gabby (KNUST-Maths) Functions 18 / 41
  • 18. Properties of Functions Maxima and Minima of Functions Maxima and minima −5 −4 −2 2 4 −3 −1 1 2 3 0 global min local max local min h(x) = (x −1)(−x +3)2 (x), Dr. Gabby (KNUST-Maths) Functions 19 / 41
  • 19. Inverse Function Inverse Functions 1 An inverse function is a function that undoes the action of the another function. 2 A function g is the inverse of a function f if whenever y = f (x) then x = g(y) 3 In other words, applying f and then g is the same thing as doing nothing. We can write this in terms of the composition of f and g as g(f (x)) = x 4 A function f has an inverse function only if for every y in its range there is only one value of x in its domain for which f (x) = y 5 This inverse function is unique and is frequently denoted by f −1 and called f inverse. Dr. Gabby (KNUST-Maths) Functions 21 / 41
  • 20. Inverse Function Inverse Functions Given the function f (x) we want to find the inverse function, f −1 (x) 1. First, replace f (x) with y. 2. Replace every x with a y and replace every y with an x. 3. Solve the equation from Step 2 for y. This is the step where mistakes are most often made so be careful with this step. 4. Replace y with f −1 (x). In other words, we’ve managed to find the inverse at this point! 5. Verify your work by checking that (f ◦ f −1 )(x) = x and (f −1 ◦ f )(x) = x are both true. Example Given f (x) = 3x −2 find f −1 (x) Dr. Gabby (KNUST-Maths) Functions 22 / 41
  • 21. Inverse Function 1 Given the function f f (x) = 3x −2 2 First, replace f (x) with y. y = 3x −2 3 Replace every x with a y and replace every y with an x. x = 3y −2 4 Solve the equation from Step 2 for y. y = 1 3 (x +2) 5 Replace y with f −1 (x). f −1 (x) = x 3 + 2 3 Dr. Gabby (KNUST-Maths) Functions 23 / 41
  • 22. Inverse Function We can verify the results, we check that (f ◦ f −1 )(x) = x (f ◦ f −1 )(x) = f [f −1 (x)] = f hx 3 + 2 3 i = 3 hx 3 + 2 3 i −2 = x +2−2 = x Dr. Gabby (KNUST-Maths) Functions 24 / 41
  • 23. Inverse Function Graph of a Function and Its Inverse There is an interesting relationship between the graph of a function and the graph of its inverse. Here is the graph of the function and inverse from the first two examples. In both cases the graph of the inverse is a reflection of the actual function about the line y = x . This will always be the case with the graphs of a function and its inverse. Dr. Gabby (KNUST-Maths) Functions 25 / 41
  • 24. Sequence and Series Sequence and Series Definition A sequence is an ordered set of numbers that most often follows some rule (or pattern) to determine the next term in the order. Example x,x2 ,x3 ,x4 , ... is a sequence of numbers, where each successive term is multiplied by x. Definition A series is a summation of the terms of a sequence. The greek letter sigma Σ is used to represent the summation of terms of a sequence of numbers. Dr. Gabby (KNUST-Maths) Functions 27 / 41
  • 25. Sequence and Series Sequence and Series Series are typically written in the following form: n X i=1 ai = a1 + a2 + a3 ···+ an where the index of summation, i takes consecutive integer values from the lower limit, 1 to the upper limit, n. The term ai is known as the general term. Example 5 X i=1 i = 1+2+3+4+5 = 15 (1) 6 X k=3 2k = 23 +24 +25 +26 = 8+16+32+64 = 120 (2) 4 X k=1 kk = 11 +22 +33 +44 = 1+4+27+256 = 288 (3) Dr. Gabby (KNUST-Maths) Functions 28 / 41
  • 26. Sequence and Series Finite Series Definition (Finite and Infinite Series) A finite series is a summation of a finite number of terms. An infinite series has an infinite number of terms and an upper limit of infinity. Properties of Finite Series The following are the properties for addition/subtraction and scalar multiplication of series. For some sequence ai and bi and a scalar k then: n X i=1 kai ±bi = k n X i=1 ai ± n X i=1 bi Dr. Gabby (KNUST-Maths) Functions 29 / 41
  • 27. Sequence and Series Theorems of Finite Series 1 The following theorems give formulas to calculate series with common general terms. These formulas, along with the properties listed above, make it possible to solve any series with a polynomial general term, as long as each individual term has a degree of 3 or less. n X i=1 1 = n (4) n X i=1 c = nc (5) n X i=1 i = n(n +1) 2 (6) n X i=1 i2 = n(n +1)(2n +1) 6 (7) n X i=1 i3 = µ n(n +1) 2 ¶2 (8) (9) Dr. Gabby (KNUST-Maths) Functions 30 / 41
