STUDY COURSE BACHELOR OF BUSINESS
ADMINISTRATION (B.A.)
MATHEMATICS (ENGLISH & GERMAN)
REPETITORIUM
2015/2016
Prof. Dr. Philipp E. Zaeh
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1
LITERATURE (GERMAN)

Böker, F., Formelsammlung für Wirtschaftswissenschaftler. Mathematik und
Statistik, München 2009.

Böker, F., Mathematik für Wirtschaftswissenschaftler. Das Übungsbuch, 2.
Auflage, München 2013.

Gehrke, J. P., Mathematik im Studium. Ein Brückenkurs, 2. Auflage,
München 2012.

Hass, O., Fickel, N., Aufgaben zur Mathematik für Wirtschaftswissenschaftler,
3. Auflage, München 2012.

Sydsaeter, K., Hammond, P., Mathematik für Wirtschaftswissenschaftler.
Basiswissen mit Praxisbezug, 4. Auflage, München 2013.

Thomas, G. B., Weir, M. D., Hass, J. R., Basisbuch Analysis, 12. Auflage,
München 2013.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
2
LITERATURE (ENGLISH)

Chiang, A. C., Wainwright, K., Fundamental Methods of Mathematical
Economics, 4rd Edition, Boston 2005. (McGrawHill)

Dowling, E. T., Schaum's Outline of Mathematical Methods for Business
and Economics, Boston 2009 (McGrawHill).

Sydsaeter, K., Hammond, P., Essential Mathematics for Economic Analysis,
4th Edition, Harlow et al 2012. (Pearson)

Taylor, R., Hawkins, S., Mathematics for Economics and Business, Boston
2008. (McGrawHill)

Zima, P., Brown, R. L., Schaum's Outline of Mathematics of Finance, 2nd
Edition, New York et al 2011 (McGrawHill).
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
3
REPETITORIUM - TOPICS
1. Introductory Topics
1.1. Numbers
1.2. Algebra
1.3. Sequences, Series, Limits
1.4. Polynomials
2. Linear Algebra
2.1. System of Linear Equations
2.2. System of Linear Inequalities
3. Differential Calculus
3.1. Basics
3.2. Derivative Rules
3.3. Applications & Exercises - Curve Sketching
4. Integral Calculus
4.1. Basics
4.2. Rules for Integration
4.3. Applications & Exercises
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
4
1. Introductory Topics
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
5
1.1. NUMBERS
2x  6  0
x3

x   natural numbers i.e   {1, 2, 3, 4....}
2x  6  0
x  3

x   integers i.e Z  {.... , - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, ....}
2x  5  0
5
x   2,5
2

x  Q rational numbers ( Z and fractions )
x2  2  0

x  IR real numbers (Q and irrationalnumbers, roots,  , e...)

x  C complex numbers (IR and imaginarynumbers)
x 2
x2  2  0
x   - 2   2   1   2i where i  - 1
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
6
1.2. ALGEBRA
1.2.1 Laws
Commutative law: a + b = b + a
Associative law:
Distributive law:
a∙b=b∙a
(a + b) + c = a + (b + c)
(a ∙ b) ∙ c = a ∙ ( b ∙ c)
(a + b) ∙ c = a ∙ c + b ∙ c
1.2.2 Binomial Theorems
1. Binomial theorem:
(a + b)2 = a2 + 2ab + b2
2. Binomial theorem:
(a - b)2 = a2 - 2ab + b2
3. Binomial theorem:
(a + b) (a - b) = a2 - b2
7
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.2. ALGEBRA
1.2.3 POWERS
Area of a square : A  h  h
Volume of a cube : V  h  h  h
General
We ca l l:
b
n
Shortcut
A  h2
V  h3
b n  b  b  b.... b with n factors
ba s i s
exp onent
Examples:
34  3  3  3  3
x5  x  x  x  x  x
3
1 1 1 1
    
z z z
z
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
8
1.2. ALGEBRA
1.2.3 POWERS
CALCULATION RULES FOR EXPONENTS N, M
b 3  b 2  b  b  b   b  b 
 bbbbb
Example:
a n  a m  a n m
 b5
 b 3 2
an
am
 a nm
Example:
c c c c c
c c
 c c c
c 5 : c2 
 c3
a 
n m
 c 52
 
 a nm  a
m n
Example:
d 3  2  d  d  d   d  d  d 
 d6
 d 23
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
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1.2. ALGEBRA
1.2.3 POWERS
Examples:
2 6  2 4  210
u5 : u  u4
0.5 
2 3
;
x 2  x  x 3 ; 10 4  10 3  10 7
; y 7 : y 5  y 2 ; 10 6 : 10 4  10 2
 0.56 ;
3y  
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
3 4
 
 3y 12  312  y 12 ; 10 3
4
 10 12
10
1.2. ALGEBRA
1.2.4 Roots
Rules for positive a, b and m, n
n
a  n b  n ab
n
a :
n
b 
n
a:b
a  mn a  n
mn
a  b  ab
special case
m
a n  n am
;
a

