2 Functions and Their Graphs The Algebra of Functions Functions and Mathematical Models Limits One-Sided Limits and Continuity The Derivative.

22

Functions and Their GraphsFunctions and Their Graphs The Algebra of FunctionsThe Algebra of Functions Functions and Mathematical ModelsFunctions and Mathematical Models LimitsLimits One-Sided Limits and ContinuityOne-Sided Limits and Continuity The DerivativeThe Derivative

Functions, Limits and the DerivativeFunctions, Limits and the Derivative

2.12.1Functions and Their GraphsFunctions and Their Graphs

x

y

(5, 3)

4

3

2

1

–1

–2

1 2 3 4 5 6 7 8

FunctionsFunctions

Function:Function: A function is a A function is a rulerule that assigns to each element that assigns to each element in a set in a set AA one and only one element in a set one and only one element in a set BB..

The set The set AA is called the is called the domaindomain of the function. of the function.

It is customary to denote a function by a It is customary to denote a function by a letterletter of the of the alphabet, such as the letter alphabet, such as the letter ff..

FunctionsFunctions

The element in The element in BB that that ff associates with associates with xx is written is written ff((xx)) and is and is called the value of called the value of f f at at xx..

The set of all the possible values of The set of all the possible values of ff((xx)) resulting from all the resulting from all the possible values of possible values of xx in its domain, is called the in its domain, is called the rangerange of of ff((xx))..

The output The output ff((xx)) associated with an input associated with an input xx is is uniqueunique: : ✦ EachEach xx must correspond to must correspond to one and onlyone and only oneone value of value of ff((xx))..

ExampleExample

Let the function Let the function ff be defined by the rulebe defined by the rule

✦ Find: Find: ff(1)(1)✦ Solution:Solution:

22 1f x x x

21 2 1 1 1 2 1 1 2f

Example 1, page 51

ExampleExample


✦ Find: Find: ff(( – 2)2)✦ Solution:Solution:

22 1f x x x

22 2 2 2 1 8 2 1 11f

Example 1, page 51

ExampleExample


✦ Find: Find: ff((a))✦ Solution:Solution:

22 1f x x x

2 22 1 2 1f a a a a a

Example 1, page 51

ExampleExample


✦ Find: Find: ff((a + h))✦ Solution:Solution:

22 1f x x x

2 2 22 1 2 4 2 1f a h a h a h a ah h a h

Example 1, page 51

Applied ExampleApplied Example

ThermoMaster manufactures an indoor-outdoor ThermoMaster manufactures an indoor-outdoor thermometer at its Mexican subsidiary.thermometer at its Mexican subsidiary.

Management estimates that the Management estimates that the profitprofit (in dollars) (in dollars) realizable by ThermoMaster in the manufacture and realizable by ThermoMaster in the manufacture and salesale of of xx thermometers per week is thermometers per week is

Find ThermoMaster’s weekly Find ThermoMaster’s weekly profitprofit if its if its level of level of productionproduction is: is:

a.a. 10001000 thermometers per week. thermometers per week.

b.b. 20002000 thermometers per week. thermometers per week.

20.001 8 5000P x x x

Applied Example 2, page 51


SolutionSolution We haveWe have

a.a. The weekly The weekly profitprofit by by producingproducing 10001000 thermometers is thermometers is

or or $2,000$2,000..

b.b. The weekly The weekly profitprofit by by producingproducing 20002000 thermometers is thermometers is

or or $7,000$7,000..

21000 0.001 1000 8 1000 5000 2000P =

22000 0.001 2000 8 2000 5000 7000P =

20.001 8 5000P x x x


Determining the Domain of a FunctionDetermining the Domain of a Function

Suppose we are given the function Suppose we are given the function y y = = ff((xx)).. Then, the variable Then, the variable xx is called the is called the independent variableindependent variable.. The variable The variable yy, whose value depends on , whose value depends on xx, is called the , is called the

dependent variabledependent variable.. To determine the To determine the domaindomain of a function, we need to find of a function, we need to find

what what restrictionsrestrictions, if any, are to be placed on the , if any, are to be placed on the independent variable independent variable xx..

In many practical problems, the domain of a function is In many practical problems, the domain of a function is dictated by the dictated by the nature of the problemnature of the problem..

x

16 – 2x

Applied Example:Applied Example: Packaging Packaging

An open box is to be made from a rectangular piece of An open box is to be made from a rectangular piece of cardboard cardboard 1616 inches wide by cutting away identical inches wide by cutting away identical squares (squares (xx inches by inches by xx inches) from each corner and inches) from each corner and folding up the resulting flaps.folding up the resulting flaps.

10 10 – 2x

16

x

x x



An open box is to be made from a rectangular piece of An open box is to be made from a rectangular piece of cardboard cardboard 1616 inches wide by cutting away identical inches wide by cutting away identical squares (squares (xx inches by inches by xx inches) from each corner and inches) from each corner and folding up the resulting flaps.folding up the resulting flaps.

a.a. Find the expression that gives the Find the expression that gives the volumevolume VV of the box as of the box as a function of a function of xx. .

b.b. What is the What is the domaindomain of the function? of the function?

The The dimensionsdimensions of the of the resulting box are:resulting box are:

x

16 – 2x10 – 2x


x

16 – 2x


SolutionSolutiona.a. The The volumevolume of the box is given by of the box is given by multiplying its multiplying its

dimensionsdimensions (length (length ☓☓ width width ☓☓ height) height), so:, so:

10 – 2x

2

3 2

16 2 10 2

160 52 4

4 52 160

V f x x x x

x x x

x x x



SolutionSolutionb.b. Since theSince the length length of each of each sideside of the box must be of the box must be greatergreater

than or than or equalequal to to zerozero, we see that, we see that

must be satisfied simultaneously. Simplified:must be satisfied simultaneously. Simplified:

All three are satisfied simultaneously provided that:All three are satisfied simultaneously provided that:

Thus, the Thus, the domaindomain of the function of the function ff is the interval is the interval [0, 5][0, 5]..

