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Chapter 3: Functions Lecture notes Math

Section 1.1: Definition of Functions

Definition of a function

Afunctionffrom a setAto a setB(f:A!B) is a rule of correspondence that assigns to each elementx in the setAexactlyone elementyin the setB. The setAis called thedomainof the functionf. Therangeor codomainof the function is the set of elements inBthat are in correspondence with elements inA.

In the case of functions described as equations in two variables, the variablexis theindependentvariable

and the variableyis thedependentvariable. In general a function is denoted asf(x)(readfofx), wheref is the name of the function,xis the domain value andf(x)is the range valueyfor a givenx. The process of finding the value off(x)for a given value ofxis calledevaluating a function.Ex.1

Demand function:Qd=f(P) = 152P.

Supply function:Qs=g(P) = 1 + 5P.

Cobb-Douglas production function:Q(K;L) =KL.

Cobb-Douglas utility function:U(X;Y) =alog(X) + (1a)log(Y).Ex.2

Constant functionsare functions that assign every object in the domain to the same object in the target. For

example,f(x) = 3is a constant function. Theidentity functionis the function that assigns every object in

the domain to itself, that isf(x) =xfor everyxin the domain.Ex.3 Letf(x) =x2. Find the domain and the range off(x). Compute f(3) f(2) f(2) f(3 +h)Ex.4

Find the domain and the range off(x) = 1=x.Ex.5

Find the domain and the range off(x) = 5px1.Graph of a function

Letf(x)be a function. Thegraph of the functionfconsists of those points(x;y)such thaty=f(x). Not every

curve is the graph of a function. The reason is that a function assigns to a given input a single number as

the output. A line parallel to theyaxis therefore meets the graph of a function in at most one point. Hence,

if some line parallel to theyaxis meets the curve more than once, then the curve is NOT the graph of a

function.Ex.6

Graph the functionf(x) =x2.Ex.7

Graph the functionf(x) = 1=x.

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Chapter 3: Functions Lecture notes Math

Zoo of function

In mathematics there are many kinds of functions. Here is a short list of some of them: Polynomial functions: linear (ex.f(x) = 2x1), quadratic (ex.f(x) =x2), cubic (ex. f(x) = 4x33x2+ 5)

Rational functions (ex.f(x) =2x3)

Irrational functions (ex.f(x) =p2x)

Absolute value functions (ex.f(x) =jx9j)

Exponential functions (ex.f(x) = 2x)

Logarithmic functions (ex.f(x) = log2(x))

Trigonometric functions (ex.f(x) = sin(x),f(x) = cos(x),f(x) = tan(x))Transformations of functions

Letf(x)be a function, then

y=f(x) +C -C >0moves it up -C <0moves it down y=f(x+C) -C >0moves it left -C <0moves it right y=Cf(x) -C >1stretches it in they-direction -0< C <1compresses it y=f(Cx) -C >1compresses it in thex-direction -0< C <1stretches it y=f(x)reflects it aboutx-axis y=f(x)reflects it abouty-axisEx.8

Graph the functionf(x) =1x

2+1.Definition of composition of two functions

Thecompositionof the functionsfandgis given by

(fg)(x) =f(g(x)) The domain of the composite function(fg)is the set of allxin the domain ofgsuch thatg(x)is in the domain off.Ex.9 Letf(x) = 1 + 2xandg(x) =x2. Compute(fg)(x)and(gf)(x).Ex.10 Letf(x) =pxandg(x) =x31. Compute(fg)(x)and(gf)(x).Ex.11

Letf(x) =pxandg(x) =x2. Compute(fg)(x)and(gf)(x).

2

Chapter 3: Functions Lecture notes Math

Even and odd functions

A functionf(x)such that

f(x) =f(x) is called aneven function.

A functionf(x)such that

f(x) =f(x) is called anodd function.Ex.12 The functionf(x) =x4is an even function. The functiong(x) =x3is an odd function. Section 1.2: Bijective Functions and Inverse FunctionsBijective functions A functionf:X!Ythat assigns distinct outputs to distinct inputs is called aninjective or one-to-one function. Hence, a function is injective if for everya;b2Xsuch thatf(a) =f(b), thena=b. The graph of

a one-to-one function has the property that every horizontal line meets it in at most one point and if each

horizontal line meets the graph of a function in at most one point, then the function is one-to-one.