m
n
m

a
1
n
am
SPECIAL CASE: NEGATIVE RADICAND
b  4
complex
number
(not covered
here)
11
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.2. ALGEBRA
1.2.4 Roots
1
3
125  1253  5
6
64  64 6  2
, since53  125
1
, since 26  64
1
324  2 324  3242  18
5
1
 32   32 5  2
, since  182  324
, since 25  32
2
32 5  5 322  5 1024  4
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
, since4 5  1024
12
1.2. ALGEBRA
1.2.5 Logarithms
Potential function
a  b n , if looking for a e.g 8  2 3
Square root function b  n a , if looking for b e.g 2  3 8
Logarithm
n  l ogb a
, if looking for exponent n
n equa l sl oga ri thmof a to the ba s i sb
Question: “To which power ‘n’ do you have to raise ‘b’, in order to get ‘a’ as the result”
Remark: This power ‘n’ by which ‘b’ has to be raised in order to get ‘a’ is called the
logarithm of ‘a’ when ‘b’ is the basis.
13
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.2. ALGEBRA
1.2.5 Logarithms
Examples:
2 4  16
 2  4 16
wherea  16, b  2, n  4
 4  l og2 16
4 2  16
 4  2 16
now a  16, b  4 , n  2
 2  l og4 16
Observation : log is always taken in accordance to some base.
i.e we can obtain same number 16 by taking different base
and raising it with different power.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
14
1.2. ALGEBRA
1.2.5 Logarithms
More Examples:
10 2  0.01  10  2 0.01   2  l og10 0.01
10 3  1000  10  3 1000
4
1
 3  l og10 1000
 0.25  4  1 0.25
  1  l og4 0.25
Small Exercise: Identify a, b and n in above examples. Take help from the
previous slides.
15
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.2. ALGEBRA
1.2.5 Logarithms (special cases)
Since b 0  1 
l ogb 1  0
Since b 1  b
l ogb b  1

Reciprocal value :
1
1

 b n
a bn
1
 n  l ogb     l ogb a
a
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
16
1.2. ALGEBRA
1.2.5 Logarithms
Now Consider:
a  bn
(1) and c  b m
( 2)
from the definition of log we have
n  log b a
(3) and m  log b c
( 4)
multiply (1) & (2) we get
ac  b n .b m  ac  b n  m
ac  b n  m
 n  m  log b ac
(5)
substitute (3) & (4) in (5) we get
log b (ac )  log b a  log b c
This is known as the first rule/law of logarithm. Similarly, we
also have other rules as presented in the next slide.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
17
1.2. ALGEBRA
1.2.5 Logarithms
Rules for logarithms:
l ogb a  c   l ogb a  l ogb c
a
l ogb    l ogb a  l ogb c
c
l ogb an  n  l ogb a
1
l ogb m a   l ogb a
m
l ogc a
l ogb a 
wi th a ny ba s i sc
l ogc b
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
18
1.2. ALGEBRA
1.2.5 Logarithms
Further Proofs:
l ogb a 
l ogc a
wi th a ny ba s i sc
l ogc b
Re ca l l a  b n  n  logb a
(1)
Al s ota kel ogonboths i de sof a  b n wi thba s ec
a  b n  logc a  logc b n

logc a  nlogc b  n 
logc a
logc b
(2)
From (1) a nd (2), we ge t
logb a 
logc a
logc b
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
19
1.2. ALGEBRA
1.2.5 Logarithms
Proof:
l ogb an  n  l ogb a
Re ca l l l ogb a  c   l ogb a  l ogb c
then we ha ve
l ogb (a.an1 )  l ogb a  l ogb an1  l ogb a  l ogb (a. an2 )
 l ogb a  l ogb a  l ogb an2  l ogb a  l ogb a  l ogb (a. an3 )
 l ogb a  l ogb a  l ogb a  l ogb an3  .......
 l ogb a  l ogb a  l ogb a  .....  l ogb a
n times
 n  l ogb a
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
20
1.2. ALGEBRA
1.2.5 Logarithms
Note that, we can develop associations
between power rules and laws of logarithm:
log b a  c   log b a  log b c
b a .b c  b a c
a
log b    log b a  log b c
c
ba
 b a c
c
b
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
21
1.2. ALGEBRA
1.2.5 Logarithms
Examples:
 log2 8  log2 2  4   log2 2  log2 4
 45 
 log3    log3 45  log3 3
 3
 log2 16  log2 2 4  4  log2 2
1
 log2 4  log2 2 16   log2 16
2
log2 16
 log4 16 
log2 4
Small Exercise: Compute the values of above expressions.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
22
1.2. ALGEBRA
1.2.5 Logarithms
Examples:
 abc3 
 logx  4   logx (abc3 )  logx d4
 d 
 logx a  logx b  logx c 3  logx d4
 logx a  logx b  3. logx c  4. logx d
1
1
 logb 3 x  logb (x) 3  logb x
3
3
 y z
Exercise : Solve logb  2 
 d 
23
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.2. ALGEBRA
1.2.5 Logarithms
Example:
l og3 x  5  1
(x  5) l og3  1
x l og3  5 l og3  1
x l og3  1  5 l og3
1  5 l og3
x
l og3
Example:
11 x  121. 11
1
11 x  112.11 2
5
l og11 11 x  l og11 11 2
5
x
2
Observation: In the left example above, no base is specified for the log. In
such cases, we can assume any number to be the base. However, typical
convention is to assume b=10.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
24
1.2. ALGEBRA
1.2.5 Logarithms
Example:
5.25 x  126.5 x  25
5.52x  126.5 x  25
Let y  5 x  y 2  52x
5y 2  126y  25  0
(5y  1)(y  25)  0
1
y
or y  25
5
1
 5x 
or 5 x  25
5
l og5 5 x  l og5 5 1 or l og5 5 x  l og5 52
x  1 or x  2
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
25
1.2. ALGEBRA
1.2.5 Logarithms
a  b logb a
we know a  b n  n  l ogb a
ta keboths i de si npowe rof b
n  l ogb a  b n  b logb a  a
i.e. ta ki ngboth s i de si npowe ri s Antil og
Example:
5  log2 32
2 5  2log 2 32
32  2log 2 32
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
26
1.2. ALGEBRA
1.2.5 Logarithms
ex
Special logarithms:
Natural logarithm:
loge a = ln a = n  a = en (e = 2,718…)
ln x
Common logarithm:
log10 a = lg a = n  a = 10n
27
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.2. ALGEBRA
1.2.6 Factorial, Absolute Value, Sums
n! (factorial)
n! = 1 ∙ 2 ∙ 3 ∙ … ∙ n
Example: 4! = 1 ∙ 2 ∙ 3 ∙ 4 = 24
1! = 1
Definition:
0! = 1
Absolute value:
a

