UTSC
MATA32H3
Final EXAM
STUDY GUIDE
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University of Toronto
Scarborough
MATA32H3
Calculus for Management I
Fall 2017
Term Test 1
Exam Guide
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Compound Interest
Compound Interest
Deals with money, time and interest
1 key equation:
� = �(1 + �)!
P  Principle
• Money invested now
r  Periodic interest rate
• Interest paid per compounding period
• Stated in percentage or annual rate
o NOTE: Must convert percentage into a decimal
n  Number of compounding periods
• Counts number of times interest is paid on investment
S  Compound Amount
• Amount of money we have at the end
Interest is paid exactly at the end of compounding periods
• Only at the end of compounds
Compounding periods
P$
Start
n
End
FixedKhl
Period of time
**For one compound period, the principle (P) plus the interest on that
1st Compound  � 1 + Pr = � (1 + �)
(Principle + Interest on Principle)
nd
2 Compound  � 1 + � + � 1 + � ∙ � = � 1 + � 1 + � = � (1 + �)!
3rd Compound  � (1 + �)! + � 1 + � ∙ � = � 1 + � ! ∙ 1 + � = � (1 + �)!
4th Compound  � (1 + �)!
NOTE: IN APPLICATIONS
• We often have APR ‘a’ (Annual Percentage Rate) and frequency ‘k’ of
compounding; and
o How often interest is paid in a year
• A time ‘t’ period
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Legend:
a  APR for period as follows
k  Compounding frequency per year (annually)
t  Number of years for a particular period
We have that…
!
= r and � = � ∙ �
!
Thus,
� = � (1 +
� !∙!
)
�
Example 1: Consider the following “investment scheme” invest $10,000 at 3.05% APR
compounding monthly for a period of 5 years.
a) Find compound amount
� = 10,000 1 +
X 60
!.!"!# !"
!"
≈ 11, 645.17
$11,645.17
$10,000
5 years
b) Periodic rate (interest at the end of each month)
�
�
0.0305
=
12
�=
≈ 0.002542 (0.2542%)
c) Compound Interest
Difference between beginning amount and end amount
�−�
≈ 1,645.17
Return due to interest
=
1645.17
∙ 100
10000
≈ 16.4517%
(Over 5 years accumulated)
d) How long does it take to earn $15,000
** Express terms in units of compounding (i.e. from this example, months)**
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We solve for the following the equation
15 = 10 1 +
1.5 = (1 +
ln 1.5 = ln
!.!"!# !"∙!
!.!"!# !"∙!
!"
)
1+
0.0305 !"∙!
12
ln 1.5 = 12� ∙ ln 1 +
∴�=
ln � ! = � ∙ ln �
!"
ln(� + �) ≠ ln � + ln �
!.!"!#
!"
log � = log!" �
!" !.!
!"∙!" !!
!.!"!#
!"
≈ 13.31 �����
ln � = log ! �
0.31
= 3.72 ����ℎ� (����� �� �� 4 ����ℎ�)
12
Therefore, it takes 13 years and 4 month or 160 months
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Effective Rate Concept
Effective Rate:
Consider one year and with an APR of ‘r’ compounding ‘n’ times for that year.
Simple Interest = 1 compound at the end of the year.
P  principle
Compound
S
Compound
Amount
Simple (Effective Rate ‘Re’_)
Effective rate = simple interest
Should expect the effective rate to be larger than the APR
• No intermediate compounding
Simple = compounding once at the end of the year
Equation:
�
�� = (1 + )! − 1
�
r  APR
• Not ‘a’
n  compound periods
• When n>1, then you are getting interest on your interest (compounding)
• If n = 1, the effective rate is the APR because ‘r’ is simple when n=1
Example 1:
In applications of effective rate it “comparing investments”
NOTE:
The more you compound the more you make on investments
Scheme A: 2.95% APR, semi-annually
Scheme B: 2.93% APR, bi-weekly
Which is more profitable?
The more profitable one is the
one with the higher and better
effective rate
Answer:
Find the effective rate (Re) for both schemes, then take the higher and better one.
(A is slightly higher and better)
A  n=2
B  n=26
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Present and Future Value
FV  Future Value
PV  Present Value
These are about
perspective
Start
Future Value
Present/Past Value
End
End
Start
Future Value:
If P = principle (now), then…
�
�� = � (1 + )!"
�
Present Value:
If S = amount (now), then…
�
�� = � (1 + )!!"
�
Summary
FV is the Future Value of an investment of ‘P’ now
PV is the present/past value of an amount ‘S’ now
• Amount invested to meet future investments
Can combine PV and FV in one equation
�
�� = �� (1 + )!"
�
�
�� = �� (1 + )!!"
�
Example 1:
In 5 years we need $35,000 in an account. Interest is weekly at 4.3% APR.
Solution:
(In ‘000’s)
�� = 35 (1 +
0.043 !!"∙!
)
52
0.043 !!"#
= 35 1 +
52
≈ 28,231.46
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Equations of Value
We have a “long period of time” throughout which we have interest, debt and payment.
Debt and payments are two types of simple financial transactions
An Equation of Value is an equation that describes interest, debt and payments and
allows us to calculate.
3 Steps:
1. Create a money-time diagram
2. Build an equation
3. Solve the equation
Example 1: We have a debt of $32,000 in 6 years. Pay $10,000 now, the balance at the
end of the 6 years. Interest is 3% APR, quarterly. Calculate the amounts of the balance
payment.
Solution: First, we make a money-time diagram.
$10,000
$32,000
1
6
Now
X
Latest ‘x’ can be
made
Let ‘x’ be the amount of the balanced payment
Main Concept:
At all times, the value of all payments must equal value of all debts.
• Selecting a time is called “calibrating”
At time 6 (end of 6 years), is $32,000.
What is the value of our payments (x)
Value = with interest
� + 10,000 (1 +
≈ 20,035.86
!.!" !∙!
!
)
= 32,000
Solve for ‘x’
Reducing what you were obligated to pay before
** Assume there are 365 days in a year **
Calibrate to “now”:
32,000 (1 +
0.03 !!"
0.03 !!"
)
= 10,000 + � (1 +
)
4
4
Another Equation of Value
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Example 2: More complicated… 2 debts and 3 payments
2 debts  $2000 in 3 years
• $3000 in 55 months
3 payments  $1000 now
• 2nd at the end of 20 months
• 3rd (1/2 of 2nd) at the end of 4 years
Interest is 6.6% APR, monthly. Find 2nd and 3rd payment amounts
Solution:
Cash-Time Diagram:
(All units are in ‘000s)
20 months
55 months
0
5 years
1
2
1
x
3
2
4
�
2
3
Let ‘x’ be the amount of the 2nd payment in thousands. Therefore the 3rd payment is x/2.
Let’s calibrate for time now (0):
�=
!
!
k=12
0.066
12
= 0.0055
=
Equation of Value:
1 + � 1 + 0.0055 !!" +
!
!
1 + 0.0055 !!" = 2 1 + 0.0055 !!" + 3 1 + 0.0055 !!!
Value of payments now
Value of debt now
� ≈ 2.23397
�
≈ 1.11699
2
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Interest Compounded Continuously & Net Present Value
Interest Compounded Continuously
Fix a period of time ‘t’ years and an APR, r % and a given principle (P). If the frequency
of compounding annually (k value) increases without bound, there is actually a maximum
amount at the end of time.
Maximum amount is Interest compounds continuously
NOTE: � ≈ 2.78128 …
� = �� !"
Time
K
Annual
1
Monthly
12
Weekly
52
Daily
365
Hourly
8,760
Second
31,536,000
Continuously = always compounding at all times
Continuously compounding is the opposite of simple interest. Simple interest is
compounding with interest at the end of every year while compounding continuously is
compounding with interest at every microsecond.
Equation:
� = �� !"
P Principal
r Nominal rate (also known as APR)
t  Number of years
S  Final amount
e  2.78128
If we have a given principal (P), an APR (r) and a given time (t), the least amount we’d
have at the end of ‘t’ years is the simple interest
� = �(1+�)!
The maximum amount we’d get at the end of ‘t’ years is continuously compounded
� = �� !"
Example 1: Find the present value (PV) of $100,000 at 4.2% APR at 4 years when
interest compounds continuously with interest.
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Solution: (In unit’s of 000’s)
�� = 100� ! !.!" !
≈ 84,535
Therefore, we can say that approximately $84,535 is the least amount we need to invest
at 4.2% APR for 4 years for it to grow it $100,000
•
•
When compounding continuously = the most amount of interest
The more you compound the less you would have to pay than with less
compounding
Net Present Value (NPV)
Idea: Suppose you invest ‘P’ in a “company” that “guarantees” payments back to you on
a “schedule” (Cash flows) assuming there is interest too.