8 5 0x x x =

16 2 0 10 2 0 0x x x

0 5x


More ExamplesMore Examples

Find the domain of the function:Find the domain of the function:

SolutionSolution Since the square root of a negative number is Since the square root of a negative number is undefinedundefined, it , it

is necessary that is necessary that xx – 1 – 1 0 0.. Thus the Thus the domaindomain of the function is of the function is [1,[1,)). .

1f x x 1f x x

Example 4, page 52



SolutionSolution Our only constraint is that you Our only constraint is that you cannot divide by zerocannot divide by zero, so , so

Which means that Which means that

Or more specifically Or more specifically x x ≠≠ –2–2 and and x x ≠ ≠ 22.. Thus the Thus the domaindomain of of ff consists of the intervals consists of the intervals (–(– , –2), –2), ,

(–2, 2)(–2, 2), , (2, (2, )). .

2 4 0x

2

1

4f x

x

2 4 2 2 0x x x 2 4 2 2 0x x x

Example 4, page 52



SolutionSolution Here, any real number satisfies the equation, so the Here, any real number satisfies the equation, so the

domaindomain of of ff is the set of is the set of all real numbersall real numbers..

2 3f x x

Example 4, page 52

Graphs of FunctionsGraphs of Functions

If If f f is a function with domain is a function with domain AA, then corresponding to , then corresponding to each real number each real number xx in in AA there is there is precisely oneprecisely one real real number number ff((xx))..

Thus, a function Thus, a function ff with domain with domain AA can also be defined as the can also be defined as the set of all set of all ordered pairsordered pairs ((x, fx, f((xx)))) where where xx belongs to belongs to AA..

The The graph of a functiongraph of a function ff is the set of all points is the set of all points ((xx, , yy)) in the in the xyxy-plane such that -plane such that xx is the domain of is the domain of ff and and y = fy = f((xx))..

ExampleExample

The graph of a function The graph of a function ff is shown below: is shown below:

x

y

Domain

Range

x

y(x, y)

Example 5, page 53

ExampleExample

The graph of a function The graph of a function ff is shown below: is shown below:✦ What is the value of What is the value of ff(2)(2)??

x

y

4

3

2

1

–1

–2(2, –2)

1 2 3 4 5 6 7 8

Example 5, page 53

ExampleExample

The graph of a function The graph of a function ff is shown below: is shown below:✦ What is the value of What is the value of ff(5)(5)??

x

y

(5, 3)

4

3

2

1

–1

–2

1 2 3 4 5 6 7 8

Example 5, page 53

ExampleExample

The graph of a function The graph of a function ff is shown below: is shown below:✦ What is the What is the domaindomain of of ff((xx))??

x

y

4

3

2

1

–1

–2

Domain: [1,8]

1 2 3 4 5 6 7 8

Example 5, page 53

ExampleExample

The graph of a function The graph of a function ff is shown below: is shown below:✦ What is the What is the rangerange of of ff((xx))??

x

y

4

3

2

1

–1

–2

Range: [–2,4]

1 2 3 4 5 6 7 8

Example 5, page 53

Example:Example: Sketching a Graph Sketching a Graph

Sketch the graph of the function defined by the equation Sketch the graph of the function defined by the equation

yy = = xx22 + 1 + 1SolutionSolution The The domaindomain of the function is the set of of the function is the set of all real numbersall real numbers.. Assign several valuesAssign several values to the variable to the variable x x and and compute the compute the

corresponding valuescorresponding values for for yy: :

xx yy

––33 1010––22 55––11 2200 1111 2222 5533 1010

Example 6, page 54

– – 33 – 2 – 2 – 1 – 1 11 22 33

1010

88

66

44

22



yy = = xx22 + 1 + 1SolutionSolution The The domaindomain of the function is the set of of the function is the set of all real numbersall real numbers.. Then Then plotplot these values in a these values in a graphgraph: :

xx

yy

xx yy

––33 1010––22 55––11 2200 1111 2222 5533 1010

Example 6, page 54

– – 33 – 2 – 2 – 1 – 1 11 22 33

1010

88

66

44

22



yy = = xx22 + 1 + 1SolutionSolution The The domaindomain of the function is the set of of the function is the set of all real numbersall real numbers.. And finally, And finally, connect the dotsconnect the dots: :

xx

yy

xx yy

––33 1010––22 55––11 2200 1111 2222 5533 1010

Example 6, page 54



SolutionSolution The function The function ff is defined in a piecewise fashion on the set is defined in a piecewise fashion on the set

of of all real numbersall real numbers.. In the In the subdomainsubdomain (–(– , 0), 0), the , the rulerule for for ff is given by is given by

In the In the subdomainsubdomain [0, [0, )), the , the rulerule for for ff is given by is given by

if 0

if 0

x xf x

x x

if 0

if 0

x xf x

x x

f x x f x x

f x x f x x

Example 7, page 55



SolutionSolution Substituting Substituting negative valuesnegative values for for xx into , while into , while

substituting substituting zero and positive valueszero and positive values into into we get: we get:

if 0

if 0

x xf x

x x

if 0

if 0

x xf x

x x

f x x f x x f x x f x x

xx yy

––33 33––22 22––11 1100 0011 1122 1.411.4133 1.731.73

Example 7, page 55



SolutionSolution PlottingPlotting these data and these data and graphinggraphing we get: we get:

if 0

if 0

x xf x

x x

if 0

if 0

x xf x

x x

f x x f x x

f x x f x x

– – 33 – 2 – 2 – 1 – 1 11 22 33

33

22

11

xx

yyxx yy

––33 33––22 22––11 1100 0011 1122 1.411.4133 1.731.73

Example 7, page 55

2.22.2The Algebra of FunctionsThe Algebra of Functions

1990 1992 1994 1996 1998 2000

2000

1800

1600

1400

1200

1000

t

y

Bill

ions

of D

olla

rs

Year

y = S(t)

y = R(t)

t

R(t)

S(t)

The Sum, Difference, Product and Quotient The Sum, Difference, Product and Quotient of Functions of Functions

Consider the graph below:Consider the graph below:✦ R(t)R(t) denotes the federal government denotes the federal government revenuerevenue at any time at any time tt..✦ S(t)S(t) denotes the federal government denotes the federal government spendingspending at any time at any time tt..