a  0
 a

a 0

a  0
a  0

Example:
a is always positive or 0
9  9
9 9
0 0
Therefore, the absolute value (or modulus) |a| of a real number a is only the numerical value of a
without regard to its sign.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
28
1.2. ALGEBRA
1.2.6 Factorial, Absolute Value, Sums
Sigma sign:
Examples:
n
 i  1  2  3  .........  n
i 1
5
 ai  a0  a1  a2  a3  a4  a5
i 0
6
 2 i  2 0  21  2 2  2 3  2 4  2 5  2 6
i 0
3
2  i  2 1  2 2  2  3
i 1
4
 i 2  12  2 2  32  4 2
i 1
5
 i 2  12  2 2  32  4 2  52
i 1
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
29
1.3. SEQUENCES, SERIES AND LIMITS
General Notation of a Sequence (Folge):
ak  a1 , a2 , a3 ...
e.g.
1
1 1 1
 1, , , ,...
k
2 3 4
1
1 2 3
 1   0, , , , ..........
k
2 3 4
ak 
ak
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
30
1.3. SEQUENCES, SERIES AND LIMITS
A sequence is called convergent, if it has a limit g;
i.e., if there is a limit g with:
l i mak  g
k 
Example : Convergent Sequences
ak 
1
1 1
 1, , , ...........
k
2 3
ak  1 
lim
1
k  k
0
1
1 2 3
 0, , , , ...........
k
2 3 4
 1
ak  1  
 k
k
9 64
 2, , , ...........
4 27
(zero sequence )
 1
lim 1    1
k
k  
k
 1
lim 1    2,71828182........ e (Euler' s number)
k  
k
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
31
1.3. SEQUENCES, SERIES AND LIMITS
A sequence is called divergent if no limit exists.
Examples : Divergent Sequences
ak  k  1, 2, 3, .........
ak  2k  2, 4 , 6, 8 , .........
Oscillatin
g Sequence :
ak  1,  1, 1,  1, 1,  1.........
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
32
1.3. SEQUENCES, SERIES AND LIMITS
Arithmetic Sequences
Definition: A sequence is called arithmetic if the following is valid:
ak+1 – ak = d, with d = constant
(for all elemements of the sequence k = 1, 2, 3 …)
Composition law for arithmetic sequences:
ak= a1 + (k -1) ∙ d
Example:
Consider the sequence 5, 10, 15, 20....
a1 = 5, a2 = 10, d = a2 - a1 = 10 - 5 = 5
a5 = 5 + (5-1) . 5 = 5 + 20 = 25
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
33
1.3. SEQUENCES, SERIES AND LIMITS
Geometric Sequences
Definition: A sequence is called geometric, if the following is valid:
ak
q
ak 1
with q = constant
(for all elements of the sequence k = 1, 2, 3 …)
Composition law for geometric sequences:
ak= ak-1 .q = ak-2 .q2 = ak-3 .q3 =…. ……. = a1 .q(k-1)
Example: Consider the sequence 2, 4, 8, 16....
a1 = 2, a2 = 4, q = 4/2 = 2
a5 = a4 .q = 16.2 = 32 or a1 .q4 = 2.2 4 = 2.16 = 32
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
34
1.3. SEQUENCES, SERIES AND LIMITS
Series (Reihen)
During the composition of an (infinite) series all elements of a sequence
are summed up:

R   ak
k 1
The sum up to the n-th element is called n-th partial sum:
n
S n   ak
k 1
35
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.3. SEQUENCES, SERIES AND LIMITS
Partial sums of the arithmetic series (a := a1):
Arithmetic series with d  1, a  1:
1
S n  1  2  3  .... n   n  n  1
2
General :
Sn  a  a  d  a  2d  ...  a  n  1  d
Sn  an  n  1d  an  n  2d  ......  an  2d  an  d  an
2Sn  n(a1  an )  Sn 
n(a1  an )
2
n
((n  1)nd)
 Sn  (2a  (n  1)d)  na 
2
2
Small Exercise: Substitute a=1, d=1 in the General formula of Sn and
simplify. What do you obtain?
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
36
1.3. SEQUENCES, SERIES AND LIMITS
Partial sums of the geometric series (a := a1):
Geometric series
(genera l )
S n  a  a q  a q  ..........  a qn1
2
S n .q  a q  a q2  a q3 ..........  a qn
S n .q  S n  a qn  a
S n .(q  1)  a qn  a
Sn 
a qn  a a(qn  1)

