Math Review 2012

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Math Camp 2012Department of EconomicsUmass Amherst

9am-1pm, M-F, Aug. 20-30, 2012hompson 919

Chapter 1, 2, 4-9 were prepared by Sung-Ha Hwang and Seung-Yun Oh,based on Chapter 1-8 of Matheat!"s for #"ono!sts$ by Ma%"o%&eberton ' (!"ho%as )au, Se"ond #d!t!on *Man"hester +n!ers!ty&ress, 2./0 ts a suary of Chapter 1-3, -9 of 5undaenta%Methods of Matheat!"a% #"ono!"s$, by 6%pha C0 Ch!ang, 7h!rd#d!t!on *M"raw-H!%%, n"0 1984/

Chapter 3 !s based on Chapter 1 of 5urther Matheat!"s for#"ono!" 6na%ys!s$ by nut Sydsaeter, &eter Haond, 6t%eSe!erstad and 6rne Stro *&earson #du"at!on :td, 28/ and Serge!7re!%, :!near 6%gebra ;one >www0ath0brown0edu>?tre!%>papers>:6;:6;

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Chapter 1 !inear E"uations

1.1 Linear equation

Example 1.1 Suppose the demand schedule for a good, say honey, is given by 8 4q p= where p denotes price and q denotes quantity demanded. Suppose also that the supply

schedule for honey is given by 3q p= + with the interpretation that bee-keepers wish toproduce 3p + units of honey when price of honey is p . Sketch the graph of demand andsupply function. ind the market clearing price and quantity.

Exercise 1.1 Sketch in the same diagram the graphs of the following linear relations!

"a# $y x= + , "b# 3y x= , "c# 8y x= +%hat do you notice& ind the equation of the line of slope $ which passes through thepoint "-$, '#

Exercise 1.2 Suppose the demand and supply schedules for milk are $$ 3q p= ,$ (q p= + respectively. Sketch these schedules in the pq -plane and find the market

clearing price and quantity.

1.2 Simultaneous equations

Remarks! )dditive property implies that a b= , then a c b c =

Example 1.2Solve the following simultaneous equation and graph each linear equationinxyplane.

3 ( 8( ' *x yx y

+ =+ =

Exercise 1.3 Solve the following simultaneous equations and draw the graphs for each."$# 3 ( 8

+ 4 *

x y

x y

+ =+ =

"(# 3 ( 8

+ 4 $+

x y

x y

+ =+ =

Math Cap 212 &age 2

Rules for Manipulating Equality

et a, b, c, d be real numbers.

"i# )dditive! f a b= , then a+c = b + c."ii# ultiplicative! f a=b, then a c = b c.

"iii# /ivisibility! f a = b, then a/c = b/cfor c0

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1.2.1 Two equations an two unknowns

1onsider solving the following linear equations.

$$ $( $

($ (( (

a x a y b

a x a y b

+ =

+ = "$.$#

f we manipulate "$.$#, then we have!

$$ (( $( (( $ ((

($ $( (( $( ( $(

a a x a a y b a

a a x a a y b a

+ =

+ = "$.(#

Subtracting the second equation from the first one, we have!

$$ (( ($ $( $ (( ( $(" # " #a a a a x b a b a =

2hus, we need to divide the cases depending on the values of $$ (( ($ $(" #a a a a and

$ (( ( $(" #b a b a .

!ase " $$ (( $( ($a a a a 0!ase "" $$ (( $( ($a a a a 0 and $( ( (( $a b a b 0!ase """ $$ (( $( ($a a a a 0 and $( ( (( $a b a b 0

!ase " $$ (( $( ($a a a a 0. 2hen, obviously, "$# is the unique solution.

%hen $$ (( $( ($a a a a 0, we also have to divide cases depending on the value of

$( ( (( $a b a b . 2his can be shown as follows.

n the steps of solving the above equations, we have

( ) ( )$$ (( $( ($ $( ( (( $a a a a x a b a b = "$.3#

!ase "" $$ (( $( ($a a a a 0 and $( ( (( $a b a b 02hus, if $$ (( $( ($a a a a 0 and $( ( (( $ 0a b a b = , then any value of x can satisfy theequation "(#. n this case, we have infinitely many solutions.

!ase """ $$ (( $( ($a a a a 0 and $( ( (( $a b a b 02hen, no real number can satisfy "(#. 2hus, we have no solution in this case.

Math Cap 212 &age 3

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)lso, we can verify the relation by drawing the graphs. n case , the two linear graphsintersect at the eactly one point "unique solution# and case , two graphs are parallel toeach other and coinciding with each other, thus intersect at infinitely many points.inally, in case , two graphs are parallel, but not intersect at all, resulting no solution.

Summary

$$ (( $( ($a a a a 0 $$ (( $( ($a a a a 0

$( ( (( $a b a b 0 $( ( (( $a b a b 0) 5nique Solution nfinitely any Solution 6o Solution

7raphs intersect at one point 7raphs coincide7raphs are parallel, but not

intersect.

Exercise 1.# Solve the following simultaneous equations.( 3

( 3 $3

x y

x y

+ = =

Exercise 1.$Suppose the demand and supply schedules for wine are

4 , $ 3q k p q p= = +

respectively, where kis a constant parameter. ind the equilibrium price and quantity interms of k. n each of the following cases, sketch the demand and supply schedules andwrite down the equilibrium price and quantity!

"a# 3k= , "b# 4k= , "c# (k= .%hat happens when

$

(k= &

Math Cap 212 &age 4

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1.2.2 T%ree equations in t%ree unknowns

1onsider the following a triangular system9. (x : ;y:z ( 3y $.2he strategy the solving the three equations is that transforming the given system into atriangular system9 by using the property of the equation.

Example 1.3 Suppose that we are given the following three equations system.

(x : 4y:z 'x: y:z +

(x: 3y:(z +

2he solution to the system isx $;,y -+, z -'.

Exercise 1.& Solve the system of equations( ( $

( ( (

3 3

x y z

x y z

x y z

+ + =

+ = + =

Math Cap 212 &age @

(x : 4y:z 'x: y:z +(x: 3y:(z +

(x : 4y:z '-y:z/2 ;?(

- y:z $

(x : 4y:z '-y:z/2 ;?(

z/2 -'?(

"S$# "S(#

"S$# eave the first equation alone and eliminatex from the second and third equations bysubtracting from those equations suitable multiples of the first equation.

"S(# eave the first and second equations alone and eliminateyfrom the third equation bysubtracting multiple of the second.

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'aussian Elimination

%e can generali>e the above procedures= 7aussian @limination. 2he eliminationprocedure is carried out using a sequence of elementary operations. 2hese are two kinds!

Example 1.3

( 4 $ A '

$ $ $ A +

( 3 ( A +

( 4 $ A '

0 $ $? ( A ; ?(

0 $ $ A $

( 4 $ A '

0 $ $? ( A ; ? (

0 0 $? ( A ' ? (

( 4 $ A '

0 $ $? ( A ; ? (

0 0 $ A '

$ ( $? ( A ' ? (

0 $ $? ( A ; ? (

0 0 $ A '

$ ( $? ( A ' ? (

0 $ 0 A +

0 0 $ A '

$ 0 $? ( A (* ? (

0 $ 0 A +

0 0 $ A '

$ 0 0 A $;

0 $ 0 A +

0 0 $ A '

)dd

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1.3 "nput-utput +nalysis

Suppose an economy produces three goods C, D, and E. 2here are non-produced goodssuch as abor, and, mported raw material, etc.