Formula:
��� =
�� �� �� ��������) − �
When…
NPV > 0  Profitable Investment
NPV < 0  Lost money in investment
NPV = 0  Neutral (break even)
Example 1: Invest $35M. Interest is 6%, semi-annually.
Years
2
5
8
Payments
13
19
24
Cash Flows
^ Don’t get all of these payments at once, only at the end of year 2, 5 and 8. MUST
CONSIDER INTEREST
Solution:
��� = 13 1.03 !! + 19 1.03 !!" + 24 1.03 !!" − 35
≈ 5.6441 > 0
Therefore, the investment is profitable
**The threshold for being profitable is as soon as it goes slightly about the amount of ‘P’
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Try yourself…
Years
2
5
8
Payments
13
19
F
Cash Flows
What is the least amount that ‘F’ has to be in order to be considered profitable?
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Annuities & Limits and Continuity
Fixed payment over a fixed period of time at a fixed interval
Annuity is a sequence of payments of a fixed amount, r, paid at “regular” points in time.
There is constant interest “for all time”.
n
…
R
R
R
R
0
n  Number of payments of fixed amounts of ‘R’
R  Periodic interest rate
Ordinary Annuity  payments are at the end of payment periods.
Annuities are ordinary unless stated otherwise
Annuity Due  payments are made at the beginning of payment periods
(i.e. rent)
NOTE: Interest paid is “timed” with the payment periods
Future Value of an annuity = FV
(�� �� ��� � ��������)
Present Value of an annuity = PV
(�� �� ��� � ��������)
Equation of a FUTURE VALUE ORDINARY ANNUITY
1+� !−1
]
�� = � [
�
Equation of a PRESENT VALUE ORDINARY ANNUITY
1 − 1 + � !!
�� = � [
]
�
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Example 1: Present Value (Ordinary Annuity)
4 Payments, 1 payment of $100 at the end of the year for 4 years at 9% APR, annually.
APR is equal to the number of payment periods
�� = 100
1 − 1.09 !!
0.09
≈ 323.97
Year One: (��)(1.09) ≈ 353.13 − 100 = 253.13
Year Two: (253.13)(1.09) ≈ 275.91 − 100 = 175.91
Year Three: (175.91)(1.09) ≈ 191.74 − 100 = 91.74
Year Four: (91.74)(1.09) ≈ 100 − 100 = 0
PV = present value of $ in account for payments can be made (least value) with interest
considered.
Example 2: How to approach annuity due using ordinary annuity concepts
An annuity due of 5 payments (n=5), r = periodic rate
Present Value:
1 − 1 + � !!
�� = �
+�
�
=�
!! !!! !!
!
+ �
Ordinary for last 4
Future Value:
�� = �
=�
1+� !−1
1+�
�
1 + � !! − 1
(1 + �)
�
** Take sum of ordinary annuity and add one more term (1+r)
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Limit and Continuity
Calculus is applied in economics, finance and management science (operations
research [OR]). Calculus has 2 main parts: (1) differential and (2) integral. Foundation =
LIMITS
Main definition 
LIMIT CONCEPT
Let y= f(x) be a function and let ‘a’ be a real number (a is a real number). We write the
following notation lim!→! � � = �.
To mean the following 2 things:
1. L is a real number (I.e. L is a number called the “value of the limit”)
2. The function values (i.e. “output” of the function)
F(x) gets closer to L and stays close as the input values ‘x’ approaches ‘a’.
However, ‘x’ will never reach ‘a’.
When we say lim!→! � � = �, also means the “limit exists” ‘L’ is a number.
If lim!→! � � = ���� ��� ����� there is no value of ‘L’
Example 1: let � � =
! ! !!
Rational function = polynomial/polynomial
!!!
Domain � = ��ℝ � ≠ −3} �� (−∞, −3) ∪
(−3, ∞)
U  union symbol (this or this)
0/0 = undefined
for � ≠ −3, � � =
!!! !!!
!!!
= � + 3 �� [ � + 3 ≠ 0]
lim � � = lim � − 3 = −6
!→ !!
!→!!
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Continuing With Limits
Example 1: Let � � = −2� ! + 4� − 1
Polynomial degree = 3
Cubic Function
lim!→! � � = lim!→! −2� ! + 4� − 1
= lim −2� ! + lim 4� − lim 1
!→!
!→!
!→!
!
= −2 [lim � ] + 4 [lim �] − 1
!→!
lim �(�) → ����� ��������
lim
!→!
!→!
(−2� !
!→!
!→!
!
!lim �! → ��� ��������� �����
!→!
!
= −2 lim �
+ 4� − 1) → ������������ ��������
+ 28 − 1
= −2 7 ! + 28 − 1
= −659
Thus, as ‘x’ gets closer & closer to 7, but x ≠ 7, the function values (outputs) p(x) gets
closer & closer to -659
NOTE: That � 7 = −659 [���������� ��������]
If y=q(x)  any polynomial and c is a real number
Then lim!→! � � = �(�) thus, the limit exists and is obtained by substituting
(evaluating)
CANNOT ALWAYS EVALUATE FOR ALL FUNCTIONS
Works for properties of continuity
Limits, Left & Right
Let y=f(x) be a MATA32 function and let c be a real number
** MATA32 function = functions we study in this course
Left:
We write lim!→!! � � = �! to mean that…
1. L1 is a real number
2. The function values f(x) approaches ‘L1’ as ‘x’ approaches ‘c’ where x<c
Right:
We write lim!→!! �(�) = �! to mean that…
1. L2 is a real number
2. The function values f(x) approaches ‘L2’ as ‘x’ approaches ‘c’ where x>c
CONNECTION between Limit, Left & Right
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Let y=g(x) be a MATA32 function and let c be a real number. We then have a “Limit
Principle”
lim!→! �(�) = � <=> (�� ��� ���� ��)
1. lim!→!! � � = �!
2. lim!→!! �(�) = �!
3. lim!→!! �(�) = lim!→!! �(�)
∴ �! = �!
We often use the principle above (Limit Principle) when the given function [g(x)] is
defined and has two or more formulas based on cases
Example 1: Let � � =
�
�� � > 1
−� + 1 �� � < 1
Breaking point is at 1 (^)
lim!→! �(�) = ? (Must use one sided limits)
lim �(�) = lim − � + 1 = 0
!→!!
!→!
lim �(�) = lim
!→!!
!→!!
�=1
Values are not equal
Therefore, property 3 fails
Therefore THE LIMIT DOES NOT EXIST
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The Limit Properties, Limits at ±∞ & Continuity and Continuous
Functions
The limit properties are a list of “tools” for finding many limits but not all.
Key problem: find the limit lim!→! �(�) where ‘f’ is a MATA32 function and ‘a’ is a real
number
Sometimes recognizing the limit properties are not always immediate
Of special importance are limits of the form 0/0. All derivatives contain the form 0/0
Example 1: lim!→!
! ! !!
�(�) = � ! − 1
�(�) = � ! − 1
lim �(�) = 0
! ! !!
� − 1 �! + � + 1
= lim
!→!
�−1 �+1
�! + � + 1
= lim
!→!
�+1
=
Polynomials
!→!
lim �(�) = 0
!→!
!"#!→! ! ! !!!!
Sub x=1
!"#!→! !!!
3
=
2
Example 2: lim!→!!
Finding the limits as is will result in
the form 0/0.
!! !
!!!
= lim x-2 [2-(-x)/x+2]
=lim x-2 [2+x/x+2]
=1
Absolute value |x| =
!
! !" !!!
!
!! !" !!!
Limits at ±∞
These kinds of limits also involve “horizontal Asymptotes”
B
A
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Example 1: lim!→!
!! ! !!!!
!! ! !!"""!!!"!
5� !
!→! 3� !
5
=
3
= lim
OR
! !
! !
!""" !"!
! ! !!
! !
!
!
! ! !! ! !
= lim!→!
1
2
+ lim !
�
�
!→!
!→!
= !→!
5000
10!
lim 3 + lim
+ lim
�
!→!
!→!
!→! � !
5
=
3
lim 5 + lim
Example 2: lim
!→!
�! + � − �
�! + � − �
∙
!→!
1
�! + � − �!
= lim
!→! � ! + � + �
!
= lim
= lim
!→! !
= lim
Since limit of 1/x, 2/x2,
5000/x and 109/x2 all go off
to 0.
�! + � + �
�! + � + �
!
!
!! !!
1
!→! 1 + 1
=
1
2
Continuity and Continuous Functions
Main idea: a MATA32 function y=f(x) is “continuous at a number ‘c’” means that the
graph of our ‘f’ has NO: holes, gaps, jumps, etc.