The function issurjectiveorontoif every element of the codomain is mapped to by at least one element of

the domain. Hence, a functionf:X!Yis surjective if the range offisY. A function isbijectiveif it is BOTH injective and surjective.Monotonic functions Iff(x1)< f(x2)wheneverx1< x2, thenf(x)is anincreasing function. Iff(x1)> f(x2)wheneverx1< x2, thenf(x)is adecreasing function. These two types of functions are calledmonotonic.Inverse functions Letf(x)be a one-to-one function. The functiong(x)that assigns to each output offthe corresponding unique input is called theinverseoff. The symbolf1denotes the inverse function.Ex.1 Determine the inverse of the following functions and then graph them. f(x) = 2x f(x) =x3 f(x) = 3x+ 2 3

Chapter 3: Functions Lecture notes Math

Section 1.3: Limits

Definition of limit 1

Thelimitoff(x)asxapproachesx0is the numberLif given any radius" >0aboutLthere exists a radius >0aboutx0such that for allx,

0 impliesjf(x)Lj< ". In other words, if the values of a functionf(x)approach the valueLasxapproaches x

0, we say thatfhas limitLasxapproachesx0and we write

lim x!x0f(x) =L: The limit off(x)asxapproachesx0from the rightis the numberLif given any radius" >0aboutLthere exists a radius >0aboutx0such that for allx, x

0< x < x0+

impliesjf(x)Lj< ". We write lim x!x+

0f(x) =L:

The limit off(x)asxapproachesx0from the leftis the numberLif given any radius" >0aboutLthere exists a radius >0aboutx0such that for allx, x

0 < x < x0

impliesjf(x)Lj< ". We write lim x!x

0f(x) =L:

A function has a limit asxapproachesx0if and only if the right-hand and left-hand limits atx0exist and

are equal.Ex.1 Find lim x!2x

24x2Ex.2

Let f(x) =2ifx3

1ifx <3

Findlimx!3f(x).Ex.3

Find lim x!53x5 4

Chapter 3: Functions Lecture notes Math

Ex.4 Show that the functiony= sin(1=x)has no limit asxapproaches zero from either side.

Proof: Asx!0, its reciprocal1x

becomes infinite and the value ofsin(1=x)cycles repeatedly from1to1. Thus there is no single numberLsuch that the function"s values get close to a single value whenx!0.

This is true even if we restrictxto positive values or to negative values, therefore the function has neither a

right-hand limit nor a left-hand limit asxapproaches zero. In conclusion, the functiony= sin(1=x)has no

limit from either side asx!0.Properties of limits

Iflimx!x0f(x) =L1andlimx!x0g(x) =L2, then

Sum rule:

limx!x0[f(x) +g(x)] = limx!x0f(x) + limx!x0g(x) =L1+L2

Difference rule:

lim x!x0[f(x)g(x)] = limx!x0f(x)limx!x0g(x) =L1L2

Product rule:

limx!x0[f(x)g(x)] = limx!x0f(x)limx!x0g(x) =L1L2

Constant multiple rule:

lim x!x0[kg(x)] =klimx!x0g(x) =kL2 for any numberk.

Quotient rule:

lim x!x0f(x)g(x)=limx!x0f(x)lim x!x0g(x)=L1L 2 ifL26= 0.Ex.5

Prove:

Iflimx!x0f(x) =L1andlimx!x0g(x) =L2, then

lim x!x0[f(x) +g(x)] =L1+L2 limx!2x+ 5 = 7 limx!5px1 = 2 5

Chapter 3: Functions Lecture notes Math

Ex.6

Compute the following limits:

limx!3x2(2x) limx!2x2+2x+4x+2 limx!5x253(x+5) limt!3+sin(t)1cos(t) limt!3p3t+7p7 2 limx!5x2252(x+5) limx!2(x3+ 3x22x17) limx!1+x+3x

3+3x+1

limx!2x+3x+6 limy!3y23ylimx!68(x5)(x7) limx!3px+ 7 limx!05p5x+4+2 limu!1u41u 31
limv!2v38v

416Definition of limit 2

Letf(x)be a function defined on an interval that containsx0, except possibly atx0. Then we say that lim x!x0f(x) = +1 if for everyM >0there is some number >0such thatf(x)> Mfor allxsuch that00such thatf(x)< Nfor allxsuch that0Compute the following limits:

limx!01x 2 limx!0+1x limx!01x

Definition of limit 3

Letf(x)be a function defined onx > Kfor someK. Then we say that lim x!+1f(x) =L if for every" >0there is some numberM >0such thatjf(x)Lj< "for allxsuch thatx > M. Letf(x)be a function defined onx < Kfor someK. Then we say that lim x!1f(x) =L if for every" >0there is some numberN <0such thatjf(x)Lj< "for allxsuch thatx < N.6

Chapter 3: Functions Lecture notes Math

Definition of limit 4

Letf(x)be a function defined onx > Kfor someK. Then we say that lim x!+1f(x) = +1 if for everyN >0there is some numberM >0such thatf(x)> Nfor allxsuch thatx > M. Letf(x)be a function defined onx < Kfor someK. Then we say that lim x!1f(x) = +1 if for everyN >0there is some numberM <0such thatf(x)> Nfor allxsuch thatx < M. In a similar way we can definelimx!+1f(x) =1andlimx!1f(x) =1.Ex.8