wi th (q  1)
(q  1)
(q  1)
e.g. : q  2, a  1 :
S 5  1  2  4  8  16 
1(2 5  1)
 31
21
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
37
1.3. SEQUENCES, SERIES AND LIMITS
Example: Arithmetic Series
Suppose that BMW offers you to purchase a new model car in monthly
instalments. The first instalment amounts to € 500 and then every month the
amount of instalment is increased by € 25. The total payback period is 2 years.
What is the total price of the car? The amount of the last instalment?
The resulting arithmatic sequence 500, 525, 550, 575....
We note that, a1 = 500, a2 = 525, d = 525 - 500 = 25, n = 12.2 = 24 (for two years)
Use general Sn formula to compute price of the car and composition formula to
compute the value of last instalment (the last instalment is last term in series).
Total price of the car: S24 = 24(500) + (24-1)(24)(25)/2 = 12000 + 6900 = 18900
Amount of last instalment: T24 = 500 + (24-1)(25) = 1075
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
38
1.3. SEQUENCES, SERIES AND LIMITS
Exercise:
Refer to the data in last example, what is the amount of money you paid in
one year (Year 1 and Year 2)?
Suppose that the payback time has been increased to 2.5 years (hint: n=30).
What is the amount by which each instalment should be decreased?
(hint: you have to calculate d now, given a = 500, Sn = 18900)
What is the amount of instalment that you made at the end of the payback
period? (hint: you are required to calculate the last term in the sequence).
Exercise:
A company incurs a cost of € 5.50 for producing a unit of some product. Due
to increased tax, for each successive unit, the associated cost is increased by €
1.0. How many units can be produced in total of € 1000?
(hint: use the Sn formula, you need to calculate n, a = 5.5, d = 1)
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
39
1.3. SEQUENCES, SERIES AND LIMITS
Example: Geometric Series
The current price of some model of an Apple MacBook is € 2000. It is
expected to lose its value by 25% every year. What would be its value in
10th year?
Note that the value is lost by 25% i.e. the value of computer at the end of each year
is 75% of previous year (q = 75% = 0.75).
The resulting geometric sequence 2000, 1500, 1125.....
a1 = 2000, a2 = 1500, q = 1500/2000 = 0.75
The tenth term in the sequence is the value of computer in the 10th year
a10 = a1 .q9= 2000(0.75) 9 = 150.17
Small Exercise: What is the value of computer at the end of 15 years?
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
40
1.3. SEQUENCES, SERIES AND LIMITS
Example: Geometric Series
Suppose that you invest in a real estate and purchase a land near Hamburg. A
renewable energy business firm takes it on rent. According to terms, the lease can
be done for every six months, the amount of rent is subject to 10% increase in
each subsequent lease. The first amount of rent is € 6000. Calculate the total
amount of rental payments that would be received for the period of 5 years.
The resulting geometric sequence 6000, 6600, 7260.....
a1 = 6000, a2 = 6600, q = 6600/6000 = 1.1, n=10
6000(1.110  1)
 95624.54
1.1  1
Small Exercise: What is the value of rental payments received at the end of 3
years?
S10 
41
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.3. SEQUENCES, SERIES AND LIMITS
Exercise:
Suppose that a production plant produces 100 chocolate bars in the first day. The
production capacity can be increased by 5% every day. Calculate the total output of
the production plant for two weeks.
(hint: you have to use Sn formula where a=100, q=1.05 and n=14)
Now also calculate the number of chocolate bars produced on the 10th day.
A super market places an order of 800 bars at the start of the first day of
production. How many days will it take to fulfill the order?
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
42
1.4. POLYNOMIALS
Definition : A function with the form
n
y  a0  a1 x  a2 x 2  a3 x 3  ......  an x n   ai x i
i 0
is called a n - th degreepolynomial function.
The real numbers ai are called coefficients.
43
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.4. POLYNOMIALS
1.4.1. 1st and 2nd degree Polynomials
Which different mathematical ways can be used to describe geometric
objects?
1st degree polynomials describe (straight lines) :
y
Two - point - form ( from the slope
)
x
y  y1 y2  y1

(  tanα )
x  x1 x2  x1
General form
y  a0  a1  x
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
with
and
y
y2  y1
 x  x 1   y 1 respectively
x2  x1
a0 : y  intercept
a1 : slope
44
1.4. POLYNOMIALS
1.4.1. 0th, 1st and 2nd degree Polynomials
Examples: The equation of a straight line (n=0, n=1)
n  0 : y  a0 for all x
y
n  1 : y  a0  a1  x
P2
y2
y
P
y
y1
a0
0
P1

a0
x
0
x1
x
x2
x
45
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.4. POLYNOMIALS
1.4.1. 0th, 1st and 2nd degree Polynomials
Example: Formulation of linear Polynomial in Business Application
A firm has € 24000 budget to produce two different products, namely product x and
product y. Each unit of product x costs € 30 and each unit of product y costs €
40. Express this information in first degree polynomial equation.
Let “a” be the number of units of product x to be produced.
Let “b” be the number of units of product y to be produced.
Therefore, the cost incurred to produce “a” units of product x is 30a and vice versa.
30 a + 40 b = 24000
Small exercise: Suppose that the firm can afford 1550 hours of labor. Each unit of product
x requires 5 hours of labor and each unit of product y requires 2 hours of labor. Express
the information in linear equation. Try to sketch the line.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
46
1.4. POLYNOMIALS
The equation of a parabola (n=2)
Which different mathematical ways can be used to describe geometric obects?
2nd degree polynomials describle parabolas
Two ways of describing a parabola:
Sta nda rdform a ndvertex form of the pa ra bol a(wi thx s , ys vertex - coordi na tes ):
y  a0  a1x  a2x 2
a nd
y  as x  x S 2  y S res pecti vel y
Vertex form of
the parabola
Expa ndi ng the vertex form y  as x S 2  y S  2x SaS x  aS x 2
a0
a1
y
a2
results in the standard form.
yS
xS
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
x
47
1.4. POLYNOMIALS
The equation of a parabola (n=2)
From s ta nda rdform to vertex form:
y  as x S 2  y S  2x SaS x  aS x 2
y  as (xS 2  2x S x  x 2 )  y S
y  as x  x S 2  y S res pecti vel y
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
48
1.4. POLYNOMIALS
The equation of a parabola (n=2)
Example:
Example:
From vertex form to standardform :
From s ta nda rdform to vertex form :
considery  x  32  4
with vertex form parameters :
x s  3, as  1, y s  4
expand, we get
y  x 2  6x  10
s ta nda rdformpa ra meters:
a2  1, a1  6, a 0  10
y  x 2  2(3)(x)  9  1
y  x 2  2(3)(x)  32  1
y  x 2  6x  9  4
y  (x  3)2  1
wi thvertex formpa ra meters:
x s  3, as  1, y s  1
y  x 2  6x  13
standardform parameters :
a2  1, a1  6, a0  13
49
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.4. POLYNOMIALS
The equation of a parabola (n=2) General form:
Examples:
y  x2
a)
x
y  x2
a0  0
y  0,5  x 2
a0  0
a1  0
a1  0
a2  1  0
a2  0,5  0
0
1
-1
2
-2
0
1
1
4
4
y
yx
b)
y  a0  a1 x  a2 x 2
x
 0.5x 2
a0  0