Careful definition:
Let y=f(x) be a MATA32 function and let ‘c’ be a real number. We say that ‘f’ is
continoust at ‘c’ if and only if lim � � = �(�)
!→!
The continuity check is used to see if a function is continuous at ‘c’.
���� � = � � < = > 1. ���� � = � ���� ����� ∴ �� = ��
�→�
�→�
2. � � ���� ����� ������ ��� ������
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3. ���� � = � � ���� �� ����
�→�
Example 1: � � =
� �� � > 0
2 �� � = 0
−� − 1 �� � <)
= lim � � = lim (−x − 1) = −1
!→!!
!→!!
= lim � � = lim
!→!!
!→!!
� =0
∴ lim ���� ��� ����� �����, �1 ≠ �2 ��� �������� #1 �����
!→!
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Derivatives
Main definition of a derivative  Let y= f(x) be a MATA32 function.
The derivative of ‘f’ is a function (written as �′)
�′ is defined as a special limit
� �+ℎ −� �
� ! � = lim
!→!
ℎ
h  limit variable
x  function variable
^Definition forms
!
!
(undefined)
“Differentiate”  find the derivative of a given function
Given a function, y=f(x), the derivative is defined as follows
� �+ℎ −� �
!→!
ℎ
** We use the definition (^) only when we are asked to
� ! � = lim
We calculate much faster using the derivative rule
Notation: Let y=f(x) be a MATA32 function
Derivative: � ! � , � ! ,
!"
,
!
!" !"
� � , �! �
Derivative Rule:
1. The Constant Rule:
�
� = 0, �ℎ��� � �� � ���� ������ ��� � �� � ���� ������
��
A constant has a slope of 0
2. Power Rule:
� = � � = � ! , �ℎ��� � �� � ���� ������ ��� �� �ℎ� �������� �����
Then � ! = �(� !!! )
!
Example 1: � = � = � !
1 !!!
�!
2
!
1
� ! = (� !! )
2
1
1
�! = × !
2
� !!
1
�! =
2 �
�! =
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!
Example 2: � � = = � !!
!
1
��
!!
= −1� = − !
�
��
3. Constant Factor Rule:
Suppose: � � = �ℎ �
∴ � ! � = �ℎ �
!
= � ℎ! �
!
Example #1: �� � = 5 � = 5� !
1 !!
�ℎ�� � ! = 5
� !
2
5 1
= × !
2
�!
5
�! =
2 �
4. Sum-Difference Rule:
• ‘f’ and ‘g’ are functions
� � + � � ! = � ! � + �! �
� � −� �
!
= � ! � − �! �
5. Product Rule:
• ‘f’ and ‘g’ are functions
� � � � = � ! � � � + � � �! �
��ℎ ! = � ! �ℎ + ��! ℎ + ��ℎ′
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Examples from 11.1-2 & 11.4-5 & Derivatives in Economics and
Management
! !!! !! !
1. Definition of derivative  lim!→! [
]
!
In 11.1, there is a different way to write �′(�).
� � −� �
!→!
�−�
The two different �′(�) formulas compute the same thing they get the derivative.
� ! � = lim
Why doe these give the same result?
� �+ℎ −� �
!→!
ℎ
Introduce a new variable ‘t’ where � = � − ℎ
∴ℎ=�−�
lim
If we use ‘t’ and ‘h’ as “limit variables” and ‘x’ is “fixed”, then h0 if and only if
tx
� �+ℎ −� �
!→!
ℎ
∴ lim
� � −� �
!→!
�−�
= lim
= lim!→!
! ! !! !
!!!
To know when to use it depends on the actual form of f(x) and the context of the
question. Whether it suggests one formula over the other.
Example 1: Find lim!→!
! !
!
! !
!!!
Solution: Is a “0/0 form”. The limit can actually be viewed as the 1st principle’s
definition of derivative of some function! Let � � =
1 1
−
� 2
∴ lim
= �! 4
!→! � − 4
1 !
� ! � = − � !!
2
∴ �! 4 = −
!
!
!
= � !! , � 4 =
!
!
=
!
!
!
1
1
4 !! = −
2
16
Example 2: Let � = � ! + 1 get y’
Solution: “MUST” use the chain rule [� � �
�=
!
= �′(� � )×�′(�)]
!
�! + 1 !
“Inner Function”  � � = � ! + 1
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!
“Outer Function”  ℎ � = � !
∴�=ℎ � � =� �
Chain rule says  � ! � = ℎ! � � ×�! �
!
1
=
� � !! × 2�
2
!
1 !
= � + 1 !! 2�
2
!
= � ! + 1 !! �
�
�! � =
!
� +1
NOTE: That we can write our answer as � ! =
!
!
11.3 – Derivatives in Economics and Management
The “Marginal” concept = Derivative
Let � = � � be a MATA32 function. The derivative is
� �+ℎ −� �
≈� �+1 −� �
!→!
ℎ
∴ �! � ≈ � � + 1 − � �
� ! � = lim
�� ��� ℎ = 1
Should not expect them to be equal
I.e. The derivative is approx. ≈ to the difference [f(x+1)-f(x)]
Apply as follows:
Let ‘q’ be quantity and suppose � ! = � � = ����� ���� ��������
To make, market and sell, ‘q’ units it costs and c=f(q)
q-axis
0
q
q+1
What is the cost of (q+1)st unit?
Answer is f(q+1) – f(q) … the Marginal cost [q+1] item
We have that � ! � = � � + 1 − � �
Therefore, the derivative � ! � ≈ �ℎ� ���� �� ���� # (� + 1)
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Example 1: Let � = � ! + 38� + 1500 be a cost function where ‘q’ is quantity
� = �(�)
Marginal cost function is � ! = � ! � =
!"
!"
= 2� + 38
!
We’ll see an example  � � ≈ � � + 1 − � �
Let q=1250
� ! � = � ! 1250 = 2538
� � + � + � � ≈ �′(�)
� 1251 − � 1250 = 2539
[Is not the limit definition; therefore is not the same answer as the limit]
The exact cost to produce unit # 1251 is 2539 and this is well approximated by c’(1250)
Example 2: See page 526 #72
Involve marginal cost and average cost and notation
Let � = �(�) be a cost function where ‘q’ = quantity
!!
Assume the following: |� = �0 = 0 for some “special quantity”, ‘q’
!"
Average cost function = � =
! !
!
Prove the following:
!!
!"
|� = �0 = � � ∘
Solution: � =
! !
!
� � �
�
� =
�� �
��
Sub in � = � ∘
��
� −� �
��
=
�!
��
|� = � ∘ (� ∘) − �(� ∘)
��
=0
�∘ !
��
∴
|� = � ∘ � ∘ − � � ∘ = 0
��
A little algebra shows that…
!!
!"
|� = � ∘ =
!(!∘)
!∘
Marginal Cost = Average Cost
∴
!!
!"
|� = � ∘ = �(� ∘ )
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Derivatives of Logarithms and Exponential Functions
Facts:
!
!
ln � =
1.
2.
!"
!
!"
!
�! = �!
Example 1: Find the equation of the tangent line to the graph of y=ln(x) at x=3
Solution: Point of tangency is (x,y) = (3, ln(3))
NOTE: We leave ln(3) as is … no calculator use, unless asked to approximate
Slope  � =
!
!"
ln �
�ℎ�� � = 3 =
!
!
=
!
!
Point – slope form  � − � ∘= � � − � ∘
1
� − ln 3 = (� − 3)
3
Slope – Intercept form  � = �� + �
1
� = � − 1 + ln 3
3
1
� = � + ln 3 − 1
3
Example 2: Differentiate …
(1) � = �� !!
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� ! = 1� !! + �� !! −1
= � !! 1 − � �������� ����
!
(2) � � = �
Let � = � !
��
= �!
��
!! ! !!
��� � = 3� ! +
1
�
�� �� ��
=
∙
�� �� ��
1
= � ! (6� − ! )
�
!!
!
(3) � � = ln [( ! )! ]
! !!
To find f’(x), it is easier to simplify first
ln �! = � ln �
�
ln
= ln � − ln �
�
�! � =
1
�!
ln !
� +2
3
1
(ln � ! − ln � ! + 2
3
1
= [2 ln � − ln � ! + 2
3
1 2
2�
�! � =
− !
3 � � +2
=
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Elasticity of Demand (EOD)
Let ‘q’ represent quantity (I.e. demand)
We assume p=f(q) as a demand function. Where p = unit price (i.e. price/unit)
Elasticity of Demand is a mathematical function that shows how percentage changes
in ‘p’ and ‘q’ happen when the demand function holds.