  • 28. Sequence and Series Types of Sequences There are two main types of sequences. 1 An arithmetic sequence is one in which successive terms differ by the same amount. For example, {3, 6, 9, 12, ···}. Note each term is obtained by adding 3 to the previous term. This is called the common difference denoted as d = 6−3 2 A geometric sequence is one in which the quotient of any two successive terms is a constant. For example, {3, 9, 27, 81, ···}. Note each term is obtained by multiplying the previous term by 3. This is called the common ratio denoted by r = 9 3 . Similarly, there are also arithmetic series and geometric series, which are simply summations of arithmetic and geometric sequences, respectively. Dr. Gabby (KNUST-Maths) Functions 31 / 41
  • 29. Sequence and Series Theorems for Arithmetic and Geometric Series Arithmetic Series Suppose we have the following arithmetic series, {a +(a +d)+(a +2d)+···+(a +(n −1)d} Then, n−1 X k=0 a +kd = n 2 (2a +(n −1)d) (10) Geometric Series Suppose we have the following geometric series, {a + ar + ar2 +...+ ar(n−1) } Then, n−1 X k=0 ark = a µ rn −1 r −1 ¶ (11) Dr. Gabby (KNUST-Maths) Functions 32 / 41
  • 30. Sequence and Series Example Solve the series 20 X i=1 2i2 +7i 20 X i=1 2i2 +7i = 20 X i=1 2i2 + 20 X i=1 7i = 2 µ 20(20+1)(2(20)+1) 6 ¶ +7 µ 20(20+1) 2 ¶ = 2 µ 17220 6 ¶ +7(210) = 5740+1470 = 7210 Dr. Gabby (KNUST-Maths) Functions 33 / 41
  • 31. Sequence and Series Example Find the sum of the geometric series: 2+8+32+128+...+8192 1 We know that 8 2 = 4 and that 32 8 = 4, so r = 4 for this geometric series. 2 The initial value represents our a value, so a = 2. 3 Before we can write the series in summation notation, we must determine the upper limit of the summation. 8192 = arn−1 = (2)(4n−1 ) 4096 = 4n−1 46 = 4n−1 6 = n −1 n = 7 Dr. Gabby (KNUST-Maths) Functions 34 / 41
  • 32. Sequence and Series So, we can rewrite the series as P6 k=0 2(4k ). From the formula for the sum of a geometric series, 6 X k=0 2(4k ) = 2 µ (4)7 −1 4−1 ¶ = 2(5461) = 10922 Therefore, the sum of the series, 2+8+32+128+...+8192 is 10922 Dr. Gabby (KNUST-Maths) Functions 35 / 41
  • 33. Sequence and Series Binomial Series In this final section of this chapter we are going to look at another series representation for a function. Before we do this let’s first recall the following theorem. Binomial Theorem If n is any positive integer then, (a +b)n = n X i=0 à n i ! an−i bi = an +nan−1 b + n (n −1) 2! an−2 b2 +···+nabn−1 +bn where, à n i ! = n (n −1)(n −2)···(n −i +1) i! ; i = 1,2,3,...n à n 0 ! = 1 Dr. Gabby (KNUST-Maths) Functions 36 / 41
  • 34. Sequence and Series Example (2x −3)4 = 4 X i=0 Ã 4 i ! (2x)4−i (−3)i = Ã 4 0 ! (2x)4 + Ã 4 1 ! (2x)3 (−3)+ Ã 4 2 ! (2x)2 (−3)2 + Ã 4 3 ! (2x)(−3)3 + Ã 4 4 ! (−3)4 = (2x)4 +4(2x)3 (−3)+ 3 2 (2x)2 (−3)2 +4(2x)(−3)3 +(−3)4 = 16x4 −96x3 +216x2 −216x +81 Dr. Gabby (KNUST-Maths) Functions 37 / 41
  • 35. Sequence and Series Binomial Series If k is any number and |x| < 1 then, (1+ x)k = ∞ X n=0 Ã k n ! xn = 1+kx + k (k −1) 2! x2 + k (k −1)(k −2) 3! x3 +··· where, Ã k n ! = k (k −1)(k −2)···(k −n +1) n! ; n = 1,2,3,... Ã k 0 ! = 1 Example Write down the first four terms in the binomial series for p 9− x Dr. Gabby (KNUST-Maths) Functions 38 / 41
  • 36. Sequence and Series We have to rewrite the terms into the form (1+ xk)k required, that is p 9− x = 3 ³ 1− x 9 ´1 2 = 3 ³ 1+ ³ − x 9 ´´1 2 So, k = 1 2 and xk = −(x/9) Then the binomial series is given, p 9− x = 3 ³ 1+ ³ − x 9 ´´1 2 = 3 ∞ X n=0 Ã 1 2 n ! ³ − x 9 ´n = 3 " 1+( 1 2 )(− x 9 )+ 1 2 (−1 2 ) 2 (− x 9 ) 2 + 1 2 (−1 2 )(−3 2 ) 6 (− x 9 ) 3 +··· # So the first four terms are = 3− x 6 − x2 216 − x3 3888 −··· Dr. Gabby (KNUST-Maths) Functions 39 / 41
  • 37. Sequence and Series Exercise 1 Find the period of the following functions 1) f (x) = sin(2x), 2) f (x) = cos(−2x +π/3), 3) f (x) = x −sin(x). 2 Find the domain of: 1)f (x) = 1 2x −6 2)f (x) = 1− x 1+ x 3)f (x) = x3 −2x x(−x −6) 4)f (x) = 3x −1− 1 2x −6 5)f (x) = x 1−2x + x2 6)f (x) = x2 −2x (x −3)(1− x2) 3 Determine whether the functions below are even, odd or neither. 1) f (x) = ex2 −1 +ln(|x|+1), 2) f (x) = x2 −2 x(1−x2) , 3) f (x) = x2 sin(x) 4) f (x) = x p |x|−1, 5) f (x) = ln ¡ tanx −e|x| ¢ , 6) f (x) = x −1. Dr. Gabby (KNUST-Maths) Functions 40 / 41
  • 38. END OF LECTURE THANK YOU Dr. Gabby (KNUST-Maths) Functions 41 / 41