'ross output of goo /0 the total amount produced, including that fed back into thesystem as industrial input.et output of goo /! 2he amount of C produced and not fed back into the system,being therefore available for consumption, accumulation, and eport.

7ross output Feproduction purpose "the portion fed back into the system as input# :1onsumption

Tec%nology! nput requirements per unit of gross output of each produced good are givenin the following table, known as an input-output ta,le. 2his says for eample that eachunit of gross output of good D requires the input of 0.( units of good C, 0.' units of good

D and 0.3 units of good E.

Butput

C D E

6G52C 0.( 0.( 0.(

D 0.4 0.' 0.$E 0.4 0.3 0.3

Relation ,etween new an gross output of goo /!6et output of C gross output of C < quantity of C required as input in industry C

< quantity of C required as input in industry D < quantity of C required as input in industry E

Example 1.# et the gross outputs of C, D, E bex, y, zand the net outputx*, y*, >H.%rite down the relation between net and gross output of good C. )ssuming that theeconomy is closed and no accumulation happens. 2hus, the final output can only be usedeither as reproduction purpose or consumption. f the demand of consumption is given as$0 for C, $' for D, ; for E respectively, what is the amount of outputs of C, D, E requiredfor this economy&

2he relations are given by

H 0.8 0.( 0.(

H 0.4 0.' 0.$

H 0.4 0.3 0.;

x x y z

y x y z

z x y z

= = + = +

Math Cap 212 &age .

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Since net output should be equal to the demand of consumption for each output by

assumptions, setting H $0, H $'x y= = , and H ;z = and solving the linear equations gives'', *0, 80x y z= = = . "5se 7aussian elimination#

Exercise 1. or the eample of the input-output model given in the tet, find the grossoutputs corresponding to net outputs of $(, $0, $0 of C, D, E respectively.

Math Cap 212 &age 8

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Chapter 2 !inear #ne"ua$ities

2.1 "nequalites

nequalities in thexyplane

Exercise 2.1 Shade the region in thexy-plane which satisfiesx+(y 3, (x-3y I $3

Math Cap 212 &age 9

x

y

L

3

(

( x

y

L

3

(

(

3x+4y + 3x+4y J +

y4

y=4x=-3x-3

Rules for Manipulating "nequality

et a, b, c, d be real numbers.

"i# )dditive! f a b< , then a+c < b + c.

"ii# Gositive ultiplicative! f a0, then a c < b c .

"iii#6egative ultiplicative! f a

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2.2 Economic +pplications

T%e ,uget set an t%e ,uget line

Suppose that there are only two goods labeled $ and (, and that an consumes quantities

$x of good $ and (x of good (. 2he prices of the goods are $p and (p . f an9s income is

m , then the statement that he cannot consume more than his income may be written as$ $ ( (p x p x m+ . Since goods can be consumed only in non-negative quantities, $ 0x

and ( 0x . 2he set of points in the $ (x x -plane satisfying these two inequalities and thebudget constraint is called the budget set.

Exercise 2.2Sketch the budget line and budget set. Ke sure to label every point ofinterest. %hat happen to the budget set if the price of good $ increases& )nd, if the priceof good ( decreases& %hat happen to the budget set if the income increases&

Exercise 2.3 Lenry has an income of $8 which he can spend on two goods labeled $ and(, with prices $ and 3 respectively. Sketch the budget set.Sketch also the budget set in each of the following cases!"a# ncome 3+, prices ( and + "b# ncome *, prices $ and 3 "c# ncome $8, prices ( and 3

rouction possi,ilities

Suppose a firm manufactures two products C and D. et the production process involvethree departments ), K and 1, with time "in minutes# required in each department perunit output of each product given by the table below.

/epartment) K 1

Groduct C (0 30 4'Groduct D 40 30 30

Suppose that /epartments ) and K are available for 8 hours per day and that /epartment1 is available for $$ hours per day. etxandybe the amounts of C and D produced eachday by the firm. Sketch the feasible set for the firm9s production plan inxyplane.

Exercises 2.# ) firm produces two products C and D using a production processinvolving two departments ) and K. 2he time in minutes required in each department perunit output of each product is given by the following table.

/epartment) K

Groduct C $+ $0Groduct D 8 (0

/epartment ) is available for 4 hours per day and /epartment K is available for ' hoursper day. Sketch the feasible set.

Math Cap 212 &age 1

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2.3 Linear rogramming

inear Grogramming is an optimi>ation problem where a linear obMective function isbeing maimi>ed subMect to linear inequalities.

1onsider the following problem. Suppose that a firm produces two products C and D,using a production process involving two departments ) and K. 2he time in minutesrequired in each department per unit of output of each product is given by the followingtable. "6ote that this firm has the same technology ecept the fact that department 1 hasbeen deleted#

/epartment) K

Groduct C (0 30Groduct D 40 30

Suppose that products C and D yield profit of N30 and N40 per unit respectively and firmmaimi>es its profit. et x andybe the amount of the products C and D produced perday respectively. n this case, the maimi>ation problem is given as follows.

,ma 30 40

. . ( (4

$+

0, 0

x yx y

s t x y

x y

x y

+

+ +

%hat would be the optimal amount of C and D that the firm chooses to maimi>e the

profit&

2o solve the above the linear programming, it is particularly helpful to use a graphicalapproach. 5sing the production constraint, we can draw the production feasibility set.)lso, for the obMective function, we can consider the different output combination which

yield a given level of profit, say . 2hus, we have 30 40x y+ = . Ky manipulatingthis equation, we see that this equation is a straight line with a slope of

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6ow, as is clear from the above graph, the firm9s profit is maimi>ed at x 8 andy8."%hat is the maimi>ed profit. 6ote that this result critically depends on the slopes ofthe budget constraints and isoprofit line.

Exercise 2.$6ow suppose that production constraint as the above eample. Lowever, theprofit function is given as generic form= x y + for 0, 0 > > . )t which level ofoutput, the firm can maimi>e the profit& %hat is the maimi>ed profit&

"Lint= n this case, depending on the slope of isoprofit curve, you should divide cases. oreach case, you can use the same approach as the above. inally, make sure that your

answer should depend on the parameters " , #.#

Math Cap 212 &age 12

x

y

$+

$(

$+ (4

"8,8#

!onstraint Set

x

y

ncrease inGrofit

(4

,4ecti5e 6unction

3 $

4 40y x= + (

40

0

40

$

40

$+y x +

$$(

(y x +

x

y

$+

$(

$+ (4

"8,8#

!onstraint Set 7

,4ecti5e 6unction

Grofit aimi>ation Goint

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Exercise 2.& 5tility aimi>ation problem.

2here are two goodsx, y. ary9s utility is described as linear utility, " , #U x y x y = + .)lso, the price ofx goods and the price ofygoods are given as xp and yp respectively.

inally, ary9s income is given as m . Set up the maimi>ation problem and find the

optimal choice in this problem by following the below steps.