Mathematical assumptions: ‘p’ and ‘q’ are as above and p=f(q) is a demand function
where ‘f’ is a differentiable and decreasing.
When ‘q’ is a special amount,
p = special amount when on
the curve
We always have � ≥ 0 and sometimes q > 0
Differentiability  Continuity
Therefore, ‘f’ is also continuous wherever it’s defined
Definition of Elasticity of Demand function is..
�
�
� �
� �
��
�
�=
= !
=
��
� �
� �! �
� = �������� ��� (���)
A.K.A. point of elasticity of demand
NOTE: That η is actually a function of ‘q’
We see that η<0 for q> 0 as follows: f(q) > 0 and q<0 and f’(x)<0
Here is an “approximation relationship” seeing elasticity of demand
Given a specific point, (q0,p0), we have that..
%�ℎ���� �� � �� � ∘
�≈
% �ℎ���� �� � �� � ∘
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∴ % �ℎ���� �� �� � �� � ∘ ×� ≈ % �ℎ���� �� � �� � ∘
� ∘= �(� ∘)
There are “3 Types” of elasticity:
1. If |η|>1  we have an “elastic demand”
2. If |η|=1  we have a “unit elasticity”
% �ℎ���� �� � �� � ∘ ≈ % �ℎ���� �� � �� � ∘
∵�=1
3. If |η| <1  we have an “inelastic demand”
We calculate |η| like this: given q0, we find η(q0) <0
Thus we actually find |η(q0)|
Implicit Differentiation
Main Idea:
Assume we have 2 variables ‘x’ and ‘y’ and they appear in a given equation. We assume
the given equation implicitly defines ‘y’ as some function of ‘x’/
Implicit differentiation helps us find the derivative without knowing ‘y’
Example #1: Assume the equation xy+x2 = 2+y
Defines ‘y’ implicitly as some function of ‘x’ (i.e. y =f(x) for some function of ‘f’ that we
may nor may not be able to actually find)
Use implicit differentiation to find the derivative of ‘y’ without respect to ‘x’
Find dy/dx. It’s helpful to write dy/dx=y’
Step 1: d/dx of both sides of the equation
�
�
�� + � ! =
2+�
��
��
Step 2: Perform differentiation while assuming ‘y’ is a function of ‘x’
!
!
�� + � ! =
2+�
!"
�
�
�� +
�! = �!
��
��
��
�+ �
+ 2� = �′
��
‘x’ and ‘y’ is a product of g(x)-x&y
y=n(x)
�
(��) = � + ��′
��
!"
Step 3: Solve for y’
∴ � + �� ! + 2� = � !
�� ! − � ! = −2� − �
� ! � − 1 = −2� − �
−2� − � 2� + � ��
=
=
→ ��� ������� ����������
∴ �! =
�−1
1−�
��
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NOTE: We still do not know y=f(x) yet we have y’ depending on both ‘x’ and ‘y’
Example #2: Assume the equation x3+y3=6xy defines ‘y’ implicitly as a function of ‘x’.
Thus, y=f(x) for some function ‘f’
a) Find y’ (i.e. dy/dx)
b) Find the point(s) on the curve given by the equation where the tangent line is
horizontal
Solution A:
� ! + � ! ! = 6�� !
� ! ! + � ! ! = 6 �� !
3� ! + 3� ! � ! = 6 � + �� !
� ! + � ! � ! = 2� + 2�� !
� ! � ! − 2�� ! = 2� − � !
� ! � ! − 2� = 2� − � !
2� − � !
∴ �! = !
� − 2�
Solution B:
We solve for y’=0 for ‘x’ and ‘y’
2� − � ! 0
= → 2� = � !
� ! − 2� 1
!
Differentiate both sides
using (‘) with respect to
‘x’
Reminder:
y=f(x) for some function that
we do not know
y’ is a function of
both ‘x’ and ‘y’
y’|y=0,x=0 = undefined
(x,y) = (0,0) makes y’ undefined
!
Answer:
16, 2!
^ Sub in equation and solve for ‘x’
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Logarithmic Differentiation
Logarithmic differentiation uses properties of logarithms to simplify first before
differentiation.
Use mostly with complicated products, quotients, or “generalized exponential functions”
Generalized exponential functions  � = �(�)!(!) where a(x) and b(x) are functions
Example 1: Let � = � ! find y’ (=dy/dx)
Logarithm Properties:
ln(�! ) = � ∙ ln(�)
ln(��) = ln(�) + ln(�)
�
ln ! ! = ln(�) − ln(�)
�
Solution: ln � = ln � !
!
ln � = � ! ∙ ln �
!
!"
!!
!
∙ ln � =
!
= �
!
�! = �
!
!
!
!
!"
!
� ! ∙ ln �
!
!
∙ ln � + � ∙
ln �
�
ln(�)
��
��
�
= ! ln(�)! ∙
��
��
1 !
= ∙�
�
= �!
�
!
!
1
+
2 �
�
1
ln
�
∴ �! = � !
+
2 �
�
NOTE: We do NOT do the following:
� ! = � ∙ � !!!  This derivative rule does not apply here
Example 2: Find where � =
! !/! ! ! !!
!!!! !
Solution: Using Logarithm properties 
3
1
ln � + ln � ! + 1 − 5 ln 3� + 2
4
2
3
2�
�!
3
=
+ !
−5
�
4� 2� + 1
3� + 2
3
�
15
=
+
−
4� � ! + 1 3� + 2
ln � =
!
�! �! + 1 3
�
15
∴� =
+ !
−
!
3� + 2
4� � + 1 3� + 2
!
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Example 3: � � = � � !(!) where a(t) and b(t) are functions
Solve for u’(t) (=du/dt)
Solution:
ln � � = ln � � ! !
ln � � = � � ∙ ln � �
�
�
ln � � =
� � ∙ ln � �
��
��
�! �
�! �
= � ! � ∙ ln � � + � � ∙
� �
� �
∴ � ! � = � � !(!) � ! � ∙ ln � �
+� � ∙
�! �
� �
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Newton’s Method, Derivation of Newton Method and Higher Order
Derivatives
Newton’s Method:
Newton’s Method is a technique for approximating a root or a zero of a function (i.e. ‘c’)
The real number ‘c’ is a root/zero of
our function.
- Means that f( c)= 0
Finding ‘c’ is often important, but often impossible exactly
Newton Method Formula:
�!!! = �! −
� �!
� ! �!
Given a function y=g(x), a root (or zero) of ‘g’ is a real number ‘c’ such that g( c)=0
There is a small issue of verifying that a function actually does have a root.
- We use the Intermediate Value Theorem (IVT)
Assume we have a continuous function y=h(x) and ‘a’ & ‘b’ are real numbers, such that
h(a) and h(b) have opposite signs (+/-). The ‘h’ has a root ‘c’ lying strictly between ‘a’ and
‘b’
- Better to try to get ‘a’ and ‘b’ “close together”
We use IVT to get a “start values ‘x’”, for Newton Method as follows:
- Assume that h(a) and h(b) (two numbers) have opposite signs
- Let x1 (start) = ‘a’ or ‘b’, depending which one, h(a) or h(b), is closer to 0
o If h(a) is closer to 0, use ‘a’ and vice versa
o I.e.
Therefore, h(a) is closer to zero than h(b)
Therefore, we can let the starting value be
X1=a (x1=0=start)
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Example 1: Let � � = 2� ! + � − 1, where ‘f’ is a polynomial and a function
Using IVT to verify ‘f’ has a root
f(0)= -1<0 , f(1)= 2>0
f(0)= negative output
f(1) = positive output
By IVT, ‘f’ has a root ‘c’ is between (0,1)
�!!! = �! −
! !!
! ! !!
2�!! + �! − 1
= �! −
6�!! + 1
4�!! + 1
�!!! = !
6�! + 1
�(�) = 2� ! + � − 1
�(�� ) = 2�3� + �� − 1
� !(�) = 6� ! + 1
� ! (�! ) = 6�!! + 1
Start: x1= 0
�! =
4 0 3 +1
=1
6 0 2 +1
4 1 !+1 5
= ≈ 0.71428571 …
6 1 !+1 7
5 !
4
+1
7
≈ 0.60517
�! =
!
5
6
+1
7
�! ≈ 0.59002
�! =
Get’s closer
and closer to
‘c’
Therefore, we can say that � ≈ 0.59002
Example 2: Sometimes in application of Newton Method is needed, but we have to
“work to get a function”.