Compute the following limits:

limx!+11x+3 limx!171x limx!111x+22x31 limx!+12x235x+4 limx!2+7xx2 limx!1+x2+533x limx!2+53xx 26x+8
limx!0px+934x limx!1p8x143x5+x3x2+25x4x2+x5 limx!+15x24x2px+32x2x+px limx!1px1x1Sandwich Theorem Suppose thatg(x)f(x)h(x)for allx6=x0in some open interval aboutx0and that lim x!x0g(x) = limx!x0h(x) =L: Then limx!x0f(x) =L:Ex.9

Compute the following limits:

limx!0sin(x) limx!0cos(x) limx!0tan(x)Theorem

Ifis measured in radians, then

lim !0sin() = 1:7

Chapter 3: Functions Lecture notes Math

Ex.10

Compute the following limits:

limx!0sin(7x)7x limx!0sin(x=2)x=2 limx!0sin(6x)x limx!0tan(2x)5x limx!0sin(5x)sin(2x) limx!0xsin(1=x)Standard Limits

Limits to remember:

limx!0sin(x)x = 1 limx!01cos(x)x 2=12 limx!0tan(x)x = 1 limx!0ln(1+x)x = 1 limx!0loga(1+x)x =1ln(a), (a >0) limx!0ex1)x = 1 limx!0ax1x = ln(a), (a >0) limx!1(1 +1x )x=e limx!0(1+x)c1x =c, (c2R)Ex.11

Compute the following limits:

lim x!0log

3(1 + 3x)e

2x1 lim x!0sin(x)ln(1 +x) lim x!+1

1 +12x

3x lim x!0tan(2x)x lim x!0(1 +x)41x 8

Chapter 3: Functions Lecture notes Math

Section 1.4: Continuous Functions

Definition of continuity

A functionf(x)iscontinuous atx0if and only if it meets all three of the following conditions: f(x0)exists; limx!x0f(x)exists; limx!x0f(x) =f(x0).

Continuity at an endpoint:

A function is continuos at a left endpointaof its domain iflimx!a+f(x) =f(a). A function is continuos at a right endpointbof its domain iflimx!bf(x) =f(b).

A function iscontinuousif it is continuous at each point of its domain. If a functionfis not continuous at a

pointc, we say thatfisdiscontinuousatcand callca point of discontinuity off.Ex.1 Sine and Cosine are continuous atx= 0.Properties of continuous functions

Iffandgare continuous functions atx=c, then

Sum:f+g

Difference:fg

Product:fg

Constant multiple:kf, for any numberk.

Quotient:f=g, providedg(c)6= 0.

are continuous functions atx=c. Moreover, iffis continuous atcandgis continuous atf(c), thengfis continuous atc.Ex.2

The following functions

f(x) = 3x5x2+1x

2+2f(x) = 4xcos(x)

f(x) = tan(x) are continuous.Removable and non-removable discontinuities

One single type of discontinuity, called aremovable discontinuity, occurs wheneverlimx!cf(x)6=f(c). We

remove the discontinuity by definingf(c)to have the same value aslimx!cf(x)6=f(c). The removability of a discontinuity of a function at a pointx=crequires the existence oflimx!cf(x) =

f(c). Without it, there is no way to fulfill the conditions of the continuity test, and the discontinuity is

non-removable.Ex.3

The function

f(x) =x2+x6x 24
is not defined atx= 2. Isx= 2a removable discontinuity? If so, how can you extend the function to make it continuous atx= 2? 9

Chapter 3: Functions Lecture notes Math

Ex.4

Solve the following problems:

Compute

lim x!3x

27x+ 12x3

Compute

lim x!4x

2+x20x4

Compute

lim t!1t

23t+ 2t1

Let f(x) =1 +x2ifx <2 x 3ifx2 Findlimx!2f(x)andlimx!2+f(x). Doeslimx!2f(x)exist? Let f(x) =5x+ 7ifx <3 x

216ifx3

Doeslimx!3f(x)exist?

Let f(t) =tift <1 t 2ift1

Doeslimt!1f(t)exist?

Suppose the total costC(Q)of producing a quantityQof a product equals a fixed cost of$1000plus $3times the quantity produced. (1) WriteC(Q). (2) Find the average cost per unit quantityA(Q). (3) Compute lim

Q!0+A(Q)

10

Chapter 3: Functions Lecture notes Math

Section 1.5: The Intermediate Value Theorem for Continuous Functions

Intermediate Value Theorem

A functiony=f(x)that is continuous on a closed intervalI= [a;b]takes on every value betweenf(a)and f(b).Connectivity Suppose we want to graph a functiony=f(x)that is continuous throughout some intervalIon thex-axis. The Intermediate Value Theorem tells us that the graph offoverIwill never move from oney-value to

another without taking on they-values in between. The graph offoverIwill be connected, that is it will

consist of a single, unbroken curve.Root finding

Suppose thatf(x)is continuous at every point of a closed interval[a;b]and thatf(a)andf(b)differ in sign.