Step $! %rite down the relevant maimi>ation problem.Step (! /raw ary9s budget constraint inxyplane.

Step 3! 7iven a certain level of utility "say, U #, find the straight line on which every

point gives the same utility and draw this line in xyplane. "2his is called an indifferencecurve#Step 4! /raw the budget constraint and indifference curve at the samexyplane and findthe optimal value "6ote that you have to divide cases similarly to @ercise (.'#Step '! nterpret your solution. "2his solution is called a demand function#

Exercise 2.(1ost inimi>ation GroblemSuppose that a firm uses two inputs "L,K# for the production "y#. 2hat is, the production

function is given as minO , Py L K = for 0, 0 > > . )lso, the firm faces the price ofinputL as wand the price of inputKas r. Suppose that, given a certain output level y ,

the firm minimi>ed its cost wL rK + . %rite down the relevant minimi>ation problem andsolve this problem. "6ote that you should develop the iso-quantity curve for the firm#

Exercise 2. 6ancy erner is trying to decide how to allocate her time in studying forher economics course. 2here are two eaminations in this course. Ler overall score forthe course will be the minimum of her scores on the two eaminations. She has decidedto devote a total of $,(00 minutes to studying for these two eams, and she wants to getas high an overall score as possible. She knows that on the first eamination if she doesnQtstudy at all, she will get a score of >ero on it. or every $0 minutes that she spendsstudying for the first eamination, she will increase her score by one point. f she doesnQt

study at all for the second eamination she will get a >ero on it. or every (0 minutes shespends studying for the second eamination, she will increase her score by one point./raw a Rbudget lineR showing the various combinations of scores on the two eams thatshe can achieve with a total of $,(00 minutes of studying. Bn the same graph, drawRindifference curvesR for 6ancy. ind her optimal allocation of time in each eam.


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Chapter % &ets an' Functions

#.1 Sets an num,ers

) collection of obMects viewed as a single entity is called aset.Sets are described either by listing their members in braces or by defining their property.2hus, the setof all positive whole numbers which are divisible by ( can be written as@ O(, 4, +, 8, Por@ Ox!xis a positive whole number divisible by (P

2he obMects in the collection are called t!e e"eme#ts $r membersof the set. or any set %, we write a % to indicate that ais a member of set ), and a % to indicate that ais notin the set%.

7iven two sets%and&such that every member of% is also a member of&, we say that% is as'bsetof K, and write % & .

2he set which contains no elements is called as empty set or null set and is denoted by .

Bperation with sets.

"$# % & ! O ! or P% & x x % x & "(# ! O ! and P% & % & x x % x & "3# ! O ! and P% & % & x x % x &

f it is clear that all sets are subsets of some "universal# set U, U % is often written as(% , and called the complement of%"in U#.

um,ers

2he set of numbers used for counting O$,(,3,P is called the set of #at'ra" #'mbersanddenoted by )

O$,(,3,4,P

2he set of #teers, denoted by 8 consists of the natural numbers, their negatives

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1an every number be written as the quotient of two integers& n other words, is everynumber a rational number& )lthough it is not readily apparent, some important numbers,

like ( , e , and cannot be written as quotients of integers. 6umbers that cannot be

written as ratios or quotients of integers are called irrational numbers.

2he set Rof rational and irrational numbers is called the set of rea" #'mbers.

2he following gives the useful set in R.

2he set O ! , FPx a x b x< < is called an $pe# #tera"and denoted as " , #a b .Similarly, the set O ! , FPx a x b x is called a c"$sed #tera"and denoted as Ta, bU.

T%e plane as a set

)n important set related to Rwhose members are not themselves numbers is denoted byR2 and consists of all ordered pairs of real numbers!

R2 O" , # ! F and FPx y x y

Vust as the members of R can be thought as points on a line, the members of R2 can bethought as points in a plane on which two aes have been drawn.

Example #.1a. 2he straight line ( +y x= + is the set= (O" , # ! ( +Px y . y x = +b. ) subset of (. which is particularly relevant for economic applications is the set

(O" , # ! 0and 0Px y . x y known as the non-negative quadrant.

Math Cap 212 &age 1@

perations in t%e Real um,er

"$# )!losure*f a and b are in F:so are a b+ and a b"(# )!ommutati5e*or any ,a b F, a b b a+ = + and a b b a = "3# )+ssociati5e*f , ,a b cF, " # " #a b c a b c+ + = + + and " # " #a b c a b c = "4# );istri,uti5e* or all , ,a b c F, " #a b c a b a c + = +"'# )"entity*2here is an element 0F such that 0a a+ = for all a F. 2here is an

element $F such that $a a = for all a F."+# )"n5erse*or any a F, there is an element b F such that 0a b+ = = such a b isusually written as a . or any non>ero a F, there is an element c F such that

$a c = = such a c is usually written as $? a .

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Exercises #.1

or the following pairs of sets +and

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Math Cap 212 &age 1.

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#.3 Some simple alge,ra

et a, b, cbe a real number.

Exercise #.$ Ky using the operation rules of real number, derive the following relation.ake sure which properties you use.

" #

" #

" #" #

a b c ab ac

a b c ac bc

a b c d ac bc ad bd

+ = ++ = ++ + = + + +

Exercise #.&Ky using the above manipulation, derive the followings.( ( (

( ( (

( (

(

" # (

" # (

" #" #

3 "( 4# + $(

"( 3#"4 '# 8 ( $'

a b a ab b

a b a ab b

a b a b a b

x y xy x

x x x x

+ = + +

= +

+ = =

+ = +

!ompleting t%e square

7iven any quadratic equation, we can complete t%e square.

Suppose that we have (3 $0 38x x + .2hen we can manipulate as follows.

( ( ($0 (' (' $0 (' ('3 $0 38 3 38 3 38

3 * * 3 * 3x x x x x x

+ = + + = + +

( (' (' ' 8*

3 38 33 3 3 3

x x = + = +

ore generally, we have( (

( 4

( 4

b b acax bx c a x

a a

+ + = +

Exercise #.(1heck the above claim.

Exercise #.1omplete the square of followings.

"a# (3 30 (;x x+ "b# ( + 8x x + "c# (( ' $x x +

Exercise #.?Solve the following quadratic equations.


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"a# (3 30 (; 0x x+ = "b# ( + 8 0x x + = "c# (( ' $ 0x x + =6actori@ation

1reatest ($mm$# act$r2he first method of factoring is called factoring out the greatest common factor.

actor the followings"a# ' 'x y+ "b# 3 ((4 $+ 8x x x +

ere#ce $ 4q'ares( ( " #" #a b a b a b = +

"a# ( $+x "b# ( ( ((' 3+x y z "c# ( (" # " #a b c d + "d# (* 3+x "e# (8" # $8x y+

4'm $r ere#ce $ ('bes3 3 ( (" #" #a b a b a ab b+ = + +3 3 ( (" #" #a b a b a ab b = + +

"a# 3 $('x + "b# 38 (;x "c# 3 3( $(8x y+ "d# + +x y

5r#$ma"s $ t!e $rm (x bx c+ +

"a# ( 8 $(x x+ + "b# ( ; $8x x "c# ( + *x x +

5r#$ma"s $ t!e $rm (ax bx c+ +( " # " #" #acx ad bc x bd ax b cx d + + + = + +

"a# (( ' $(x x "b# (8 (+ (0x x +

act$r# by rer$'p#

"a# ay az by bz + + + "b# ( ( ((x xy y z+ +


" # " #" #x a b x ab x a x b+ + + = + +

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2he relation between the factori>ation and the completing the square can be shown asfollows

Example #.#Solve the following quadratic equation by completing the square.