19901990 19921992 19941994 19961996 19981998 20002000

20002000

18001800

16001600

14001400

12001200

10001000

tt

yy

Bil

lion

s of

Dol

lars

Bil

lion

s of

Dol

lars

YearYear

y y = = SS((tt))

y y = = RR((tt))

tt

RR((tt))

SS((tt))

DD((tt) = ) = RR((tt) ) – – SS((tt))


Consider the graph below:Consider the graph below:✦ The difference The difference RR((tt) ) – S– S((tt)) gives the gives the budget deficitbudget deficit (if negative)(if negative)

or or surplussurplus (if positive)(if positive) in billions of dollars at any time in billions of dollars at any time tt..

19901990 19921992 19941994 19961996 19981998 20002000

20002000

18001800

16001600

14001400

12001200

10001000

tt

yy

Bil

lion

s of

Dol

lars

Bil

lion

s of

Dol

lars

YearYear

y y = = SS((tt))

y y = = RR((tt))

tt

RR((tt))

SS((tt))


The budget balanceThe budget balance D D((tt)) is shown below: is shown below:✦ DD((tt)) is also a is also a function function that denotes the federal government that denotes the federal government

deficit (surplus)deficit (surplus) at any time at any time tt. . ✦ This function is the This function is the differencedifference of the two function of the two function R R and and SS..✦ DD((tt) ) has the has the same domainsame domain as as RR((tt) ) andand S S((tt))..

19921992 19941994 19961996 19981998 20002000

400400

200200

00

–– 200200

–– 400400

tt

yy

Bil

lion

s of

Dol

lars

Bil

lion

s of

Dol

lars

YearYear

tt DD((tt))

y y = = DD((tt))


Most functions are built up from other, generally Most functions are built up from other, generally simpler functions.simpler functions.

For example, we may view the function For example, we may view the function ff((xx) = 2) = 2xx + 4 + 4 as the as the sumsum of the two functions of the two functions gg((xx) = 2) = 2xx and and hh((xx) = 4) = 4..

( )

( )

f f xx

g g x

( )

( )

f f xx

g g x

The Sum, Difference, Product and Quotient of FunctionsThe Sum, Difference, Product and Quotient of Functions

Let Let ff and and gg be functions with domains be functions with domains AA and and BB, respectively., respectively. The The sumsum f + gf + g, the , the differencedifference f – gf – g, and the , and the productproduct fgfg of of ff

and and gg are are functionsfunctions with domain with domain A A ∩∩ B B and rule given byand rule given by

((ff + + gg)()(xx) = ) = ff((xx) + ) + gg((xx)) SumSum((ff –– gg)()(xx) = ) = ff((xx) ) –– gg((xx)) DifferenceDifference

((fgfg)()(xx) = ) = ff((xx))gg((xx)) ProductProduct

The The quotientquotient ff//gg of of f f and and gg has domain has domain A A ∩∩ B B excluding all excluding all numbers numbers xx such that such that gg((xx)) = 0= 0 and rule given by and rule given by

QuotientQuotient

ExampleExample

LetLet and and gg((xx) = 2) = 2xx + 1 + 1. . Find the Find the sumsum ss, the , the differencedifference dd, the , the productproduct pp, and the , and the

quotientquotient qq of the functions of the functions ff and and gg..

SolutionSolution Since the domain of Since the domain of ff is is AA = [–1, = [–1,)) and the domain of and the domain of gg

is is BB = (– = (– , , )), we see that the , we see that the domaindomain of of ss, , dd, and , and pp is is A A ∩∩ B B = [–1,= [–1,))..

The rules are as follows:The rules are as follows:

( ) 1f x x ( ) 1f x x

( ) ( )( ) ( ) ( ) 1 2 1

( ) ( )( ) ( ) ( ) 1 2 1

( ) ( )( ) ( ) ( ) 2 1 1

s x f g x f x g x x x

d x f g x f x g x x x

p x fg x f x g x x x

( ) ( )( ) ( ) ( ) 1 2 1

( ) ( )( ) ( ) ( ) 1 2 1

( ) ( )( ) ( ) ( ) 2 1 1

s x f g x f x g x x x

d x f g x f x g x x x

p x fg x f x g x x x

Example 1, page 68

ExampleExample

LetLet and and gg((xx) = 2) = 2xx + 1 + 1. . Find the Find the sumsum ss, the , the differencedifference dd, the , the productproduct pp, and the , and the

quotientquotient qq of the functions of the functions ff and and gg..

SolutionSolution The The domaindomain of the of the quotientquotient function is function is [–1,[–1,)) together together

with the restriction with the restriction xx ≠≠ –– ½½ ..

Thus, the domain is Thus, the domain is [–1, –[–1, – ½) ½) (– (– ½,½,)).. The rule is as follows:The rule is as follows:

( ) 1f x x ( ) 1f x x

( ) 1( ) ( )

( ) 2 1

f f x xq x x

g g x x

( ) 1( ) ( )

( ) 2 1

f f x xq x x

g g x x

Example 1, page 68


Suppose Puritron, a manufacturer of water filters, has a Suppose Puritron, a manufacturer of water filters, has a monthly monthly fixed costfixed cost of of $10,000$10,000 and a variable cost of and a variable cost of

–– 0.00010.0001xx22 + 10 + 10xx (0 (0 xx 40,000) 40,000)

dollars, where dollars, where xx denotes the number of denotes the number of filters filters manufacturedmanufactured per month. per month.