Consider two functions: � = � � = � ! ��� � = ℎ � = −� + 3 (can represent the supply
and demand functions)
(r, g( r)) = (r, h( r))
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Verify that by IVT that the graphs of ‘g’ and ‘h’ cross
We seek to “solve” the equation g(x)=h(x) for ‘x’
Consider the difference function: f(x)=g(x) – h(x) [Apply IVT, Newton Method to ‘f’]
Then, (r is a root of) if and only if f( r)=0 and g( r)=h( r)
Derivation of Newton Method Formula
How and Why the Newton Method “works”
Let y=f(x) have a root ‘c’ and therefore f( c)=0
Suppose we have a value �!
Geometric Explanation of how to get �!!!
Point: (�! , � (�! ))
Tangent Line to: � = �(�) �� (�! , � (�! ))
Intercept: �!!!
Slope is  � ! �!
Tangent line is  � − � �! = � �! � − �!
! !
Intercept at 6+y=0  − ! ! = � ! �! � − �!
! !!
� �!
� ! �!
Therefore, we let this (^) be �!!!
∴ � = �! −
Higher Order Derivatives
(i.e. 2nd and 3rd derivatives)
Let y=f(x) be a MATA32 function
F’(x) (or y’ or dy/dx) is the first derivative
NOTE: that f’(x) is also a function, so it can be derived
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We have “higher – order” derivatives as follows:
2nd derivatives  f’’(x), y’’ or d2y/dx2 or f(2)(x), y(2)
3rd derivatives  put 3’s everywhere, where there is a 2 indicating the derivative
number
And so on…
Nth derivative  f(n)(x) or y(n) or d(n)y/dx(n)
Example 1: let � !!! = �, find f(100)(0)
� � � !!!
� ! � = −2� !!!
� !! � = −2 ! � !!!
� ! � = −2 ! � !!!
The pattern suggests that f(100)(0)= -2100
** When the derivative number (n) is too big, look for a pattern
Example 2: Find all values of the constant λ for which the function � = � !" satisfies the
equation 2� !! = 7� ! + 3� = 0
1. Find 1st and 2nd derivative
� ! = �� !" ��� � !! = �! � !"
2. Sub in y, y’ and y’’ into the equation and see where this takes us
2 �! � !" = 7 �� !" + 3 � !" = 0
� !" 2�! + 7� + 3 = 0
Forces 2�! + 7� + 3 = 0
Therefore, we must have that 2�! + 7� + 3 = 0
We get � = −3 ��� � = −
!
!
!
Our functions are � = � !!! ��� � = � !!!
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Graph Sketching and Optimization
We have two main ideas:
1. Use the tools of calculus to provide a good graph of a given function y=f(x)
2. Apply the ideas of calculus to solving real-world optimization problems (max/min)
NOTE: The 5 number ‘ci’ give 6 open-intervals:
(−∞, �1), (�1, �2), (�3, �4), (�4, �5) ��� (�5, ∞)
ℎ! �1 = ℎ! �2 = ℎ! �4 = 0 �ℎ����������� �� �ℎ� ℎ��������� ��������
ℎ! �3 = ℎ! �5 = ���������
We call the numbers ‘ci’, i=1-5, critical numbers or values
Critical Numbers
Definition  y=f(x) be a MATA32 function. A real number ‘c’ is a critical number of ‘f’ iff
‘c’ is in the domain of ‘f’ and either:
a) � ! � = 0 ��
b) � ! � = ���������
Thus, to find all critical numbers of ‘f’
1. Find the derivative
2. Take derivative and solve for ‘x’ when f’(x)=0
a. Factor equation
3. Determine if the derivative is undefined
Regarding critical numbers and max/min:
If y=f(x) has a local max/min at a point ‘a’ is a real number then ‘a’ is a critical number of
‘f’
Important Note: The converse statement is false (switching the order)
Intervals of monotonicity:
• Monotonicity  same direction
• (open) intervals
• Increasing part = f(x) is increasing on an open interval
o (a,b) if and only if f’(x)>0 for all ‘x’ (a,b)
• Decreasing part = f(x) is decreasing on an open interval
o (a,b) if and only if f’(x)<0 for all ‘x’ (a,b)
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−∞, �1 �2, �3 = � � ����������
�1, �2 �3, �4 �4, �5 ��� �5, ∞ = � � ����������
When first derivative changes signs tells us we have a max/min
Good Luck!!
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University of Toronto
Scarborough
MATA32H3
Calculus for Management I
Fall 2017
Final Exam
Exam Guide
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CONTENTS
Exam Guide Covers the Following Topics:
• Deri ati es a d Co a it
• A solute E tre a o a Closed a d Bou ded I ter al
• Horizo tal &Verti al As
ptotes
• Applied Opti izatio Pro le s
•
I tegratio
• I tegratio
Parts
• The Defi ite I tegral
• The Fu da e tal Theore
of Cal ulus
• Properties of the Defi ite I tegral
• Appli atio s of I tegral to Area
• Areas Bou ded Bet ee Cur es
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Derivatives and Concavity
1st Derivative Test
Idea: given y=f(x), the 1st derivative is used to determine whether a critical number of a
function is a local max, min, or neither
*** Critical number is 0 or undefined
Let y=f(x) be a function and assume ‘c’ is a critical number of ‘y’. Assume (I) is an open
interval
1. Max: if f’(x) changes signs from + to -, as ‘x’ moves through I, the ‘f’ has a local
max at ‘c’
2. Min: If f’(x) changes sign from + to -, as ‘x’ moves through I, the ‘f’ has a local
min at ‘c’
3. Neither: if f’(x) does not change sign for ‘x’ to the left side of ‘c’ in I, then ‘f’ has
no local max/min at ‘c’
2nd Derivative
Gives information about the bending upwards/downwards of a function
(i.e. concavity (point of inflection)
Concavity Test
Suppose we have a function y=f(x) so the f’’(x) is defined on a given open interval (a,b)
I
1. if f’’(x) > 0 for all x of I, the f(x) is concave up on I
2. If f’’(x) <0 for all x of I, then f(x) is concave down on I
We obtain intervals I, as in the concavity test
1. Find f’’(x)
2. Solve f’(x)=0
3. See if f’’(x) is ever undefined
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Definition: we have/say (c,f( c)) is a point of infection of y=f(x) where concavity changes
2nd Derivative test
Assume y=f(x) is a function such that f’( c)=0 (thus, ‘c’ is a critical number)
1. Min: f’’( c) > 0, then ‘f’ has a local min at ‘c’
2. Max: f’’( c)<0, then ‘f’ has a local max at ‘c’
3. Not helpful: if f’’( c) = 0 or undefined, then the 2nd derivative is not used to give
any info about extrema at ‘c’ max at ‘c’
NOTE: 1st and 2nd derivative can often give information about extrema (max/min)
1st Derivative
• Need only f’(x)
• Need to analyze sign of f’(x)
• Can always work
• Can always be used
2nd Derivative
• Must get f’’(x)
• Only evaluate f’’( c)
• Might not be helpful
• Applicability is limited
NOTE: A function may not have any absolute extrema (even if it has a local)
We have to study a special situation where absolute extrema ‘s are guaranteed to exist
and we can find them
13.2  absolute extrema on a closed interval [a,b]
The main theory tool/concept is:
Extreme Value Theorem (EVT)
Assume y=f(x) is a MATA32 function that is continuous for xe[a,b] (‘a’ closed interval)
then ‘f’ has an absolute max and absolute min on [a,b]
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Absolute Extrema on a Closed and Bounded Interval
!
!
Example 1: Discuss the absolute extrema of the function � � = � ! + �ℎ��� � ∈
!
−9, 1
Solution: We are given a closed and bounded interval [-9, 1]
Our function ‘g’ is continuous on [-9, 1] as follows:
!
� � = �! � + �! � where �! � = � ! =
also continuous
!
�
!
is continuous on [-9, 1] and �! � =
!
!
is
Therefore by EVT, ‘g’ has both an absolute max and min on [-9, 1]
Now we find the absolute extrema using the closed interval method (CIM)
Step #1: Evaluate function at end points
� −9 ≈ 1.327
4
� 1 = ≈ 1.333
3
Step #2: Find all critical points
!
!
!
� ! � = ∙ � !! +
!
!
2 1 1
= ∙ !+
3 ! 3
�
g’(x) is undefined when x=0,
since � ∈ [−9, 1] → ������ ��
Consider g’(x)=0
2 1 1
∙
+ =0
3 !! 3
�
2
! = −1
�!
!
� ! = −2
� = −8
∴ � = −8 ∈ −9, 1
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Step #3: Evaluate the function at critical points that are in the interval [-9, 1]
� 0 =0
!
−8 4
� −8 = −8 ! +
=
3
3
Step #4: Absolute max. is the largest function values from steps 1 and 3
Therefore, absolute max is 4/3 at x=-8 and x=1
Absolute min. is 0 at x=0
*** 1st and 2nd derivative test is not necessary for closed intervals***
- Not interested in local max/min
NOTE: A closed and bounded interval may have to be found from a particular
question/application
!