(

( ' $($( ' $( (4 8

x x x =

( ( (' $($ ' $$

( (4 $+ 4 4

x x =

( )' $$ ' $$ 3

( ( 4 " 4#"( 3#4 4 4 4 (

x x x x x x = + = + = +

Bf course, the 4x= and 3(

x = are solutions to (( ' $( 0x x = .

Ky this way, we can always factori>e trinomials by using the completing the square.

7enerally, we have,

( (( 4 4

( (

b b ac b b acax bx c a x x

a a

+ + + = + +

rom here, we can also infer that the solutions to the quadratic equation ( 0ax bx c+ + =

are( (4 4

,( (

b b ac b b ac

a a

+ .

Math Cap 212 &age 2

9uaratic polynomial

"i# 1ompleting the square( (

( 4

( 4

b b acax bx c a x

a a

+ + = +

"i# actori>ation

( (( 4 4

( (

b b ac b b acax bx c a x x

a a

+ + + = + +

"ii# 7eneral ormula for the solutions of ( 0ax bx c+ + =( 4

(

b b acx

a

=

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#.# "nices

et a, b,x, b be the real number. /efine 0 $x = . 2hen, the following properties hold.

F$. a b a bx x x + =F(. " #a b abx x= forx Y0F3. " #

a a axy x y= forxY0,y Y0

Exercise #.?

"a# rom 0 $x = and F$, derive$a

ax

x

= for ,x a .

"b# )lso, from F$, derive $? (x x= , $? 3 3x x= , thus conclude $? # #x x=

Exercise #.1ASimplify the followings.8 * 4" # ?x x x , $?3 $? + 3? (" # ?x x x , ( )

$? 4( (x y y

, ( ) ( )

$? 3 $? (++4 *x x

"nices an inequalities

et c be a real number such that , then and for every natural

number #. ore generally, if , .

"a# f ', , aare positive numbers such that , then and .

"b# f ', a, b are positive numbers such that , then if and

if .


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Chapter ( )raphs

$.1 ;efinition of grap%

2he graph of a functionfromR to R is the set (O" , # ! " #Px y . y / x =2hus the graph of /can be depicted in thexyplane by Moining up the points " , " ##x / x for

all values of x .

6ote that sketch does not mean plot.

Example $.1 /raw the following graph.

"a# " # 4 A A/ x x= , "b# (

" # 4/ x x=

1an you eplain what happened to the original graph&

$.2 S%ift of 'rap%

Suppose that we know the shape of the graph " #y / x= already. 2hen, as we already sawin the previous eample, we can draw the graph " #y / x a b= + by shifting the originalgraph.

T%e s%ift of grap%.

)ssume , 0a b> . 2hen the graph of " #y b / x a = is obtained by shifting" #y / x= alongxais by a andalong y ais by b

2o see why this is really the case, let 0 0" , #x y be on the graph of " #y / x= . 2hen we have

0 0" #y / x= "--"$##. Ky shifting this point alongxais by a andalong y ais by b, we can


x

y

(y x=

x

y

A Ay x=

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get new point " 0 0,x a y b+ + #. 6ow, if we show that " 0 0,x a y b+ + # is on the graph of" #y / x a b= + , by repeating the argument again, we can generate the graph of" #y / x a b= + by shifting all points on " #y / x= .2hus, the above claim follows. 2o

show that " 0 0,x a y b+ + # is on the graph of " #y / x a b= + , by plugging 0x a+ into ,x

we have 0 0 0" # " #/ x a a b / x b y b+ + = + = + "by "$##. 2herefore " 0 0,x a y b+ + # is on thegraph of " #y / x a b= + .

2he similar argument as the above shows the followings.

Example $.2


" , #0 0x y

" , #0 0x a y b+ +

Shift by a

Shift by b

" #y / x=

" #y / x a b= +

" #y / x=

" #y / x=

Feflecting againstx ais

T%e reflection of t%e grap%.

2he graph of " #y / x= can be obtained by reflecting " #y / x= againstx ais.2he graph of " #y / x= can be obtained by reflecting " #y / x= againstyais.

" #y / x=

Feflecting againsty ais

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$.3 'rap%s of 5arious functions

9uaratic function

n section 3.( we sketched the graph of the function(y x= . 6ow consider the function

(y ax= where ais a real number such that 0a .

Exercise $.1 /raw the following graph.

"i#((y x= , "ii# (3y x= , "iii# (

$

(y x= , "iv# (

$

3y x=

"v#((y x= "vi# (3y x= "vii# (

$

(y x= "viii# (

$

3y x=

7iven(y ax= , as we saw in the above eercise, the magnitude of a "or A Aa # determines

the width "or si>e# of parabola and the sign of a decides whether graph will be 5-shapedor inverted 5-shaped. ")lso, all the verte of these function are "0,0##

Ky this, we now eplain how to sketch the graph of any quadratic function. 2he essentialtrick is to complete the square.

Example $.3Sketch the graph of the function(" # ( 8 */ x x x= + + .

( (" # ( 8 * (" (# $/ x x x x= + + = + +2hus, by the above rule ", we know that this graph can be drawn by shifting the graph

of((y x= by

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n this case,xais is called hori>ontal asymptote since whenxgoes to the graph getcloser to thex ais. Similarly, theyais is called vertical asymptote.

6ow consider how to draw the3

(

xy

x

+=

+.

f we manipulate the above equation, then we have

3 ( $ $ $( ( (

x xyx x x

+ + += = = ++ + +

2hus, by the same reasoning, we can infer that this graph is obtained by shifting the graph

of$

yx

= along x ais by

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Bseful grap%s

Exercise $.&Ky plugging the relevant value into the equations, verify the above graphs.

Math Cap 212 &age 2

x

$

(y

x

=

y

x

y

3y x=

x

( ?3y x=

x

y

y x=y

x

3y x=y

x

3? (y x=

y

x

( ($x y+ =

y

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Example $.$aimi>ing and minimi>ing quadratic function

Suppose a monopolist can produce quantity xof a product at total cost ;:+x. et thedemand function for the product be given by

$$'

(

x p=

where p is the price charged. ind the profit-maimi>ing price and output, and themaimal profit.

irst epress the profit as a function of output x. 2hen the revenue for the firm is givenby= ree#'e = px="30 < (x#x.

2hus we have the following profit maimi>ation problem.

ma "30 ( # "; + #

. . 0

xx x x

s t x

+

)s is clear, the obMective function is quadratic. 2hus, we can graph the obMective functionand verify that the obMective function is maimi>ed at its verte.

Exercise $.(5sing the graph, solve the following optimi>ation problem.