a1  0  vertex at 0
a 2  0 a2  0


2
upwards
0
0
2
-2
- 0,5 - 0,5 -2
1
-2
y
-1
1
x
opening
downwards
1
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
50
1.4. POLYNOMIALS
Quadratic equations (n=2)
Zeros of a quadratic function, i.e. intersection with the x-axis
Or the points where the value of the quadratic function is exactly equal to zero.
a x2  bx  c  0  x 1,2 
 b  b 2  4a c
2a
2
x  px  q  0  x 1,2
2
p
p
    q
2
2
51
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.4. POLYNOMIALS
2nd degree polynomials (n=2):
It has to be differentiated between 3 cases of discriminants:
b2 - 4·a·c > 0 => 2 real zeros
y
y
x1
x2
x
x1
x2
x
Observation: Curves cross the x-axis at
exactly two points
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
52
1.4. POLYNOMIALS
2nd degree polynomials (n=2):
b2 - 4·a·c = 0 => one zero
y
y
x1
x
x1
x
Observation: Curves touch the x-axis at
exactly one point.
53
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.4. POLYNOMIALS
2nd degree polynomials (n=2):
b2 - 4·a·c < 0 => no zero
y
y
x
x
Observation: Curves do not cross x-axis
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
54
1.4. POLYNOMIALS
1.4.2 Polynomials of a higher degree
3rd degree polynomial
f x   a3  x 3  a2  x 2  a1  x  a0
4th degree polynomial (Special biquadratic equation)
f x   a  x 4  b  x 2  c (coefficients of x and x 3 arezero)
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
55
1.4. POLYNOMIALS
1.4.2 Polynomials of a higher degree
N-th Degree Polynomials
y  a0  a1 x  a2 x 2  ......  an x n
n
  ai x i
i 0
This is a polynomial function.
A rational function is defined through the ratio of two polynomials Z(x)
and N(x):
Z (x)
N( x )
where Z(x) is an n-th degree polynomial and N(x) is an m-th degree polynomial, for all x
 IR and N(x) ≠ 0
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
56
1.4. POLYNOMIALS
1.4.2 Polynomials of a higher degree
Comparing coefficients
Polynomials are identical if they have the same degree and the respective
coefficients are identical.
n
m
i 0
i 0
 ai x i  bi x i

ai  bi and n  m
 a1  b1
a2  b2
a3  b3
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
57
1.4. POLYNOMIALS
1.4.2 Polynomials of a higher degree
Comparing coefficients
Example:
4x 2  3x  7  a (x  1)(x  3)  b(x  1)  c
4x 2  3x  7  a (x 2  3x  x  3)  bx  b  c
4x 2  3x  7  ax2  2ax  3a  bx  b  c
4x 2  3x  7  ax2  2ax  bx  3a  b  c
comparing the coefficients we get
a  4 , 2a  b  3,  3a  b  c  7
 b  5, c  0
Therefore a  4 b  5, c  0
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
58
1.4. POLYNOMIALS
1.4.3 Determination of the zeros of a polynomial
Decompositition into linear factors
Theorem: If x0 is the (e.g. guessed) zero of a n-th degree polynomial P(x) , the linear
factor (x – x0 ) can be seperated from the polynomial:
P(x) = u(x)(x – x0). In this case u(x) is a polynomial with the degree (n – 1).
The coefficients of the remaining polynomial u(x) can be determined through either
Horner‘s method (not covered here) or polynomial long division.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
59
1.4. POLYNOMIALS
Polynomial long division
If, in a rational function, the degree of the numerator is higher or equal to the
degree of the denominator, a polynomial long division can be conducted.
Through polynomial long division the zeros of a polynomial can be determined, if
one zero is already known. In this case the original polynomial is divided by the
respective linear factor.
Example: x3 – 3x2 + 4 has a zero at x = 2.
Further zeros can then be determined through polynomial long division.
In addititon to this slant asymptotes can be determined, if the degree of the
numerator is exactly one higher than the degree of the denominator (also see
chapter 2, differential and integral calculus).
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
60
1.4. POLYNOMIALS
Polynomial long division
Example:
(x3 – 3x2 + 4) : (x – 2) = (x2 – x – 2)(x – 2)
x2  x  2
x  2 x  3x 2  4
3
 x 3  2x 2
 x2  4
 x 2  2x
 2x  4
 2x  4
0
Therefore, x-2 is the factor of (x3 – 3x2 + 4)
61
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1.4. POLYNOMIALS
Polynomial long division
Example:
(x4 + x2 – 2) : (x + 1) = (x3 – x2 + 2x – 2) (x + 1)
x 3  x 2  2x  2
x  1 x 4  x2  2
 x4  x3
 x3  x2  2
 x3  x2
2x 2  2
Small Exercise:
(x2 - 9x – 10) / (x+1)
 2x 2  2x
 2x  2
 2x  2
0
We know
(x2 - 9x – 10) = (x+1))(x-10)
Verify it with long division.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
62
1.4. POLYNOMIALS
Properties of polynomials
An n-th degree polynomial has a maximum of n real zeros.
The addition, subtraction, multiplication and linking of polynomial functions
(polynomials) always results in polynomials again.
The division, on the other hand, results in rational functions.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
63
2. LINEAR ALGEBRA
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
64
2.1. SYSTEM OF LINEAR EQUATIONS
Linear Equations:
Recall that we can represent a straight line algebraically by an equation of the form
a1 x  a2 y  b
where

a1 , a2 and b are real constants

a1 and a2 are not both zero.

x and y are the variables of the equation
(often representing two different products x and y of a firm).