Example 2: Let � = � � = �� !! > 0
Use “basic properties of ‘f’ and the derivative to produce a good graph of y=f(x)
‘f’ is positive when x=positive
‘f’ is negative when x=negative
�
�(�) = ! ! → 0 �� � → ±∞
�
!
� ! � = � !! 1 − 2� !
!
� ! � = � �� ��� ���� �� � !! 1 − 2� ! = 0
!
!
Since � !! > 0 therefore, 1 − 2� ! = 0 �� ��� ���� �� � = ±
!
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Horizontal &Vertical Asymptotes and Applied Optimization Problems
Horizontal Asymptotes:
Y=k is a horizontal asymptote of y=f(x) and y=g(x) because
lim �(�) = � ��� lim �(�) = �
!→!
!→±!
Definition of horizontal asymptote (HA): let y=f(x) be a MATA32 Function
For a real number ‘k’ the (horizontal) line y=k is a horizontal asymptote of ‘f’ iff
lim!→! � (�) = �, lim!→!! � (�) = � �� ���ℎ
Example 1: � = � � =
!! ! !!
!! ! !!!!
 Rational function
Horizontal Asymptotes:
1
�! 4 + !
4� ! + 1
�
lim � � = lim
= lim
=2
1 1
!→!!
!→!! 2� ! − � + 1
!→!! !
� 2− + !
� �
lim �(�) = 2
!→!
Vertical Asymptotes:
For vertical asymptotes (VA): we check to see if a given function approaches a vertical
line of the form x=a
(a=constant) where xa+ or xa- or both
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Two vertical asymptotes: x=a1 and x=a2
Function is y=h(x)
lim!→!!! ℎ(�) = ∞, ���. This is sufficient to say x= a1 is a vertical asymptote of y=h(x)
We also have lim!→!!! ℎ(�) = −∞, ���, enough to say x=a1 is a vertical asymptote of
y=h(x)
For the line x=a2, lim!→!!! ℎ(�) = −∞, ���. This is enough to say x=a2 is a VA of y=h(x)
lim!→!!! ℎ(�) = ∞, ���. This is enough to say x=a2 is a VA of y=h(x)
NOTE: Read carefully about vertical asymptotes for rational functions on page 601
!"#
Example 1: Let � = � � =
; ��� ���, � > 0 & � (���, � ≠ 0)
!
The domain of ‘f’ (only where the bottom and top are both defined) is � = ��� � > 0
Asymptotes:
(I) VA, the only possibility is perhaps
���
lim � � = lim
= −∞, ��� ������� lim ln � = −∞, ���
!→!!
!→!! �
!→!!
Therefore, ln(x) dominates over ‘x’. Therefore, x=0 is a vertical asymptote of ‘f’
(II) Horizontal Asymptote (HA)
���
lim � � = lim [ ]
!→! �
!→!
=0
Since ln(x) grows much slower than ‘x’
Can’t take the limit of the top and bottom
because both limits do not equal numbers
and do not exist.
Therefore, we get y=0 as a horizontal asymptote
because y=lnx grows much slower than the function
y=x.
!"#
= 0, � > 0
F(x)=0 if and only if
!
F x = 0 if and only if lnx = 0
F x = 0 if and only if x = 1
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Graph crosses x-axis at the point (1,0)
! !
!
!!!"#
!
!!
=
: �
1 − lnx
=
�!
∴ �� ��� �ℎ�� � ! � = 0 �� ��� ���� �� � = � ln � = 1
F(x) is increasing when ‘x’ is between (0,e) because f’(x)>0 when 0<x<e
F(x) is decreasing when ‘x’ is between (e,∞) because f’(x)<0 when x>e
Applied Optimization Problems
Idea: Use calculus (and pre-calculus) to formulate real-world problems and solve/find
the optimal solution
Example 1: (minimum material cost)
A rectangular storage container has a volume of 10m3 and no top. The length of the
base is 2 times its width. Base unit cost is $10/m2 and side unit cost of $6/m2. Find the
dimensions (w,l,h) that minimizes the total material cost.
Solution:
Let ‘x’ represent the width x>0
Let ‘y’ represent the height y>0
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10m3 volume  2� ! � = 10 → � ! � = 5 � "����������"
We must next get the “total cost function” ‘C’
C= total cost
C = (base cost) + (front & back cost) + (left & right cost)
C= (2x2)(10) + (2xy)(6)(2) + (xy)(6)(2)
C = 20x2 + 36xy
�
Use x2y=5 to write � = � and sub into ‘C’
�
Therefore, C is a function of ‘x’
!"#
Therefore, � = 20� ! +
,� > 0
!
Critical numbers are needed to minimize ‘C’
180
� ! = 40� − ! , � > 0
�
!"#
Solve � ! = 0 <=> �� ��� ���� �� 40� − ! = 0
!
180
<=> 40� = !
�
180
9
<=> � ! =
=
40
2
!
∴�=
9
2
∴ �ℎ�� �� ��� ���� �������� ����� �ℎ��� � > 0
2nd Derivative Test is easiest to test for minimum
� !! = 40 +
! 9
360
>
0
�ℎ��
�
=
�!
2
Local min. when � =
�
�
�
Therefore, also it is also the absolute min because there is only one critical point within
the domain.
!
∴ � = 2� = 2� =
9
2
5
∴ℎ=�=
!
∴�=
!
9
2
9
2
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Integration
Calculus has two main parts
1. Derivatives
2. Integrals
Anti derivatives (AD) also known as indefinite integrals
Definition of Anti Derivative: Let y=f (x) be a MATA32 function
An Anti derivative of f (x) is a function F’ (x) with the property F’ (x)=f (x) or [d/dx F (x)=f
(x)]
To find an AD of ‘f’ is called anti differentiation
There are an infinite number of different possibilities for an anti derivative of a
function
Example 1: Let f (x)=x2
�!
�1 � =
+ 32
3
!
�
�2 � =
− 2
3
�1! � = � !
�2! � = � !
We see that F1 (x) and F2 (x) are anti derivatives of f (x)=x2. It’s easy to see that if c is a
!!
real number and is a constant and � � = + �, the F’ (x)=x2=f (x), so is an Anti
!
derivative of f (x)=x2. Are there any other kinds of functions that are also AD’s of f (x)=x2?
!!
Say we use �1 � = + �
!
�1! � = � ! + 1 ≠ � ! ≠ � �
Therefore, G1 (x) is not an anti derivative of f (x)=x2
Therefore, the answer is NO!
We write
� ! �� =
!!
!
+ �, � �� � ��������
→ �������� ����
� ! → ���������
�� → ������� �� �ℎ�� �ℎ� ���������� �� ���� ���ℎ �������� �� �
� → �������� �� �����������
This means that the most general AD of f (x)=x2 is x3/3+c
Example 2: Let u(t)=ex
�1 � = � ! + 5
�2 � = � ! − 2
�1! � = � �
�2! � = �(�)
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We often use the following FACT about general ADs:
• Let y=f (x) be a MATA32 function and suppose F1 (x) and F2 (x) be AD’s of f(x)
• Then there is some constant k is a real number such that F1 (x) - F2 (x)=k
F1’ (x) = f (x) and F2’ (x)= f (x)
Therefore, F1’ (x)= F2’ (x)  at each value of ‘x’, the slope of F1’ (x) = slope of F2’ (x)
Let y=f (x) be a MATA32 function
1.
2.
!
!"
[ � � ��] = � �
!
!"
� �
�� = � � + �, �ℎ��� � �� �� �������� ��������
These show how differentiation and AD are almost inverse processes.
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Integration (Continued)
We consider techniques for solving the following main problem: Given a MATA32
function y=f(x), find � � ��
(i.e. the anti derivative of y=f(x))
** This means, find the “most general” function f(x) such that F’(x)=f(x)
There are 4 methods/techniques:
1. Basic Integration Method
2. Simplify/manipulate f(x) first
3. The Substitution Rule
• Chain rule in reverse
4. Integration by parts
The following table of basic integration formulas is very useful and easy (Table 14.1
page 633)
1.
��� = �� + �
k  is a constant
C  Constant of integration
��!�
�� �� =
+�
�!�
a constant, � ≠ −1
C  Constant of integration
2.
�
3.
�� = �� � + �
�≠0
C  Constant of integration
�
4.
�� �� = �� + �
5.
�� � �� = � [ � � ��]
� � → ����32 ��������
6.
� � ± � � �� =
� � ��
� � �� ±
WARNING: If you have 2 functions and they are multiplied
� � and same with division
Example 1:
=
=5
=5
2��� +
� ! �� + 2
3
! !