Find a function Find a function CC that gives the that gives the total monthly costtotal monthly cost incurred by Puritron in the manufacture of incurred by Puritron in the manufacture of xx filters. filters.



SolutionSolution Puritron’s monthly Puritron’s monthly fixed costfixed cost is always is always $10,000$10,000, so it can , so it can

be described by the constant function: be described by the constant function:

FF((xx)) = 10,000= 10,000 The The variable costvariable cost can be described by the function: can be described by the function:

VV((xx)) = –= – 0.00010.0001xx22 + 10 + 10xx The The total costtotal cost is the is the sumsum of the fixed cost of the fixed cost FF and the and the

variable cost variable cost VV::

CC((xx)) = = VV((xx) + ) + FF((xx) )

= –= – 0.00010.0001xx22 + 10 + 10x x + 10,000+ 10,000 (0 (0 xx 40,000)40,000)



Lets now consider Lets now consider profitsprofits Suppose that the Suppose that the total revenuetotal revenue RR realized by Puritron realized by Puritron

from the sale of from the sale of xx water filters is given by water filters is given by

RR((xx)) = –= – 0.00050.0005xx22 + 20 + 20x x (0 (0 ≤≤ xx ≤≤ 40,000) 40,000) FindFind

a.a. The total profit function for Puritron.The total profit function for Puritron.

b.b. The total profit when Puritron produces The total profit when Puritron produces 10,00010,000 filters per filters per month.month.



SolutionSolution

a.a. The The totaltotal profitprofit PP realized by the firm is the difference realized by the firm is the difference between the between the total revenuetotal revenue R R and the and the total costtotal cost CC::

PP((xx)) = = RR((xx) – ) – CC((xx) )

= (–= (– 0.00050.0005xx22 + 20 + 20xx)) – (–– (– 0.00010.0001xx22 + 10 + 10x x + 10,000)+ 10,000)

= –= – 0.00040.0004xx22 + 10 + 10x x – 10,000– 10,000

b.b. The The totaltotal profitprofit realized by Puritron when realized by Puritron when producingproducing 10,00010,000 filters per month isfilters per month is

PP((xx)) = –= – 0.0004(10,000)0.0004(10,000)22 + 10(10,000) + 10(10,000) – 10,000– 10,000

= 50,000= 50,000

or or $50,000$50,000 per month. per month.


The Composition of Two FunctionsThe Composition of Two Functions

Another way to Another way to build a functionbuild a function from other functions is from other functions is through a process known as the composition of functions.through a process known as the composition of functions.

Consider the functions Consider the functions ff and and gg: :

Evaluating the function Evaluating the function gg at the pointat the point ff((xx)), we find that:, we find that:

This is an entirely This is an entirely new functionnew function, which we could call , which we could call hh::

2( ) 1f x x 2( ) 1f x x ( )g x x

2( ) ( ) 1g f x f x x 2( ) ( ) 1g f x f x x

2( ) 1h x x 2( ) 1h x x

The Composition of Two FunctionsThe Composition of Two Functions

Let Let ff and and gg be functions. be functions. Then the Then the compositioncomposition of of gg and and f f is the function is the function

ggff (read “(read “gg circlecircle f f ”)”) defined by defined by

((ggff )()(xx) = ) = gg((ff((xx))))

The The domaindomain of of ggff is the set of all is the set of all xx in the domain in the domain of of ff such that such that ff((xx)) lies in the domain of lies in the domain of gg..

ExampleExample

LetLet Find:Find:

a.a. The rule for the composite function The rule for the composite function ggff..

b.b. The rule for the composite function The rule for the composite function ffgg..

SolutionSolution To find To find ggff, evaluate the function , evaluate the function gg at at ff((xx))::

To find To find ffgg, evaluate the function , evaluate the function ff at at gg((xx))::

2( ) 1 ( ) 1an d .f x x g x x 2( ) 1 ( ) 1an d .f x x g x x

2( )( ) ( ( )) ( ) 1 1 1g f x g f x f x x 2( )( ) ( ( )) ( ) 1 1 1g f x g f x f x x

2 2( )( ) ( ( )) ( ( )) 1 ( 1) 1

2 1 1 2

f g x f g x g x x

x x x x

2 2( )( ) ( ( )) ( ( )) 1 ( 1) 1

2 1 1 2

f g x f g x g x x

x x x x

Example 4, page 70


An environmental impact study conducted for the city of An environmental impact study conducted for the city of Oxnard indicates that, under existing environmental Oxnard indicates that, under existing environmental protection laws, protection laws, the level of carbon monoxidethe level of carbon monoxide (CO) (CO) present in the air due to pollution from automobile present in the air due to pollution from automobile exhaust will be exhaust will be 0.010.01xx2/32/3 parts per million when the parts per million when the number ofnumber of motor vehiclesmotor vehicles isis xx thousand. thousand.

A separate study conducted by a state government A separate study conducted by a state government agency estimates that agency estimates that tt years from now the years from now the number of number of motor vehiclesmotor vehicles in Oxnard will be in Oxnard will be 0.20.2tt22 + 4 + 4tt + 64 + 64 thousand. thousand.

Find:Find:

a.a. An expression for the concentration of CO in the air due An expression for the concentration of CO in the air due to automobile exhaust to automobile exhaust tt years from now. years from now.

b.b. The level of concentration The level of concentration 55 years from now.years from now.


Applied ExampleApplied ExampleSolutionSolution Part Part (a)(a)::

✦ The level of CO is described by the function The level of CO is described by the function

gg((xx) =) = 0.010.01xx2/32/3

where where xx is the number (in thousands) of motor vehicles. is the number (in thousands) of motor vehicles.

✦ In turn, the number (in thousands) of motor vehicles is In turn, the number (in thousands) of motor vehicles is described by the functiondescribed by the function

ff((xx) = 0.2) = 0.2tt22 + 4 + 4tt + 64 + 64

where where tt is the number of years from now. is the number of years from now.