( )

( (

ma

. . $

0, 0

xa x

s t x y

x y

+

,

(

" # ma (

. . 4 0

0, 0

x yb x y

s t x x y

x y

+

+

( (

," # min

. . (

0, 0

x yc x y

s t x y

x y

+

+

," # ma

. . ( $0

0, 0

x yd xy

s t x y

x y+


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Chapter * &e"uences an' &eries

&.1 Sequences

et ) be the set of natural numbers= we define a sequence of real number to be afunction from ) to . f ' is such a function, we may write the values it takes as

"$#, "(#, "3#' ' ' and so on. t is conventional to denote these values by

$ ( 3, , ,...' ' '

and to refer to the sequence as O P#' .

Exercise &.1ind the #t!term for each sequence. 1an you tell where the sequence tends&

"a# (, 4, +, 8, "b# (0, $$, (, -;,

"c#

"d# 8, -(,

"e# $

Limits

n this case, we say that the sequence c$#erestozand we write

lim #' z= or lim ## ' z = or simply #' z

%hat this definition means, roughly speaking, is that we can make #' as close as we like

tozfor all sufficiently large #.


;efinition.et O P#' be a sequence and letzbe a real number. %e say that z be the"mt of

this sequence if for any "small# positive number , there is a positive integer ) such that for

all # ) , #' is in the -interval aboutz= that isA A#' z < .

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2here is a small person "/ /e W# and a large person ")S#. / /e W chooses a small

positive number, say 0.0$. )S then looks for a natural number $) such that

A A 0.0$#' z < for all $# )> . Suppose he finds such an $) . 2hen / /e W choosesan even smaller number, say ;$0 , and )S tries to find an integer () such that #'

is squee>ed between ;$0z and ;$0z + for all (# )> . f this process can berepeated indefinitely, lim #' z= = if not, not.

Example &.1 1laim$lim lim8 " 4# 0##

# #'

= = .

et 0> be given. 1hoose a positive integer ) Y( )log 4 log ? 8

log 4

.

2hen, for # )> , we havelog4 log"

log4#

/8)> --- "$#

Ky manipulating "$#,

$ $log log 4 log 4 4 8 48

# ##

> > > 8

----"(#

2hus, for # )> , then $ $ $A 8 " 4# A 8 A " 4# A 8 4# # # = = < "by "(##.

2herefore, given any small , we can always choose)which satisfies the definition.

"6ote that "(# holds since the logarithm is increasing function #

Exercise &.2 Show that$ $

lim( (#

#

#

+=

Lere, we notice that it is difficult to calculate the limit by the direct method. 2hus, thefollowing results would be useful.


T%eorem &.1. 7iven a real numberx, lim 0#

#x

= if and only if $ $x < e theorem, we have$

lim sin" # 0#

#x#

= .

T%eorem &.#0 Limit T%eorem " )Sequence*

et O P#x and O P#y be a sequence and suppose that #x x and #y y .2hen,

"i# ( )lim # ##

x y x y

+ = +

"ii# ( )lim for a real number##

cx cx c

=

"iii# ( )lim # ##

x y xy

=

"iii# n addition, 0 and 0 for all#y y # , then lim #

##

x x

y y

=

Exercise &.3 n each of the following cases, state whether the sequence O P#x converges,

and find the limit if it eists.

"a#$

($#x #= + "b#

$

($#x #= "c# ( )$

(

#

#x = "d# ( )$(#

#x =

&.2 Series

2he series generated by a sequence O P#' is defined to be the sequenceO P#s , where

$ $s '= , ( $ (s ' '= + , 3 $ ( 3s ' ' '= + + and so on.

2hus we have$

#

# *

*

s '=

=

Exercise &.#

"a# Show that$

" $#

(

#

*

# #*

=

+=

"b# ind the sum of the first $' terms of the sequence 48, 44, 40, 3+,

"c# Show that$

$

$

$

##*

*

rar a

r

=

=

for $r"d# ind the sum of the first 8 terms of the geometric progression $0, '0, ('0, $('0,

Math Cap 212 &age 3

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!on5ergence of series

et O P#' be any sequence and let O P#s be the series generated by it. f #s approaches a

limit 4 as # , we say that the series O P#s is convergent and we write

$

r

r4 '

== ) series which is not convergent is said to be dere#t $r per$dc.rom the result of the limit, we have the following results=

$

$

*

*

ar

= converges if and only if $ $r < < , in which we have $

$ $

*

*

aar

r

=

=

@ample= ind the sum of the series$ $

8 ( ...( 8

+ +

Ky applying the above formula, we see that the sum of series is $4

8 3(

'$

=

+

&.3 +pplication in economics

et an amount of money6be invested at the beginning of year $. et the rate of interestrate be r. %e also suppose that interest is paid once a year at the end of the year. 2hus, atthe beginning of year (, the investment is worth6"$:r#.

Example&.3f N'0 is invested at 4Z per annum, the value in N at the beginning of the ; th

year is +'0 "$.04# +3.(; = .

Example&.#n this eample we allow compounding of interest to take place more oftenthan once a year. et interest be compounded quarterly= then with a rate of interest of 4Zper annum, $Z of the value of the investment accrues as interest at the end of eachquarter. %ith N'0 invested at the beginning of the first year, the value in N of the

investment after + years "(4 quarters# is (4'0 "$.0$# +3.4* = .

7enerally, if a sum6is invested at a rate of interest of rper year, compounded mtimesper year, the value of the investment after #years is

$

m#r

6m

+

f $m> , the amount of interest earned in a year per dollar invested at the beginning ofthe year is given as

Q $ $

mr

rm

= +

.

%hen r and r7are epressed as percentages, ris called the flat rate and r7 t!e a##'a"perce#tae rate 8%69.


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Exercise &.$ 6ow suppose that an amount of money %is placed in an account at thebeginning of eac! year, starting at the beginning of year $. 2hen, how much is theamount of money in the account at the beginning of year #"Must after the #th investmenthas been made#

Exercise &.&f N+0 is invested at a constant rate of interest r, what value of ris requiredfor the investment to be worth N*0 at the end of the si years&

Exercise &.(f N80 is invested in an account once per year, and the rate of interest is 8Zper annum, after how many payments will there be at lease N;00 in the account&

resent Calue

Suppose that a sum of money: is going to be available 3 years from now. ts presentvalue6is the amount of money which, if invested now, would compound to in the

three years time. )ssuming a constant rate of interest rate r , we have 3"$ #6 r :+ = , sothat 3"$ #6 : r

= + . 2he process of finding present values from future ones is calleddsc$'#t#. %hen the interest rate is r, the discount factor is $?"$:r#

Similarly, the present value #6 of an annual income stream of %per year for # years,

starting one year from now, is given by

( ...

"$ # "$ # "$ ## #

% % %6

r r r

= + + +

+ + +) security which gives this income stream is called annuity.

Exercise &.1alculate the above equation.

Exercise &.?Suppose that N('0,000 is borrowed to be repaid over (0 years. Suppose alsothat the "flat# rate of interest is +Z per annum, and that repayments, and compounding ofinterest, take place monthly. )ssuming that each month9s repayment is a constant amountofz, what mustzbe&

Exercise &.1A)ssume a constant rate of interest r. 2hen how much is the value of theperpetuity giving its holder the right to an annual income of %per year, starting one yearfrom now&"Gerpetuity= a security which gives its holder the right to an annual income of%per yearforever, starting one year from now#


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Chapter + Continuit

(.1 Limit of a function

nformally speaking, the limit of " #/ x as xapproaches ais the numberL, if " #/ x gets

arbitrarily close toLfor allxsufficiently close to a.