Then zero lies betweenf(a)andf(b)differ in sign, so there is at least one numbercbetweenaandbwhere

f(c) = 0. In other words, iff(x)is continuous andf(a)andf(b)differ in sign, then the equationf(x) = 0

has at least one solution in the open interval(a;b). A pointcwheref(c) = 0is called azeroorrootoff. Hence, the zeros offare the points where the graph offintersects thex-axis.Ex.1 Is any real number exactly1less than its cube?Ex.2 Show thatx3x1 = 0has a root somewhere in the interval[1;2]. 11

Chapter 3: Functions Lecture notes Math

Section 1.6: Extreme Value Theorem

Maxima and Minima

Suppose thatfis a function which is continuous on the closed interval[a;b]. Then there exist real numbers

canddin[a;b]such that We say thatf(x)has anabsolute (or global) maximumatx=cif for everyxin the domain we are working on we havef(x)f(c). We say thatf(x)has arelative (or local) maximumatx=cif for everyxin some open interval around x=c,f(x)f(c). We say thatf(x)has anabsolute (or global) minimumatx=dif for everyxin the domain we are working on we havef(x)f(d). We say thatf(x)has arelative (or local) minimumatx=dif for everyxin some open interval around x=d,f(x)f(d). A functionfdefined onXis calledbounded, if there exists a real numberMsuch thatjf(x)j Mfor allx inX. A function that is not bounded is said to beunbounded. Iff(x)Afor allxinX, then the function is

said to bebounded abovebyA. Iff(x)Bfor allxinX, then the function is said to bebounded belowbyB.Extreme Value Theorem

Suppose thatfis a function which is continuous on the closed interval[a;b]. Then there exist real numbers

canddin[a;b]such that fhas a maximum value atx=cand fhas a minimum value atx=d.Section 1.7: Piecewise and Uniform Continuous Functions

Piecewise continuity and uniform continuity

A function or curve ispiecewise continuousif it is continuous on all but a finite number of points at which

certain matching conditions are sometimes required.

A functionfisuniformly continuousif it is possible to guarantee thatf(x)andf(y)are as close to each other

as we please by requiring only that x and y are sufficiently close to each other: for every" >0there is >0

such that for everyx;y2Iwithjyxj< , thenjf(x)f(y)j< ".

Every uniformly continuous function is continuous, but the converse does not hold. Consider for instance

the functionf:R!R,x7!x2. Given an arbitrarily small positive real number", uniform continuity requires the existence of a positive numbersuch that for allx1;x2withjx1x2j< , we have jf(x1)f(x2)j< ". But f(x+)f(x) = 2x+2=(2x+); and for all sufficiently largexthis quantity is greater than".Ex.1

The function

f(x) =8 :x+ 4ifx <0 x

2if0< x <5

7ifx5 is piecewise continuous.Ex.2

The functionf(x) = 4x1is uniformly continuous.

12

Chapter 3: Functions Lecture notes Math

Section 1.8: Economic Applications of Continuous and Discontinuous Functions

Introduction

There are many natural examples of discontinuities from economics. In fact, economists often adopt conti-

nuous functions to represent economic relationships (that is, they build a continuous model) when the use

of discontinuous functions would be a more literal interpretation of reality. It is important to know when

the simplifying assumption of continuity can be safely made for the sake of convenience and when it is

likely to distort the true relationship between economic variables too much.Ex.1

reality. The first step in modeling the decisions of a firm is usually the analysis of the available technology.

This relationship between inputs used and outputs generated is generally presumed to be represented by

some production function:y=f(x). What does it mean to say that this function is continuous on some domain (usuallyx0)? To assume thatf(x)is continuous at a pointx=cimplies thatf(x)is defined on

some open interval of real numbers containingc. This meansxmust beinfinitely divisible: one can choose

xto be a value that deviates even by infinitesimal amounts fromx=c.

An example of input that would not be infinitely divisible would be bolts used in the production of a

car. Since one would not use a fraction like a half of a bolt, it would only make literal sense to treat bolts

as integer valued. Therefore, it does not make sense to contemplate an open interval of points including

some valuex=cbolts. However, if we denote byxthe number of bolts used and byythe number of cars produced, we have y=x1;050

then using the closest value that is a multiple of1;050would probably be reasonably accurate. Thus, even

if a commodity is not infinitely divisible, we may often assume that it is, without distorting realty very

much. Draw the graph of this liner function considering the domain of real numbersx0.Ex.2quotesdbs_dbs14.pdfusesText_20

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