✦ Therefore, the concentration of CO due to automobile Therefore, the concentration of CO due to automobile exhaust exhaust tt years from now is given by years from now is given by

((ggff )()(tt) = ) = gg((ff((tt)) = 0.01(0.2)) = 0.01(0.2tt22 + 4 + 4tt + 64) + 64)2/32/3


Applied ExampleApplied ExampleSolutionSolution Part Part (b)(b)::

✦ The level of CO The level of CO five yearsfive years from now is: from now is:

((ggff )(5))(5) = = gg((ff(5)) = 0.01[0.2(5)(5)) = 0.01[0.2(5)22 + 4(5) + 64] + 4(5) + 64]2/32/3

= (0.01)89= (0.01)892/32/3 ≈≈ 0.20 0.20

or approximately or approximately 0.200.20 parts per million. parts per million.


2.32.3Functions and Mathematical ModelsFunctions and Mathematical Models

5 10 15 20 25 30

6

4

2

t (years)

y ($trillion)

Mathematical ModelsMathematical Models

As we have seen, mathematics can be used to solve real-As we have seen, mathematics can be used to solve real-world problems.world problems.

We will now discuss a few more examples of real-world We will now discuss a few more examples of real-world phenomena, such as:phenomena, such as:✦ The solvency of the U.S. Social Security trust fund (p.79)The solvency of the U.S. Social Security trust fund (p.79)

✦ Global warming (p. 78)Global warming (p. 78)

Mathematical ModelingMathematical Modeling

Regardless of the field from which the real-world problem Regardless of the field from which the real-world problem is drawn, the problem is analyzed using a process called is drawn, the problem is analyzed using a process called mathematical modelingmathematical modeling..

The The four stepsfour steps in this process are: in this process are:

Real-world problemReal-world problem Mathematical Mathematical modelmodel

Solution of real-Solution of real-world problemworld problem

Solution of Solution of mathematical modelmathematical model

FormulateFormulate

InterpretInterpret

SolveSolveTestTest

Modeling With Polynomial FunctionsModeling With Polynomial Functions

A A polynomial functionpolynomial function of degree of degree nn is a function of the form is a function of the form

where where n n is a nonnegative integer and the numbers is a nonnegative integer and the numbers aa00, , aa11, …. , …. aann are are constantsconstants called the called the coefficientscoefficients of of

the polynomial function.the polynomial function. Examples:Examples:

✦ The function below is polynomial function of The function below is polynomial function of degree 5degree 5::

1 21 2 1 0 (( ) 0) n n

n n nf x a x a x a x aa x a 1 21 2 1 0 (( ) 0) n n

n n nf x a x a x a x aa x a

5 4 3 212( ) 2 3 2 6f x x x x x 5 4 3 212( ) 2 3 2 6f x x x x x

Modeling With Polynomial FunctionsModeling With Polynomial Functions

A A polynomial functionpolynomial function of degree of degree nn is a function of the form is a function of the form

where n is a nonnegative integer and the numbers where n is a nonnegative integer and the numbers aa00, , aa11, …. , …. aann are are constantsconstants called the called the coefficientscoefficients of of

the polynomial function.the polynomial function. Examples:Examples:

✦ The function below is polynomial function of The function below is polynomial function of degree 3degree 3::3 2( ) 0.001 0.2 10 200g x x x x 3 2( ) 0.001 0.2 10 200g x x x x

1 21 2 1 0 (( ) 0) n n

n n nf x a x a x a x aa x a 1 21 2 1 0 (( ) 0) n n

n n nf x a x a x a x aa x a

Applied ExampleApplied ExampleMarket for Cholesterol-Reducing DrugsMarket for Cholesterol-Reducing Drugs

In a study conducted in early In a study conducted in early 20002000, experts projected a , experts projected a rise in the market for cholesterol-reducing drugs.rise in the market for cholesterol-reducing drugs.

The U.S. market (in billions of dollars) for such drugs The U.S. market (in billions of dollars) for such drugs from from 19991999 through through 20042004 was was

A A mathematical modelmathematical model giving the approximate U.S. giving the approximate U.S. market over the period in question is given by market over the period in question is given by

MM((tt) = 1.95) = 1.95tt + 12.19 + 12.19

where where tt is measured in years, with is measured in years, with t t = 0= 0 for for 19991999..

YearYear 19991999 20002000 20012001 20022002 20032003 20042004

MarketMarket 12.0712.07 14.0714.07 16.2116.21 18.2818.28 20.0020.00 21.7221.72



MM((tt) = 1.95) = 1.95tt + 12.19 + 12.19

a.a. Sketch the Sketch the graphgraph of the function of the function MM and the given data and the given data on the same set of axes.on the same set of axes.

b.b. Assuming that the Assuming that the projectionprojection held and the trend held and the trend continued, what was the market for cholesterol-continued, what was the market for cholesterol-reducing drugs in reducing drugs in 20052005 ((tt = 6 = 6))??

c.c. What was the What was the rate of increaserate of increase of the market for of the market for cholesterol-reducing drugs over the period in question?cholesterol-reducing drugs over the period in question?

YearYear 19991999 20002000 20012001 20022002 20032003 20042004

MarketMarket 12.0712.07 14.0714.07 16.2116.21 18.2818.28 20.0020.00 21.7221.72



MM((tt) = 1.95) = 1.95tt + 12.19 + 12.19SolutionSolution

a.a. Graph:Graph:

YearYear 19991999 20002000 20012001 20022002 20032003 20042004

MarketMarket 12.0712.07 14.0714.07 16.2116.21 18.2818.28 20.0020.00 21.7221.72

1 2 3 4 5

25

20

15

t

y

Bil

lion

s of

Dol

lars

Year

MM((tt))




b.b. The projected market in The projected market in 20052005 for cholesterol-reducing for cholesterol-reducing drugs wasdrugs was

MM(6) = 1.95(6) + 12.19 = 23.89(6) = 1.95(6) + 12.19 = 23.89

or or $23.89$23.89 billion. billion.