Example (.1


doesn9t eist

a

lim " #x a

/ x b

=

a

b

;efinition et " #/ x be defined on an open interval about a, ecept possibly at aitself. %e write

lim " #x a

/ x L

= ,if, for every > 0 , there eists a > 0 such that for allx,A " # A/ x L whenever 0 A Ax a <

T%eorem (.1 ) function " #/ x has a limit as xapproaches a if and only if it has

left-hand and right-hand limits there and these one-sided limits are equal

lim " # lim " # lim " #x a x a x a

/ x L / x L a#d / x L +

= = =

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%e have the limit theorem of a function.

T%eorem (.2 Limit T%eorem "" )6unction*

et lim " #x a

/ x = and lim " #x a + x =2hen,

"i# ( )lim " # " #x a

/ x + x

+ = +

"ii# ( )lim " # for a real numberx a

c x c c

=

"iii# ( )lim " # " #x a

/ x + x

=

"iii# n addition, 0 and " # 0 for+ x x in the domain of, then" #

lim" #x a

/ x

+ x

=

Exercise (.1 ind the limits of the function(8 *y x x= +

"a# )s 0x "b# )s 3x "c# )s $x

Exercise (.2ind the limits of the function "3 '# ?" (#y x x= + +"a# )s 0x "b# )s 'x "c# )s $x

Exercise (.37iven( (0

" #4

x xy / x

x

+ = =

"a# s it possible to apply the quotient limit theorem to find the limit of this function as

4x &"b# s this function continuous at 4x= & %hy&"c# ind a function which, for 4x , is equivalent to the above function, and obtain fromthe equivalent function the limit of y as 4x .

(.2 !ontinuity

Ky the definition of the limit of a function, lim " # " #x a

/ x / a

= if for every > 0 , there

eists a > 0 such that for allx, A " # " # A/ x / a whenever 0 A Ax a <

is said to be c$#t#'$'sif it is continuous at every ain a domain of.


;efinition " ) function"x# is continuous at a point a in a domain of if and only if

lim " # " #x a

/ x / a

=

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%e can summari>e continuity at a point in the form of a test.

Example &.2 ind the points at which is continuous or discontinuous.

2he following gives the useful results.

T%eorem (.3

et and be continuous functions and let ;be the intersection of the domain ofand the domain of.

"a# / ++ is continuous on ;"b# or any constant c, the function cis continuous on the domain of"c#is continuous on;

"d#/

+is continuous at allx ; such that " # 0+ x

"g# / +o is continuous where it is defined.

2he above result easily follows from the limit theorem of the sequence.

Exersice (.#Grove 2heorem "a#["d# using the limit theorem of the sequence.


$ 43(0

(

$

!ontinuity Test

) function"x# is continuous at x a= if and only if it meets the following threeconditions$. " #/ a eists "alies in the domain of#

(. lim " #x a

/ x eists " has a limit as x a #

3. lim " # " #/ x / a=

"the limit equals the function value#

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(

$ ($

2here is another definition of continuity.

Remark 2he above definition suggest that ( )lim " # lim# ## #

/ x / x

= . 2hat is, you can

change the order of limit and function.

6ote that the second definition corresponds to the first definition of continuity of our tet.%e can start from one definition and derive the other definition as a theorem.

%e have lim " #x a

/ x L

= if and only if lim " ###

/ x L

= for all sequences O P#x in the domain

of , with #x a that converge to a. 2hus we can always check lim " #x a / x by using the

sequence of O P#x which is sometimes easier to check than the definition of lim " #x a/ x

.

Example (.31onsider the function(" #/ x x= whose domain is the whole real line R. et

a Rbe given and suppose .#x a 2hen,

( )(

( (lim " # lim lim " ## # ## # #

/ x x x a / a

= = = = ,

where we used the limit theorem of sequence in the second step. 2hus, is continuous ata. Since ais arbitrary,is continuous on R.

Example(.#1onsider the following function defined on T0,(U =

($

(

( if 0 $" #

( if $ (

x x/ x

x x

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2he following theorem is one of the most important theorems which we will use.

2he following eample shows that why the hypothesis of the continuity and closedinterval is required.

Example (.$ 1onsider the following function defined on T0,$U( $

(

$

(

4 4 if

" #$

if(

x x x

/ xx

=

=

n this case, the maimum of the function doesn9t eist.

Example (.& 1onsider a function y x= defined on "0,$#. n this case, the maimum ofthe function doesn9t eist.

Exercise (.$ Ky drawing the graphs of eample +.4 and eample +.', verify the aboveclaim by the graph.

Exercise (.& 1onsider a function$

xy= defined on "0.$U. /oes this function has a

maimum in "0.$U.

Exercise (.( 1onsider a function$

xy= defined on Tt, $U where "0,$#t . /oes this

function has a maimum in Tt, $U&


T%eorem (.# )Maximum an Minimum* et be a continuous function on a closedinterval Ta,bU. 2hen there eist points cand din Ta, bU such that

" # maO " # ! T , UP/ c / x x a b= and " # minO " # ! T , UP/ d / x x a b= .

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Chapter Dierentiation

.1 ;ifferentiation

%e will use the standard notation Q" #d/

/ xdx

=

Example .1 2o illustrate the definition, we9ll check that(" #/ x x= is differentiable

everywhere. etxbe given, then

( (" # " # " #(

/ x ! / x x ! xx !

! !

+ + = = +

Since the limit of "( #x !+ eists as 0! and equals (x, we conclude from thedefinition that Q/ eists atxand Q" # (/ x x=

2he difference quotient is the slope of the straight line through the point " , " ##x / x and

" , " ##x ! / x !+ + on the graph of. 2hus, Q" #/ x is the limit of these slopes, that is theslope of the line tangent to the graph of at " , " ##x / x .

Example .21onsider the function " # A A/ x x= . Lere, we will show that this function isnot differentiable atx0.

irst, we can write the difference quotient as follows="0 # "0# A A/ ! / !

! !

+ =

rom this, we know that

0 00 0

"0 # "0# "0 # "0#lim $ and lim $! !! !

/ ! / / ! /! !

> ero by the limit theorem. 2hus,the left-hand side converges to >ero too, which shows that is continuous atx.

%e have the following rules of differentiation.

T%eorem .2 ;ifferentiation T%eorem

Suppose that"x# and"x#are differentiable atx. 2hen,

"i# or any constants and , " # " #/ x + x + is differentiable at x and" " # " ## Q Q Q/ x + x / + + = +

"ii# 2he product " # " #/ x + x is differentiable atxand

" " # " ## Q Q" # " # " # Q" #/ x + x / x + x / x + x= +

"iii# f " # 0 for + x x in the domain of, then" #

" #

/ x

+ xis differentiable atx and

Q

(

" # Q" # " # " # Q" #

" # " #

/ x / x + x / x + x

+ x + x

=


T%eorem .1 f is differentiable atx, then is continous atx.