Equation of this form is called Linear Equation i.e. the one in which variables have the
power exactly equal to 1 and no variables are multiplied to other variables e.g. if the
term like “xy” appears in equation, then it will not be a linear equation.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
65
2.1. SYSTEM OF LINEAR EQUATIONS
Linear Equations:
Before, we formulated a linear equation for a firm taking decision on producing number
of units of two different products. Often real life situations are complex, nowadays an
ordinary supermarket has thousands of products. To describe such situations we can
extend our idea of linear equations. For example, consider a situation if the firm has to
decide on production of units of ‘n’ different products, then we can write the linear
equation as
a1x 1  a2 x 2  a3x 3  ......  anx n  b
where (as a convention)
x1 , x2 , x3 ,……, xn are all the variables of the equation and a1, a2, a3, …, an are coefficients.
These variables are also called as unknowns.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
66
2.1. SYSTEM OF LINEAR EQUATIONS
System of Linear Equations:
A finite set of linear equations is called a system of linear equations or a linear system.
For example,
System of two equations:
2x  y  3
5x  y  4
System of three equations:
x y-z5
2x  y  z  3
-x  y  z9
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
67
2.1. SYSTEM OF LINEAR EQUATIONS
System of Linear Equations:
System of linear equations (generalizing the notion):
An arbitrary system of “m” equations with “n” variables can be given by:
a11 x 1  a12 x 2  a13 x 3  ......  a1n x n  b1
a21 x 1  a22 x 2  a23 x 3  ......  a2n x n  b 2
am1 x 1  am2 x 2  am3 x 3  ......  amn x n  bm
In order to learn the Algorithm (step by step procedure) used to solve linear systems, we
will go through some preliminary concepts on Matrices.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
68
2.2. SYSTEM OF LINEAR INEQUALITIES
Definition (cp. Opitz, p. 252): A system of inequalities of the form
a11x1 + a12x2 + a13x3 + ... + a1nxn ≤ b1
a21x1 + a22x2 + a23x3 + ... + a2nxn ≤ b2
...
am1x1 + am2x2 + am3x3 + ... + amnxn ≤ bm
Is called linear system of inequalities with n unknowns (variables) x1, ..., xn and m
equations.
The values aij and bi (i = 1, ..., m, j = 1, ..., n) are given and they are called the
coefficients of the system of inequalities.
Sought after are the values for the variables x1, ..., xn , so that all inequalities are
fulfilled simultaneously.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
69
2.2. SYSTEM OF LINEAR INEQUALITIES
Every system of inequalities with relationships of inequality (≤ , ≥) and equality can
be rearranged to the above form.
Possible rearrangements are:

Multiplication by -1

Decompose one equation into two inequalities
Definition:
The set of all assignments of values (vectors x  IRn), which fulfill a certain
system of inequalities, is called solution set or admissible range of the system of
inequalities.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
70
2.2. SYSTEM OF LINEAR INEQUALITIES
Example:
A manufacturer produces two types of mountain bikes, type „Sport“ (S) and type
„Extra“ (E).
During the production every bike goes through two different workshops. In shop 1:
120 working hours are available each month; in shop 2: 180 hours are available.
To produce a type S bike six hours are needed in shop A and three hours are
needed in shop B. For a type E bike four and ten hours are needed respectively.
a) What is the respective system of inequalities?
b) Show the admissible range graphically!
71
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
2.2. SYSTEM OF LINEAR INEQUALITIES
Decision variables: A decision has to be made about the amount which should be
produced of both types of bikes.
S = Monthly amount produced of type S bikes.
E = Monthly amount produced of type E bikes.
Condition for producer 1:
required time <= available time
6S + 4E <= 120
Condition for producer 2:
required time <= available time
3S + 10E <= 180
Non-negativity:
S >= 0, E >= 0
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
72
2.2. SYSTEM OF LINEAR INEQUALITIES
E
producer 1
feasible region
20
producer 2
20
40
60
S
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
73
2.2. SYSTEM OF LINEAR INEQUALITIES
Three cases have to be distinguished:

Normal case:
The space of admissible solutions Z is bounded and not empty.
Then it is a convex polyhedron (as in the above example).

There is no solution for the system of inequalities. The space of admissible
solutions Z is empty.

The space of admissible solutions is unbounded.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
74
2.2. SYSTEM OF LINEAR INEQUALITIES
Convexity of the admissible range:
not convex
convex
Convexity means that every connecting line between two point of the admissible
range completely lies inside the range.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
75
3. Differential Calculus
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
76
3.1. BASICS
The (constant) slope of a straight line is equal to the derivative of the
corresponding function
f(x) = y = ax + b.
The situation for non-linear functions (curves) is more complicated, since the
slope is not equal at every point, but changes depending on x.
Slope of a curve: Slope at a point P
Therefore, to determine the slope at a certain point x, you have to look at the
tangent of the curve in this point.
Tangent: A straight line, which touches a curve at a certain point, but which
does not intersect the curve.
77
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
3.1. BASICS
EXAMPLE : TANGENT TO A CURVE
y
The slope of a tangent at a point
gives the change in y with
respect to a (marginal) change in
x
Tangent at point x*
x
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
78
3.1. BASICS
Examples: Different tangents to a curve
y = x3
y
slope at x* = 2:
dy/dx = 12
8
slope at x* = 1:
dy/dx = 3
1
x
2
1
79
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
3.1. BASICS
a) Power Function
Examples:
y  xn
y  nx n1
y  x9
y  x2
y   9 x 91  9 x 8
y   2 x1  2 x
y  x(  x 1 )
y   1x11  x 0  1
y  1( x 0 )
1
y  ( x 1 )
x
y   0 x 1  0
1
y  x ( 2 x  x 2 )
5
y  7 x5  x 7
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
1
x2
1
1
1 1 1 
1
y  x 2  x 2  1
2
2
2x 2
2