YearYear 19991999 20002000 20012001 20022002 20032003 20042004

MarketMarket 12.0712.07 14.0714.07 16.2116.21 18.2818.28 20.0020.00 21.7221.72




c.c. The function The function MM is is linearlinear, and so we see that the , and so we see that the rate of rate of increaseincrease of the market for cholesterol-reducing drugs is of the market for cholesterol-reducing drugs is given by given by the slope of the straight linethe slope of the straight line represented by represented by MM, , which is approximately which is approximately $1.95$1.95 billion per year.billion per year.

YearYear 19991999 20002000 20012001 20022002 20032003 20042004

MarketMarket 12.0712.07 14.0714.07 16.2116.21 18.2818.28 20.0020.00 21.7221.72


Modeling a Polynomial Function of Degree 2Modeling a Polynomial Function of Degree 2

A polynomial function of A polynomial function of degree 2degree 2 has the form has the form

Or more simply, Or more simply, yy = = axax22 + + bxbx + + cc, and is called a , and is called a quadratic functionquadratic function..

The graph of a quadratic function is a The graph of a quadratic function is a parabolaparabola::

22 1 20( ) ( 0) f x a x a x a a 22 1 20( ) ( 0) f x a x a x a a

x

y

x

yOpens upwards if

a > 0

Opens downwards if

a < 0

Applied ExampleApplied ExampleGlobal WarmingGlobal Warming

The increase in The increase in carbon dioxide (COcarbon dioxide (CO22)) in the atmosphere is in the atmosphere is

a major cause of global warming.a major cause of global warming. Below is a table showing the Below is a table showing the average amount of COaverage amount of CO22, ,

measured in measured in parts per million volume (ppmv)parts per million volume (ppmv) for various for various years from years from 19581958 through through 20072007::

YearYear 19581958 19701970 19741974 19781978 19851985 19911991 19981998 20032003 20072007

AmountAmount 315315 325325 330330 335335 345345 355355 365365 375375 380380



Below is a Below is a scatter plotscatter plot associated with these data: associated with these data:

YearYear 19581958 19701970 19741974 19781978 19851985 19911991 19981998 20032003 20072007

AmountAmount 315315 325325 330330 335335 345345 355355 365365 375375 380380

1010 2020 3030 4040 5050

380380

360360

340340

320320

t t (years)(years)

y y (ppmv)(ppmv)



A A mathematical modelmathematical model giving the approximate amount of giving the approximate amount of COCO22 is given by: is given by:

YearYear 19581958 19701970 19741974 19781978 19851985 19911991 19981998 20032003 20072007

AmountAmount 315315 325325 330330 335335 345345 355355 365365 375375 380380

1010 2020 3030 4040 5050

380380

360360

340340

320320

t t (years)(years)

y y (ppmv)(ppmv)

2( ) 0.01076 0.8212 313.4A t t t 2( ) 0.01076 0.8212 313.4A t t t



a.a. Use the model to Use the model to estimateestimate the average amount of atmospheric the average amount of atmospheric COCO22 in in 19801980 ((tt = 23= 23))..

b.b. Assume that the trend continued and use the model to Assume that the trend continued and use the model to predictpredict the average amount of atmospheric COthe average amount of atmospheric CO22 in in 20102010..

YearYear 19581958 19701970 19741974 19781978 19851985 19911991 19981998 20032003 20072007

AmountAmount 315315 325325 330330 335335 345345 355355 365365 375375 380380

2( ) 0.010716 0.8212 313.4 (1 50) A t t t t 2( ) 0.010716 0.8212 313.4 (1 50) A t t t t



SolutionSolution

a.a. The average amount of atmospheric COThe average amount of atmospheric CO22 in in 19801980 is given by is given by

or approximately 338 ppmv.or approximately 338 ppmv.

b.b. Assuming that the trend will continue, the average amount Assuming that the trend will continue, the average amount of atmospheric COof atmospheric CO22 in in 20102010 will be will be

YearYear 19581958 19701970 19741974 19781978 19851985 19911991 19981998 20032003 20072007

AmountAmount 315315 325325 330330 335335 345345 355355 365365 375375 380380

2(23) 0.010716 23 0.8212 23 313.4 337.96A 2(23) 0.010716 23 0.8212 23 313.4 337.96A

2(53) 0.010716 53 0.8212 53 313.4 387.03A 2(53) 0.010716 53 0.8212 53 313.4 387.03A

2( ) 0.010716 0.8212 313.4 (1 50) A t t t t 2( ) 0.010716 0.8212 313.4 (1 50) A t t t t


Applied ExampleApplied ExampleSocial Security Trust Fund AssetsSocial Security Trust Fund Assets

The The projected assetsprojected assets of the Social Security trust fund of the Social Security trust fund (in (in trillions of dollars)trillions of dollars) from from 20082008 through through 20402040 are given by: are given by:

The scatter plot associated with these data is:The scatter plot associated with these data is:

YearYear 20082008 20112011 20142014 20172017 20202020 20232023 20262026 20292029 20322032 20352035 20382038 20402040

AssetsAssets 2.42.4 3.23.2 4.04.0 4.74.7 5.35.3 5.75.7 5.95.9 5.65.6 4.94.9 3.63.6 1.71.7 00

55 1010 1515 2020 2525 3030

66

44

22

t t (years)(years)

y y ($trillion)($trillion)



The The projected assetsprojected assets of the Social Security trust fund of the Social Security trust fund (in (in trillions of dollars)trillions of dollars) from from 20082008 through through 20402040 are given by: are given by:

A A mathematical modelmathematical model giving the approximate value of assets giving the approximate value of assets in the trust fund (in trillions of dollars) is:in the trust fund (in trillions of dollars) is:

55 1010 1515 2020 2525 3030

66

44

22

t t (years)(years)

y y ($trillion)($trillion)4 3 2( ) 0.00000324 0.000326 0.00342 0.254 2.4A t t t t t 4 3 2( ) 0.00000324 0.000326 0.00342 0.254 2.4A t t t t t

YearYear 20082008 20112011 20142014 20172017 20202020 20232023 20262026 20292029 20322032 20352035 20382038 20402040

AssetsAssets 2.42.4 3.23.2 4.04.0 4.74.7 5.35.3 5.75.7 5.95.9 5.65.6 4.94.9 3.63.6 1.71.7 00



a.a. The first baby boomers will turn The first baby boomers will turn 6565 in in 20112011. What will be . What will be the the assetsassets of the Social Security trust fund of the Social Security trust fund at that timeat that time??

b.b. The last of the baby boomers will turn The last of the baby boomers will turn 6565 in in 20292029. What will . What will the the assetsassets of the trust fund be of the trust fund be at the timeat the time??

c.c. Use the graph of function Use the graph of function AA((tt)) to to estimateestimate the yearthe year in which in which the current Social Security system the current Social Security system will go brokewill go broke..

4 3 2( ) 0.00000324 0.000326 0.00342 0.254 2.4A t t t t t

YearYear 20082008 20112011 20142014 20172017 20202020 20232023 20262026 20292029 20322032 20352035 20382038 20402040

AssetsAssets 2.42.4 3.23.2 4.04.0 4.74.7 5.35.3 5.75.7 5.95.9 5.65.6 4.94.9 3.63.6 1.71.7 00



SolutionSolutiona.a. The assets of the Social Security fund in The assets of the Social Security fund in 20112011 ((tt = 3 = 3)) will be: will be:

or approximately or approximately $3.18$3.18 trillion. trillion.The assets of the Social Security fund in The assets of the Social Security fund in 20292029 ((tt = 21 = 21)) will be: will be:

or approximately or approximately $5.59$5.59 trillion. trillion.

4 3 2(3) 0.00000324 3 0.000326 3 0.00342 3 0.254 3 2.4 3.18A

4 3 2(21) 0.00000324 21 0.000326 21 0.00342 21 0.254 21 2.4 5.59A

YearYear 20082008 20112011 20142014 20172017 20202020 20232023 20262026 20292029 20322032 20352035 20382038 20402040

AssetsAssets 2.42.4 3.23.2 4.04.0 4.74.7 5.35.3 5.75.7 5.95.9 5.65.6 4.94.9 3.63.6 1.71.7 00

4 3 2( ) 0.00000324 0.000326 0.00342 0.254 2.4A t t t t t



SolutionSolutionb.b. The graph shows that function The graph shows that function AA crosses the crosses the tt-axis-axis at about at about

tt = 32 = 32, suggesting the system will go broke by , suggesting the system will go broke by 20402040::

55 1010 1515 2020 2525 3030

66

44

22

y y ($trillion)($trillion)

Trust runs out of funds

YearYear 20082008 20112011 20142014 20172017 20202020 20232023 20262026 20292029 20322032 20352035 20382038 20402040

AssetsAssets 2.42.4 3.23.2 4.04.0 4.74.7 5.35.3 5.75.7 5.95.9 5.65.6 4.94.9 3.63.6 1.71.7 00

4 3 2( ) 0.00000324 0.000326 0.00342 0.254 2.4A t t t t t

t t (years)(years)Applied Example 3, page 79

Rational and Power FunctionsRational and Power Functions

A A rational functionrational function is simply is simply the quotient of two the quotient of two polynomialspolynomials..

In general, a rational function has the formIn general, a rational function has the form

where where ff((xx)) and and gg((xx)) are are polynomial functionspolynomial functions.. Since the Since the division by zero is not alloweddivision by zero is not allowed, we conclude that , we conclude that

the the domaindomain of a rational function is the set of all real of a rational function is the set of all real numbers except the numbers except the zeros zeros of of g g (the (the rootsroots of the equation of the equation gg((xx)) = 0= 0))

( )( )

( )

f xR x

g x

( )( )

( )

f xR x

g x


ExamplesExamples of rational functions: of rational functions:

3 23 1( )

2

x x xF x

x

3 23 1( )

2

x x xF x

x

2

2

1( )

1

xG x

x

2

2

1( )

1

xG x

x


Functions of the formFunctions of the form

where where rr is is any real numberany real number, are called , are called power functionspower functions.. We encountered examples of power functions earlier in We encountered examples of power functions earlier in

our work. our work. ExamplesExamples of power functions: of power functions:

1/2 22

1( ) ( ) and f x x x g x x

x 1/2 2

2

1( ) ( ) and f x x x g x x

x

( ) rf x x( ) rf x x


Many functions involve Many functions involve combinationscombinations of of rationalrational and and power functionspower functions..

Examples:Examples:2

2

2

1/22 3/2

1( )

1

( ) 3 4

1( ) (1 2 )

( 2)

xf x

x

g x x x

h x xx

2

2

2

1/22 3/2

1( )

1

( ) 3 4

1( ) (1 2 )

( 2)

xf x

x

g x x x

h x xx

Applied Example:Applied Example: Driving Costs Driving Costs

A study of driving costs based on a A study of driving costs based on a 20072007 medium-sized medium-sized sedan found the following average costs (car payments, sedan found the following average costs (car payments, gas, insurance, upkeep, and depreciation), measured in gas, insurance, upkeep, and depreciation), measured in cents per mile:cents per mile:

A A mathematical modelmathematical model giving the average cost in cents per giving the average cost in cents per mile is:mile is:

where where xx (in thousands)(in thousands) denotes the number of miles the denotes the number of miles the car is driven in car is driven in 11 year. year.