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Lere, the main idea of the product rule can be illustrated as follows. 2o do this, suppose

that " #, " #/ x + x is increasing function with respect to .x

irst note that the area of the

shaded rectangle in thediagram corresponds to

/ + . 6ow suppose that

there is a slight increase inx.2hen, by this change, will

increase by / and +by +. 2hus the area of rectanglewill increase by the areawhich is composed of ,,. %e can write this

as follows.

" # " # " # " #" #/+ / + / + / + = + + -"$#

/ivide "$# by x and taking a limit, we have

0 0 0 0

" #lim lim lim limx x x x

/+ / + /+ / +

x x x x

= + +

Since as 0x , 0+ "why, we can drop the third term in the right hand side.

2hus, we can obtain the following formula.

( )0

" #" # " # Q lim Q" # " # " # Q " #

x

/+/ x + x / x + x / x + x

x

= = +

Exercise .2Ky using theorem "ii# show that(

(dx

xdx

= . )gain by using this result and

theorem "ii#, establish3

(3dx

xdx

= and conclude that $#

#dx #xdx

= .

2his result also can etend to the case where the power is the real number.

T%eorem .30 ower rule

et c F, then we have $" #c cd

x cxdx

=

Math Cap 212 &age 4

/

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T%eorem .#0 T%e !%ain Rule

Suppose thatis differentiable atx and is differentiable at"x#. 2hen

" # " #1 x + / x= o is differentiable atxand ( )Q" # Q " # Q" #1 x + / x / x=

2o see why this is true, let " # and " #' / x y + '= = . irst note that we can write=y y '

x ' x

=

2hen,

( ) 0 0 0 0Q" # " # lim lim lim limx x ' xd y y ' y '1 x + / xdx x ' x ' x = = = = ( ) ( )Q Q" # Q " # Q" #+ ' / x + / x / x= =

n the above, we use the fact that as 0x , 0' "why and the limit theorem"where.

inally, the following definition will be used in the later.

;efinition

fis differentiable at every point of an open interval "a, b#, we say that is

differentiable on "a,b#. f, in addition, Q/ is continuous on "a, b#, we say thatis

c$#t#'$'s"y dere#tab"eon "a, b#.

Exercise .3 /ifferentiate the followings.

"a#3 + 4"+ #" 3 (#x x x x+

"b#4 ("4 ( $#" '# #x x x x+ +

Exercise .# /ifferentiate the followings.

"a#( 3" $# ?"( $#x x+ +

"b# ("3 #" #x a x b+ +

Exercise .$ 2he functions andare defined as follows!3 4" # $, " # (/ x x + x x= + =

ind epressions for ( )" #/ + x and ( )" #+ / x

Exercise .&/ifferentiate!


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"a# 3 '" $#x + "b# + (?3" $#x "c# 4 ( $? 4" 3 ' $#x x x + + "d# ( )'

8? (x+

"e#( 3 '" $#" $#x x + "f# $?3 ' 3" (# ?" (#x x


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.2 Monotone function an "n5erse function

) function is said to be strictly monotone increasing if x y< implies " # " #/ x / y< .Strictly monotone decreasing is defined similarly. f a function is strictly monotone

increasing, for any given 0y , we can always find the unique 0x such that 0 0" #y / x= ." Seethe graph#. 2hen we can define an inverse functionsuch that " #x + y= . %e denote thisinverse function as $/ .

Example .3f 3" # 4y / x x= = + then the inverse function of is given as$?3" # " 4#x + y y= = .

t is easy to see that the inverse function satisfy the following properties.

( )" #/ + y y= and ( )" #+ / x x= .

T%eorem .$

Suppose thatis a strictly monotone continuous function. 2hen,$/ is continuous.

)ssociated with inverse functions, there is another rule for differentiation.

T%eorem .&0 T%e "n5erse 6unction Rule

Suppose thatis a strictly monotone continuous function on an interval Ta, bU. fisdifferentiable atx and Q" # 0/ x , then $/ is differentiable at " #y / x= and

( )$ $

Q" #Q" #

/ y/ x

=

2o see why this is true, note the following.$y

x

x

y

=

Ky taking a limit at both sides,

( )$ 0 0 $ $Q" # lim lim Q" #y x yx

x/ yy / x

= = =

n the above, we use the previous theorem "theorem and the limit theorem.


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2he following eample illustrates how useful this theorem is.

Example .#2he inverse of " #y / x mx= = is " # "$? #x + y m y= = . 6ote that

( )

$ $ $Q" #

Q" #/ y

m / x

= =

Example .$ et$

" #$

x/ x

x

=

+. %e are interested in the value of ( )$ Q" #/ y when (x=

"or $? 3y= #. n this case, the inverse $/ sends $?3 to (. Since (Q" # ( ?" $#/ x x= + ,Q"(# ( ? */ = . 2hus, by the inverse theorem,

( )$3

$ $ *Q

Q"(# ( ?* (+

/= = =

%e can check this answer by computing directly that

( )( $3$ ( ( *

" # , Q" # , and Q$ "$ # 4 ? * (

y+ y + y +

y y

+= = = =

Exercise .(Ky the inverse function theorem, establish the following result. "which wealready saw#or any positive integer #,

( )$

$$? $# #x x

#

=

Exercise . Ky using the above results, show the following result.or any positive integer mand #,

( ) $

?

m

m # #mx x#

=

Exercise .?Show that the function3y x= is monotonic increasing. ind ?dx dy !

"a# by finding an eplicit formula for the inverse function and differentiating it.

"b# by using the inverse function rule.

Exercise .1A )re the following functions monotonic&

"a# + 'y x= + "b# + ' " 0#y x x= + > "c# ' 3* 4y x x x= + +


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Chapter 9 Ma/ima an' Minima

?.1 Bsing t%e first eri5ati5e for grap%ing

2he derivative of a function carries much information about the important properties of

the function. %e will see that the knowing Must the signs of a function9s first and secondderivatives and the location of only a few points on its graph usually enables us to drawan accurate graph of the function.

T%eorem ?.1 )T%e 6irst eri5ati5es "nformation*

et be a continuously differentiable function on domain ; R.f Q 0/ > on interval " , #a b 3 , then is increasing on " , #a b .f Q 0/ < on interval " , #a b 3 , then is decreasing on " , #a b .

f is increasing on" , #a b

, thenQ 0/

on" , #a b

.f is decreasing on " , #a b , then Q 0/ on " , #a b .

2he following steps describe how to use first derivatives to sketch graphs.

"$# irst find the points at which Q" # 0/ x = . Such points are called crtca" p$#tsof /.Lopefully, the function under consideration has only finitely many critical points

$ (, , .... kx x x .

"(# @valuate the function at each of these critical points $ (, , .... kx x x , and plot the

corresponding points on the graph

"3# 2hen, check the sign of Q/ on each of the intervals

$ $ ( $" , #, " , #,...., " , #, " , #

k k kx x x x x x .

2hen, on these intervals, the sign of Q/ is always negative or always positive "why.

"4# f Q 0/ > on interval;, draw the graph of increasing over;. f Q 0/ < on;, draw adecreasing graph over;.