5
5
y  x 7 
7
7
7 x2
y   1x 2 
y 
1
2 x
80
3.1. BASICS
y  ex
b) Exponential Function
y  e x
y  e xln a  ln a  a x ln a
y  a x  (eln a ) x  e xln a
𝐴𝑝𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛: 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑓´ 𝑥 :
𝑓 𝑥 = 𝑥𝑥
y 
y  ln x
c) Logarithmic Function
x0
1
x
1 1
y  
x ln a
y  log a x
a y  x  y  ln a  ln x  y' ln a  (ln x)' 
y  sin x
y  cos x
d) Trigonometric Function
1
x
y  cos x
y   sin x
81
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
3.2. DERIVATIVE RULES
a) Factor Rule: y  c  f (x)
Examples:
y  4x
3
y  5( 5  x )
yc
Examples:
y   4  3x  12 x
y  5  0  0
y  0
2
0
b) Sum Rule:
y   c  f (x)
y  f ( x)  g ( x )
y   f ( x)  g ( x)
y  x 4  5x 3  7 x 2  2
y  ax 3  be x  ln x
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
c  const
2
y '  4 x 3  15 x 2  14 x
1
y '  3ax 2  be x 
x
82
3.2. DERIVATIVE RULES
c) Product Rule:
y  f ( x)  g ( x)
y  f g
y  f  g h
y '  f 'g  f  g '
y '  f 'g  h  f  g 'h  f  g  h'
Example:
y  (ln x)  4 x 2
y  x e x  x 5
1
y   4  ( x 2  (ln x)  2 x)  4 x(1  2 ln x)
x
1
x
y  (
 e x  x 5  x  e x  x 5  x  e x  5x 4 )  e x  x 4 (
 x x  5 x)
2 x
2 x
83
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
3.2. DERIVATIVE RULES
d) Quotient Rule:
y
f ( x)
g ( x)
y' 
f 'g  f  g '
g2
Example:
y
x3
4  x2
y 
3x 2  (4  x 2 )  x 3  (2 x) 12 x 2  x 4 x 2 (12  x 2 )


(4  x 2 ) 2
(4  x 2 ) 2
(4  x 2 ) 2
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
84
3.2. DERIVATIVE RULES
e) Chain Rule:
z

y  f ( g ( x ))
y'  ( f  g )' ( x)  f ' ( g ( x))  g ' ( x)
y 
Example: 1)
2)
y  (a  x 3 ) 5
ye
x 2 1
df dz

dz dx
y   5(a  x 3 ) 4  (0  3x 2 )
y   15 x 2 (a  x 3 ) 4
y  ( f  g  h)' ( x)
2
1
y  e x 1 
 2x
2 x2 1
2
x
y' 
 e x 1
2
x 1
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
85
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
Example:
y  3x 4  2 x 2  7 x  5
y  12 x 3  4 x  7
y  36 x 2  4
y  72 x
y 4   72
y 5   y ( 6 )  y ( 7 )  ...  0
Derivatives of a higher order are necessary for the solution of optimization problems
(curve sketching)
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
86
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
Curve sketching gives information about the characteristics and the behaviour of
the particular function, of a profit, revenue or cost function.
a) Domain (x-value) and (if necessary) codomain (y-values)
b) Symmetry characteristics
Symmetry about the y-axis f(x)=f(-x)
Point symmetry about the origin –f(x)=f(-x)
e.g. y=x2
e.g. y=x3
c) Intersection with the axes
Intersection with the y-axis: x=0; f(0) corresponding y-value
Intersection with the x-axis: Zeros. f(x)=0; solve for x
d) Gaps and poles (vertical asymptotes) e.g. in the case of rational functions (quotient of
two polynomials Z(x)/N(x)) gaps in the definition exist in the zeros of N(x).
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
87
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
e) Asymptotical behaviour and asymptotes:
Asymptotical behaviour (horizontal asymptotes):
Behaviour of the function if x → ∞ and if x → -∞ :
Does the function tend to ∞ , -∞ or to a constant value?
Vertical asymptotes:
If the gap in the definition x0 is a pole, then the straight line x=x0 is a vertical asymptote of the
graph of f.
Slant asymptotes:
If a function approaches a straight line more and more with increasing x, then this line is called a
slant asymptote. Rational functions can have asymptotes, if the degree of the numerator is
maximum one higher than the degree of the denominator. Then the linear equation of the
asymptote can be found through polynomial long division. The term in front of the remainder
shows the linear equation of the asymptote.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
88
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
Connection between Extrema and Derivative:
Observation: If the function shows a local maximum, the slope of the curve
must be positive to the left and negative to the right of the maximum. I.e., the
function values increase before the maximum and decrease afterwards.
Conclusion: At the maximum the slope (= value of the derivative) is exactly
zero!
slope= 0
slope
positive
slope
negative
Local maximum: There is an
environment in which no
point has a higher function
value.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
89
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
f) Slope and extrema: maxima, minima and saddlepoints
f ( xE )  0 
 local minimum xE
f ( xE )  0 
extrema
(horizontal tangent)
f ( xE )  0 
 local maximum xE
f ( xE )  0 
f ( xE )  0 
 saddle point
f ( xE )  0 
The second derivative shows a change in the slope.
Second derivative positive: slope increases => local minimum
Second derivative negative: slope decreases => local maximum
Second derivative is equal to zero: neither maximum nor minimum, but saddle
point.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
90
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
Necessary and sufficient conditions for local extrema
Necessary condition for a local maximum or minimum: f’(x0) = 0
I.e., if there is a maximum or a minimum at x0, then f’(x0) = 0
Sufficient condition for a local maximum: f’(x0) = 0 and f’’(x0) < 0
I.e., if f’(x0) = 0 and f’’(x) < 0 , there is a local maximum at x0
Sufficient condition for a local minimum: f’(x0) = 0 and f’’(x0) > 0
I.e., if f’(x0) = 0 and f’’(x) > 0 , there is a local minimum at x0
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
91
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
g) Curvature and inflection points
f ( xW )  0 