≠
� � � �
≠
� � ∙
! ! !"
! ! !"
5� ! + 2� + 32 ��
5� ! �� +
�!
! !
32��
��� + 32� (����� ����������� ������� ����)
�!
+2
2
+ 32�
5 !
� + � ! + 32� + � (��� �������� �� �ℎ� ���)
3
^ Most general anti derivative of f(x)
=
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!
Example 2:
5� −
��
!!
1
5��� −
��
=
7�
!
1 1
= 5 � ! �� −
��
7 �
2 ! 1
= 5( � ! − ln � + �
3
7
2 5 ! 1
� ! − ln � + �
=
3
7
!
√5� = √5√� = √5� !
1 1
1
= ∙
7� 7 �
!!!!
Example 3:
��
!
5� 3
+ ��
=
�
�
3
=
5 + ��
�
= 5� + 3 ln � + �
Example 4:
! !! !!!! ! !! !
! ! !!
��
�! + � !
=
�! + �
��
� !! + 2� ! � + � ! = (� ! + �)!
� ! + � ��
�
= �! + + �
2
=
The Substitution Technique
Idea: Given a “hard integral”. Often a strategic substitution can transform it into much
easier one
The “strategic substitution” is based on the chain rule
Example 1:
� ! + 3 ! 2���
� ! ��
��� ��� �ℎ�� ! �! ��
�!
+�
8
�! + 3 !
=
+�
8
��� � = � ! + 3
=
=
��
= 2� → �� = 2���
��
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Example 2:
�!1
=
��
2
!
=
� ! ��
� ! + 3 ! ���
��� � = � ! + 3 → �ℎ� � − ������������
1
�� = 2��� →
= ���
2��
!
1 �!
∙
+�
2 8
1 !
� +3 !+�
=
16
=
Example 3:
� ! + 3 ! 5���
��� � = � ! + 3 �����
!"
Example 4:
→
!!!!
1
=
��
3� + 2
! !
∙ ��
=
! !
1 1
=
��
3 �
1
= ln � + �
3
1
= ln 3� + 2 + �
3
!
!!!!
�� ≠ ln 3� + 2 + �
��� � = 3� + 2
�� = 3��
1
�� = ��
3
Example 5: Let � = � ��������, � ≠ 0
� ∝! ��
=
� ! �� = � ! + �
=
1
� ∙ ��
∝
!
1
=
∝
The number on the denominator must be
the same as the value of the x in the dx
��� � =∝ �
�� =∝ ��
1
∴ �� = ��
∝
�������:
! � !!! ��
1
= − � !!! + �
5
1
1
� �� = ∙ � ! + � = ∙ � ∝! + �
∝
∝
!
The Substitution Formula in General:
!
� � � � � �� =
� � ��
(“Given” (i.e. we have exactly or can get this form) = “easier form”
� = �(�)
��
= � ! (�)
��
�� = � ! (�)��
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Harder Examples:
!"
Example 1:
=
!"#!
1 1
=
∙
��
� ���
1 1
=
∙ ��
��� �
1
=
��
�
= ln |�| + �
= ln ln � + �
Example 2:
=
!
!
!
��
!"#!
��� � = ln(�) = �(�)
1
�� = ∙ �� = � ! (�)��
�
1
�(�) =
�
������ ��:
!!√1 + √2!� ! ��
= ! �! + �!
1 + � ! � ! ��
!
� � − 1 ! ∙ ��
!
!
1
=
� ! � ! − 2� + 1 ��
2
!
!
!
1
=
� ! − 2� ! + � ! ��
2
1 2 ! 4 ! 2 !
=
∙ �! − ∙ �! + ∙ �! + �
2 7
5
3
!
!
!
1
2
1
= 1 + �! ! − 1 + �! ! + 1 + �! ! + �
7
5
3
� = 1 + �! → �! = � − 1
�� = 2���
1
�� = ���
2
� ! = � − 1 → � ! = (� ! )! = (� − 1)!
Integration By Parts (IBP)
General Formula has the Form:
��� = �� =
“Given
exact or we
can get it
��� → �� ���� �ℎ�� �� �� ������
^ We get this
� = �(�)
�� = � ! (�)��
�� = � ! (�)��
(1)
(2)
� = �(�)
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Integration by Parts (IBP)
Idea: Integration by Parts turns a given Anti Derivative into one that is hopefully easier
than the given one
- Integration by Parts is based on the product rule
IBP Formula:
� � �! � �� = � � � � −
� � � ! � ��
Here is an easier way to remember Integration by Parts:
�� = � ! � ��
�=� �
�� = � ! � ��
��� = �� −
�=� � =
�! � ��
���
** The constant ‘c’ will cancel out, so no need to include it now**
Remarks: The skill with using Integration by Parts comes from:
1. Making sure the new integral ��� is no harder than the given on
2. Selecting ‘dv’ so that ‘v’ can be found
Example 1: � !! ��� (Very standard IBP problem)
= �� !! ��
= −�� !! �� — � !! 1��
�=�
= −�� !! + � !! ��
�� = 1��
= −�� !! − � !! + �
I will try the following:
� !! ��
� = � !!
�� = −� !!
���
�� = � !! ��
� = ! � !! �� = −� !!
�� = ���
�!
�=
2
And you see that Integration by parts “fails” because I chose the wrong function for u &
dv
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Example 2:
=
�
1
ln(�) =
�
��
ln x dx
��� ! ln(�) �� =
ln � − 1��
1
�
!
= ��� � −
� ∙ ∙ ��
� = ln(�)
!
�� = ∙ ��
!
= ��� � − 1��
= ��� � − � + �
WARNING: To find
� � �� it may require a couple (or several) techniques
� ! ��
Example 3:
=2
!
��� � = √� → � ! = �
∴ 2��� = ��
�=�
�� = � !
�� = 1��
� = �!
� ! ���
= 2{�� ! −
���ℎ ���, �� ���
�� = 1��
� = ∫ 1�� = �
�^���}
= 2 �� ! − � ! + �
= 2 �� ! − � ! + �
!
��
�
!� ! !
=
�� ��
1
1
��
=
=
�� 2√� 2�
Example from 14.3: Integration with a given condition
Given that � ! � = 5� ! + 3 ����� � 1 = 2
Find the function f(x)
Solution:
� � =
� ! � �� =
5� ! + 3�� =
5 !
� + 3� + �
3
Solve for f (1)
2=� 1 =
�=−
5 !
� + 3� + �
3
8
3
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The Definite Integral and the Fundamental Theorem of Calculus
The Definite Integral
We’ll see that the definition of the definite integral is based on these simple ideas
Begin with a given interval [a,b] (a<b) and a continuous MATA32 function y= f(x) where
� ∈ [�, �]
Let �� = 1,2,3,4 … = ������� �������
(� − �)
�
=Sub-interval width
��� ∆� =
As ‘n’ gets larger and larger ∆� gets
smaller and smaller
lim ∆� = 0
!→!
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Right endpoint of the kth sub-interval is �! = � + � ∆� = �, �ℎ��� � = 0
�! = � + � ∆� = �, �ℎ��� � = �
In the nth Riemann Sum is defined as follows:
�� = � �1 ∆� + � �2 ∆� + � �3 ∆� + ⋯ + � �� ∆�
In the Summation Notation (Sigma Notation)
!
� � ∆�
�� =
!!!
** ‘k’ must start at 1
Definition: The definite integral of y=f(x) from ‘a’ to ‘b’ is:
!
!
� � �� = lim
� �! ∆�
!→!
!
!!!
!
� � �� = # is a special real number that is the limit of a certain kind of sum
!
Special, Important Case:
If � � ≥ 0 when � ∈ �, � then,
!
� � �� = �ℎ� ���� �� �ℎ� ������ ����� �ℎ� �����, ����� � − ���� �ℎ��� � ≤ � ≤ �
!
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!
In order to use the definition of ! � � �� , we need to be careful and make use of some
other formulas
!
� + � + � + � + ⋯ + � = �� =
�
!!!
!
�=
1+2+3+4+⋯+ � −1 +� =
!!!
� �+1
2
!""
�=
!!!
100 101
= 50 101 = 5050
2
!
!
!
!
!
1 +2 +3 +4 +⋯+ � −1
!
�! =
!
+� =
� � + 1 2� + 1
6
!!!
Example 1: let � ≥ 0 ��� ��� � = � � = � !
Use the definition to find:
!
� ! ��
!
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Solution:
!
!
� � �� = lim
� �! ∆�
!→!
!!!
!
!
= lim
!→!
� ! ! �
�
�
�
!!!
= lim!→!
�
�
! ! !
!