Example ?.11onsider the cubic function3" # 3/ x x x= . Bne easily computes that

(Q" # 3 3 3" $#" $#/ x x x x= = +2he critical points are -$, $ and the corresponding points on the graph of are "-$, (# and

"$, -(#. 6et, we check the sign ofQ/on the three intervals.

"a# Q" (# * 0/ = > , so Q 0/ > and is increasing on " , $# ="b# Q"0# 3 0/ = < , so Q 0/ < and is decreasing on " $, $# ="c# Q"(# * 0/ = > , so Q 0/ > and is increasing on "$, # =

Kased on the above information, we can draw the following graph. "Low to draw theeact shape will be discussion later.#


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?.2 Secon ;eri5ati5es an !on5exity

f is a differentiable function, then Q/ is a function, which may itself be differentiable.

%e denote the derivate of Q" #/ x with respect to x by QQ" #/ x and call it the second

derivative of " #/ x . %e also use the following notation for " #y / x=(

(QQ" #

d dy d y/ x

dx dx dx

n the above graph, for "0, #x a , the slope of Q" #/ x is an increasing function. Ky theabove theorem "theorem, the derivative of Q/ , QQ" #/ x , is nonnegative there! QQ" # 0/ x on "0, #a . or " , #x a , Q/ is a decreasing function ofx= so QQ" # 0/ x on " , #a . 2hefunction /for which QQ" # 0/ x on an interval ", is said to c$#ex.2he function /forwhich QQ" # 0/ x on an interval ", is said toc$#cae.

2he following is a formal definition of concave "conve# function.

Math Cap 212 &age 4

-$

$

3" # 35!e +rap! $/ / x x x=

a x

y

increasing

decreasing

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;efinition

) functionis called c$#caeon interval " if

( )$ ( $ ("$ # "$ # " # " #/ x x / x / x + +

for all " and all T0,$U

) function is called c$#exon interval " if( )$ ( $ ("$ # "$ # " # " #/ x x / x / x + +

for all " and all T0,$U

) point where the concavity "or conveity# is changed is called a point of inflection. "n

the above figure, " , " ##a / a .

Example ?.27raph the function 4 3" # 4 $0y / x x x= = + .

Step $! ind Q" #/ x and QQ" #/ x .3 ( (

(

Q" # 4 $( 4 " 3#

QQ" # $( (4 $( " (#

/ x x x x x

/ x x x x x

= =

= =

Step (! Sketch the sign pattern for Q" #/ x and QQ" #/ x and use it to describe the behavior of

" #/ x

(4 " 3#x x +

Step 3! Summary


$( " (#x x + +

x0 ( 3

03

0(

convconcconv

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Step 4! Specific points and curves. Glot the curve9s intercepts "if convenient# and the

points where Q" #/ x and QQ" #/ x are >ero.

)s the above eample reveals, if we are given any twice differentiable function, then in

principle, by the similar method, we can draw the graph of the function.

Exercise ?.1 Sketch the graphs of the following functions

"a# 3 3x x+ "b# 4 3 (8 $8 $$x x x + "c# ; ;x x "d# (? 3x

Exercise ?.2Sketch the graphs of the following functions

"a# '?3 ( ?3'y x x= "b# "xY0#

?.3 Maxima an Minima

7iven the function " #y / x= , ifyhas its greatest "smallest# value in a neighborhood ofthe point, such a point is called a local maimum "minimum#. )lso, ify has its greatest"smallest# value in the domain of , then such a point is called a global maimum"minimum#. )lso, we call a ma"or min# that occurs at a boundary point of the domain of a boundary ma "or boundary min#. %e will call a ma"or min# which is not an endpointof the domain ofan interior ma "or interior min#.

T%eorem ?.2

"a# f 0Q" # 0/ x = and 0QQ" # 0/ x < , then 0x is a local maimum of="b# f 0Q" # 0/ x = and 0QQ" # 0/ x > , then 0x is a local minimum of="c# f 0Q" # 0/ x = and 0QQ" # 0/ x = , then 0x is a local maimum, a local minimum orneither.

ote ! @amples of the ambiguous case "c#


x

y

"0,$0#nflection

point

"(,-+#nflection

point"3,-$;#ocalinimum

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"a# "b# "c#

Remarks7iven maimi>ation "or minimi>ation# problem, Q" # 0/ x = is called the first-order condition and 0QQ" # 0/ x < "or 0QQ" # 0/ x > # is called the second order condition.

Example ?.3Suppose the monopolist faces the demand function$00q p=

%here q is output and p is the price, and let the monopolist9s total cost be3 ($

3" # ; $$$ '0( q q q q= + +

$. %rite the firm9s maimi>ation problem. %hat kind of assumption you use about themonopoly in writing this problem.

(. /erive the first order condition and find the critical points.

3. 1heck the second order condition for each critical points and determine whether it islocal maimum or local minimum.

4. /raw the profit function.

'. ind the marginal revenue "defined by Q" #. q when " #. q is total revenue# and the

marginal cost "similarly, Q" #( q #.

+. /raw the marginal revenue, marginal cost and demand function in the same graph.1ompare marginal revenue with the demand curve and compare the marginal revenue

and marginal cost. ind the optimal price and quantity for the monopoly in the graph.

Example ?.#

1onsider a firm9s cost function(" # $0 3+( x x x= + + , which epress the dependence of

total cost on outputx. )verage cost is defined to be " # ?( x x . 2he firm is assumed to

minimi>e its average cost.

$. Show that the firm9s cost function is a conve function.

(. Setup the relevant minimi>ation problem

3. Solve the above minimi>ation problem. "Lere, you should find B1, SB1 and optimalchoice#

4. ind the marginal cost. @valuate these functions at the optimal level of output. %hatcan you find& 1an you Mustify your answer more rigorously&


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2he following generali>es the above observation.

T%eorem ?.2

Suppose that the cost function " #( x is continuously differentiable. 2hen,

"a# if 1 Y )1, then )1 is increasing"b# if 1 J )1, then )1 is decreasing"c# at an interior minimum of )1, )11

Exercise ?.3 5sing the definition of average cost, that is" #

" # ( x

%( xx

= and the \uotient

Fule, prove the above theorem.

rom an intuitive point of view, the theorem says that if you do better than your averagesome day, your average goes up that day. Bn days that you do worse than your average,

your average goes down. or baseball fans, a batter who goes hitless in a game will seehis batting average drop= a batter who has a ]perfect day at the plate^ will raise hisbatting average.

2aking a geometric point of view, consider the graph of cost function " #y ( x= .

Math Cap 212 &age @

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)s x increase from 0 in the upper figure, the slope of the lime from " , " ##x ( x to the

origin starts very large, decreases past $x , reaches its minimum value at 0x , and then

increases again as xpasses (x and become arbitrarily large. f we graph this slope, that is,

if we graph the average cost curve " #%( x , versus x, we find a 5-shaped curve, as

pictured in the right figure. %e have also drawn the marginal cost curve 1 in the right

figure. 2he critical property in this figure is that for 0x x< , the 1-curve lies below the)1-curve while )1 is decreasing, for 0x x> the 1-curve lies above the )1-curve as)1 increases.

Slopes of the lines from the origin ! )1

" #y ( x=

y

x

Slopes of the tangent line ! 1

$

x

0x (

x

p

x

$x

0x (x

)11

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Math Review 2012