f ( xW )  0 
Inflection Point
f ( x)  0
f ( x)  0
At the inflection point a curvature to the right
changes to a curvature to the left
f ( x)  0
f ( x)  0
At the inflection point a curvature to the left
changes to a curvature to the right
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
92
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
Example: f(x) = (-x2 – 4x – 12)·(x – 6)2
f ( x)  ( x 2  4 x  12)( x  6) 2
Domain IR
f ( x)   x 4  8 x 3  432
No symmetry
f ( x)  4 x 3  24 x 2
 4 x 2 ( x  6)
X to +/- ∞, f(x) to -∞
f ( x)  12 x  48 x
f ( x)  24 x  48
 12 x( x  4)
2
f (x)  0
1. Zeros:
(x 2  4 x  12)(x  6) 2  0  x 01 02  6
x  4 x  12
2
x 2  4 x  22
(boundary point)
0
 12  2 2
(x
 2) 2
 8
x
2
  8
Mathematics ∙ Prof. Dr. Philipp E. Zaeh

   x 03 04  
93
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
Continuation of the Example:
2. Extrema:
f (x)  0
 4x (x  6)  0
x 11 12  0
x 13  6
2
Max or Min ?
f (x 11 )  f (0)  0  NEITHER
f (x 13 )  f (6)  144  MAX at (6,0) (q.v. ZEROS!)
f (x)  0
 12x(x  4)  0
x 21  0 x 22  4
f(0)  432
f(4)  176
3. Inflection points:
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
94
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
Continuation of the Example:
Nature of the change
f (x 21 )  f (0)  48  0  r/l
f (x 22 )  f (4)  48  0  l/r
f (0)  0 

f (0)  0  SP at (0,-432), r/l
f (0)  0
WP at (4,-176), l/r
95
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
Continuation of the Example
Tangent at P(-2,-512) => t4(x)=128x-256
Tangent at IP=> t1(x)=128x-688
4. Sketch:
-2
4
6
MAX
IP
l/r
-256
x
f(x)
r/l
SP
P
- 432
- 512
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
96
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
Continuation of the Example:
4. Sketch of the derivative
-2
4
x
6
f (x)
97
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
Continuation of the Example
4. Sketch of the seconde derivative
-2
4
6
x
f (x)
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
98
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
Continuation of the Example
5. Tangent at IP:
Slope at IP:
f (4)  128
t(x)  mx  b
 176  128  4  b
 688  b
t 1 (x)  128x  688
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
99
3.3. APPLICATIONS & EXERCISES –
CURVE SKETCHING
6. Parallel line to the inflection tangent:
f (x)  128
 4x 3  24x 2  128
 4x 3  24x 2  128  0
(4x 3  24x 2  128) : (x  4) 2  (x  2)  4
x  2
Boundary point P  (2,512)
f(2)
Tangent at P(2,512)
m  128
t(x)  mx  b
 512  128  (2)  b
 256 
b
t(x)  128x  256
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
100
4. Integral Calculus
101
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
4.1.
BASICS
1. Theorem of differential and integral calculus
If f(x) is integrated, you obtain the primitive function F(x).
If F(x) is differentiated, you obtain f(x).
Integration is the reversal of differentiation.
2. Theorem of differential and integral calculus
If F(x) is the primitive function of f(x), then the definite integral is
b
 f ( x)dx  F ( x)
b
a
 F (b)  F (a)
a
Definition: the general form F(x)+ c (c = constant of integration) of a primitive function
f(x) is called indefinite integral of f(x). You write:
 f ( x)dx  F ( x)  c
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
Without limits a, b
102
4.1.
BASICS
ADVISE FOR INTEGRATING:
 If a function is continuous, it is also integrable.
 If a function is piecewise continuous, a primitive function can be determined for every
continuous interval. The definite integrals are calculated one-by-one for the partial
intervals and then added up to the complete integral.
About the calculation of area:
 During the calculation of area functions cannot be integrated beyond zeros. If there are
zeros, it has to be integrated in sections.
 In order to determine the whole area between the curve and the x-axis, it has to be
integrated over the absolute value of the function in the negative sections. Then all
values have to be added up.
103
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
4.1.
BASICS
Power Function:
f ( x)  x n
x n 1
c
 x dx 
n 1
n
n  -1
Testing by Differentiation:

x ( n 1)1
 x n 1 
 xn

  (n  1) 
n

1
n

1


Mathematics ∙ Prof. Dr. Philipp E. Zaeh
104
4.1.
BASICS
Examples:
Testing by differentiation:

 x2

x 21
  c   2 
0 x
2
 2

x2
2
x2
 xdx  2  1
x2
xdx

3

2
x2
xdx

c

2
 xdx 
105
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
4.1.
BASICS
Exponential functions:
f(x)  a x ln(a)
 (a
x
f(x)  e x
e
dx  e x  c
x
ln(a)) dx  a x  c
ln x  c für x  0
f ( x) 
1
x
1
 x dx  ln x  c 
ln(( x))  c für x  0
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
106
4.2.
RULES FOR INTEGRATION
Sum Rule:
y  f(x)  g(x)
 ( f ( x)  g ( x)) dx   f ( x) dx   g ( x) dx
Rule: A sum of functions can be integrated one by one!
Example:
y  x  sin x
2
 ( x  sin x)dx   x dx   sin x dx 
2
2
3
x
 cos x  c
3
107
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
4.2.
RULES FOR INTEGRATION
Factor Rule:
 a  f ( x) dx  a  f ( x) dx
Rule: A constant factor can be moved in front of the integral during integration!
Example:
y  3x 2
2
2
 (3x )dx  3 x dx  3
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
x3
 c  x3  c
3
108
4.3.
APPLICATIONS & EXERCISES
Example 1:
2
 (9 x
2
 4 x  2  a 2  x 1  72 x 2 )dx
1
Example 2:
2
 (3x
2
 x 2  a 2  4 x)dx
a
Example 3:
2
 (3  6 x  2
1
1
x
 e x )dx
Partial integration and integration by substitution are not covered here.
Mathematics ∙ Prof. Dr. Philipp E. Zaeh
109