!
!
!
!
!!! �
� � + 1 2� + 1
)
!→!
6
� � + 1 2� + 1
= lim � ! ∙
∙
!→!
�
6�
�
1
!
=� ∙
3
�!
=
3
= lim [
∙(
!
� ! �� =
∴
�!
= ���� �� ������
3
!
�−0 �
=
�
�
�! = � + �(∆�)
�
= 0+�! !
�
�
= ! ! (�)
�
!
�
!
�(�! ) = (�! ) = !! ! (�)!
�
� !
= ! ! (�)!
�
∆� =
The Following is TRUE:
If every definite integral can only be found by the definition, we would not go very far!
To help with this we have…
The Fundamental theorem of calculus (FTC)
The Fundamental Theorem of Calculus (FTC)
Let y=f(x) be a MATA32 continuous function where ��[�, �] for some a<b
Then,
!
� � �� = � � − � � , �ℎ��� � � �� ��� ���� ���������� �� �(�)
!
=� � |
�=�
�=�
!
Example 1: ! � � ��
�! � = �
|
=
3 �=0
� ! 0!
−
=
3
3
!
�
=
3
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!
Example 2: ! � !! �� =
! !!
!
| !!!
!!!
�! �!
−
3
3
!
�
1
=
−
3 3
1
= �! − 1
3
=
�(�) = � √!
� = 1; � = 1 → � = √1 = 1
� = 4; � = 4 → � = √4 = 2
�. ������������
!
Example 3: ! � ! ��
!
� ! ��
=2
!
!
=2
��
!
�=2
−
�=1
!
� ! ��
!
= 2 2� ! − � − � !
= 2 2� ! − � − � ! − �
�=2
�=1
��� � = √� = � !
1 !
1
�� = � !! =
2
2 √�
∴ 2√��� = ��
�. ������ ��������� ��� ������
� = 1 → � = √� = √1 = 1
� = 4 → � = √� = √4 = 2
�. ����������� �� �����
� = � �� = � ! ��
�� = �� � = ! � ! �� = � !
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Definite Integral, FTC Continued and Properties of the Definite Integral
Recall the definition of the definite integral:
Let y=f(x) be a continuous function on an interval [a,b]
The definite integral of ‘f’ over [a,b] is
!
!
� � �� = lim
� �! ∆�
!→!
!!!
!
� − �� − �
− ��������
�����ℎ
= ���
− ��������
�����ℎ
∆� = = ���
� �
��!!==��++�(∆�)
� ∆�
∴ �ℎ� �������� �������� �� � ���� ������ ��� �� �ℎ� ����� �� � ������� ���� �� ��� �������� ��������
= � ������
We have a related concept called Anti Derivative (AD)
A function ‘F’ is an AD of ‘f’ iff F’ (x)=f (x)
We write the most general AD of ‘f’ as
� � ��  Stands for some function
Definite and Indefinite integral are related through FTC
!
Usually, finding ! � � �� via the definition is “hard”
Fortunately, we have FTC
Fundamental Theorem of Calculus (FTC)
!
!!!
= � � − �(�)
If ‘F’ is an AD of ‘f’, then ! � � �� = � � | !!!
In using FTC:
1. Find � � �� = � �
!
2. � � − � � = ! � � ��
Properties of the Definite Integral
‘f’ and ‘g’ are continuous functions on [a,b]
1.
!
� � �� = 0
!
!
!
� � �� = −
!
� � ��
!
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2.
!
� �
!
!
!
± � � �� = ! � � �� ± ! � � ��
!
!
!
WARNING: ! � � ± � � �� ≠ ! � � �� ∙ ! � � �� ��� ���� ���ℎ ���������
!
� �
� �
�� ≠
!
!
� � ��
!
!
� � ��
!
3. If k= a constant, then
!
!
� � ��
�� � �� = �
!
!
4. The “Partition” Property
Let ��[�, �] then
!
!
� � �� =
!
!
� � �� +
!
� � ��
!
ℎ��� � = � � �� ��� ���������� �������� �� �, � ��� �� ��������
5. The “Area” property
Let R be the region in the xy-plane that has boundaries: the lines x=a, x=b, the xaxis and the curve y=f(x)
Area of R=
�
� � ��
�
�
|� � |�� = � + � − (�)
�
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FTC, Applications of Integral to Area, Areas Bounded Between Curves
FTC (Fundamental Theorem of Calculus)
If y=f(x) is a continuous on [a,b] and F is an Anti Derivative (AD) of ‘f’, then
!
� � �� = � � − � � ,
�! � = � �
!
A nice application of this is when we have “marginal _____”
See page 664 (top)
By FTC,
!
� ! � �� = � � − �(�)
!
** Sometimes called the ‘Net Change Theorem’
NOTE: f(x) is an AD of f’(x)
Example 1: (See example 5, page 664)
Suppose a marginal revenue function is
�! � =
� !
+
�
1
. �ℎ��� � > 0 ��� �� � ��������
�
Find the change in revenue from x=10 to x=12 (in some appropriate units)
Solution: We would like r(12)=r(10)
By FTC (NCT)
��
� ! � ��
� 12 − � 10 =
��
��
=
��
� � 1
+
�� =
�
�
!"
� !
( )�� +
�
!"
��
1
��
�
��
� = 12
� = 12
= 2� !
+ ln �
� =1−
� = 10
!"
!"
=2 �
−�
+ ln 12 − ln 10
Can leave this (^) as is
**When I integrate a derivative, I get a change in the function
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Example 2: (Page 665 #53)
NOTE: Answer in the back of the book is wrong
Let ‘x’ be income
Let ‘N’ be the number of people (in some kind of sample) that have an annual income of
≥�
∴ � = � � ��� ���� �������� �� � � > 0
We are given the following:
!"
= −�� !! , �ℎ��� �&� ��� ��������� ��� �&� > 0
�
=− !
�
��
�� ��� �ℎ��
<0
��
!"
Let 0<a<b where a&b are constants. Write an expression that gives the total number of
people with annual income in the interval [a,b]
Thus, we need to calculate N(a)-N(b)
(This difference gives the number of
people with incomes ≥ � ��� < �)
!
!
��
�� =
��
� � −� � =
!
−�� !! ��
!
!
� !! ��
= −�
Signs change
!
!!
=�
! !!
� �� =
!
� ! ! �� , �� � = 1
!
� ! � !! �� , �� � ≠ 1
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If,
!
�
1. � ln
, �=1→
�
!
�=�
�
1
= ln � − ln � = ln ( )
�� = ln � |
�=�
�
�
!
2. �
!
�
!!
�� ,
�≠1→
!
�
!!
!
� !!!!
� !!!!
�!!!!
�� =
=�
−
−� + 1 −� + 1
�+1
Applications of Integral to Area
Example 1: (See page 673 Ex. 1)
Find the area of the region bounded by x-axis, lines x=1 and x=3, and the curve
� = � ! − 6� = �(�)
Solution: We do need a “decent sketch” but does not need to be exact (only need to know
where the graph is positive and negative)
Solve f(x)=0
� � = � �! − 6 = 0
∴ � = 0 ��� � = ± 6
lim � � = ∞
!→!
����ℎ ������� ������������ ��������
lim � � = −∞
!→!!
����ℎ ������� ������������ ��������
� � = � �1 + � �2 + � �3
!
� � ��
∗� � =
!
!
!
[−� � ]�� +
� � �� +
� � =
!!
!
!!
� � �� = 14
!
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Areas Bounded Between 2 (or more) Curves
Case #1: Area is best understood using “vertical strips” or “vertical elements”
Case #2: Area is best understood using “horizontal strips” or “horizontal elements”
Case #1: Given on interval [a,b]
Width = ∆�
Height = � � − � �
!
(� �! − �(�! )(∆�)
!!!
!
!
� � − � � �� =
���� �� � =
!
��� − ������ ��
!
Case #2:
!
���� �� � =
� � − � � ��
!
!
���ℎ� �������� − ���� �������� ��
=
!
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Important Question: What happens if our region ‘R’ is seen as both vertical and
horizontal element?
Answer: Mostly happens that one of case 1 or 2 is easier/less work than the other
Example 1: Use the diagram to see the region ‘R’
Case #1:
��� �������� = � = � + 12
������ �������� = − � + 12
���� 2 ��������� ������� ������ �������� �ℎ����� �� � = −3
!!
� � = � �1 + � �2
!
� + 12 − � �� = ���� ����
� + 12— � + 12 ��+
=
!!"
!!
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Case #2:
���ℎ� �������� = � = � = � �
���� �������� = � ! = � + 12 = � ! − 12
!
� − � ! − 12 �� =
� � =
343
6
!!
Good Luck!!
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