Miles/year, Miles/year, xx 50005000 10,00010,000 15,00015,000 20,00020,000

Cost/mile, Cost/mile, yy ((¢¢)) 83.883.8 62.962.9 52.252.2 47.147.1

0.42

164.8( )C x

x 0.42

164.8( )C x

x



Below is the scatter plot associated with this data:Below is the scatter plot associated with this data:

0.42

164.8( )C x

x

55 1010 1515 2020 2525

140140

120120

100100

8080

6060

4040

2020

t t (years)(years)

y y ((¢¢))

CC((xx))





Using this model, estimate the Using this model, estimate the average costaverage cost of driving a of driving a 20072007 medium-sized sedan medium-sized sedan 8,0008,000 miles per year and miles per year and 18,00018,000 miles per year.miles per year.

SolutionSolution The The average costaverage cost for driving a car for driving a car 8,0008,000 miles per year is miles per year is

or approximately or approximately 68.868.8¢/mile¢/mile..

0.42

164.8( )C x

x

0.42

164.8(8) 68.81

8C





Using this model, estimate the Using this model, estimate the average costaverage cost of driving a of driving a 20072007 medium-sized sedan medium-sized sedan 8,000 8,000 miles per year and miles per year and 18,00018,000 miles per year.miles per year.

SolutionSolution The The average costaverage cost for driving a car for driving a car 18,00018,000 miles per year is miles per year is

or approximately or approximately 48.9548.95¢/mile¢/mile..

0.42

164.8( )C x

x

0.42

164.8(18) 48.95

18C




Some Economic ModelsSome Economic Models

People’s decision on how much to People’s decision on how much to demanddemand or purchase of a or purchase of a given product depends on the given product depends on the priceprice of the product: of the product:

✦ TheThe higherhigher the the priceprice the the lessless they want to they want to buybuy of it. of it.

✦ A A demand functiondemand function pp = = dd((xx)) can be used to describe this. can be used to describe this.


Similarly, firms’ decision on how much to Similarly, firms’ decision on how much to supplysupply or or produce of a product depends on the produce of a product depends on the priceprice of the product: of the product:

✦ TheThe higherhigher the the priceprice, the, the moremore they want to they want to produceproduce of it. of it.

✦ A A supply functionsupply function pp = = ss((xx)) can be used to describe this. can be used to describe this.


The interaction between demand and supply will ensure the The interaction between demand and supply will ensure the market settles to a market settles to a market equilibriummarket equilibrium::

✦ This is the situation at which This is the situation at which quantity demanded quantity demanded equalsequals quantity suppliedquantity supplied..

✦ Graphically, this situation occurs when the Graphically, this situation occurs when the demand demand curvecurve and the and the supply curvesupply curve intersectintersect: where : where dd((xx)) = = ss((xx))..

Applied Example:Applied Example: Supply and Demand Supply and Demand

The The demanddemand functionfunction for a certain brand of bluetooth for a certain brand of bluetooth wireless headset is given bywireless headset is given by

The corresponding The corresponding supply functionsupply function is given by is given by

where where pp is the expressed in is the expressed in dollarsdollars and and xx is measured in is measured in units of a units of a thousandthousand. .

Find the equilibrium quantity and price.Find the equilibrium quantity and price.

2( ) 0.025 0.5 60p d x x x

2( ) 0.02 0.6 20p s x x x



SolutionSolution We solve the following We solve the following system of equationssystem of equations::

SubstitutingSubstituting the the secondsecond equation into the equation into the firstfirst yields: yields:

Thus, either Thus, either xx = – = – 400/9400/9 (but this is (but this is not possiblenot possible), or ), or xx = 20 = 20. . So, the So, the equilibrium quantityequilibrium quantity must be must be 20,000 20,000 headsets.headsets.

2

2

0.025 0.5 60

0.02 0.6 20

p x x

p x x

2 2

2

2

2

0.02 0.6 20 0.025 0.5 60

0.045 1.1 40 0

45 1100 40,000 0

9 220 8,000 09 400 20 0

x x x x

x x

x x

x xx x



SolutionSolution The The equilibrium priceequilibrium price is given by: is given by:

or or $40$40 per headset.per headset.

20.02 20 0.6 20 20 40p


Constructing Mathematical ModelsConstructing Mathematical Models

Some mathematical models can be constructed using Some mathematical models can be constructed using elementary geometricelementary geometric and and algebraic argumentsalgebraic arguments..

GuidelinesGuidelines for constructing mathematical models: for constructing mathematical models:

1.1. Assign a letter to each variableAssign a letter to each variable mentioned in the mentioned in the problem. If appropriate, problem. If appropriate, draw and labeldraw and label a figure. a figure.

2.2. Find an expressionFind an expression for the quantity sought. for the quantity sought.

3.3. Use the Use the conditionsconditions given in the problem to given in the problem to write write the quantity sought as a function the quantity sought as a function ff of one variable. of one variable.

Note any Note any restrictionsrestrictions to be placed on the to be placed on the domaindomain of of f f by the nature of the problem. by the nature of the problem.

Applied Example:Applied Example: Enclosing an Area Enclosing an Area

The owner of the Rancho Los Feliz has The owner of the Rancho Los Feliz has 30003000 yards of yards of fencingfencing with which to with which to enclose a rectangular pieceenclose a rectangular piece of of grazing land along the straight portion of a river.grazing land along the straight portion of a river.

Fencing is Fencing is not requirednot required along the along the riverriver.. Letting Letting xx denote the denote the widthwidth of the rectangle, find a of the rectangle, find a functionfunction

ff in the in the variablevariable xx giving the giving the areaarea of the grazing land if she of the grazing land if she uses alluses all of the fencing. of the fencing.


Applied Example:Applied Example: Enclosing an Area Enclosing an Area

SolutionSolution This information was This information was givengiven::

✦ The The areaarea of the rectangular grazing land is of the rectangular grazing land is AA = = xyxy..✦ The amount of The amount of fencingfencing is is 22xx + + yy which must equal which must equal 30003000

(to use all the fencing), so:(to use all the fencing), so:

22xx + + y y = 3000 = 3000 Solving forSolving for y y we get: we get:

y y = 3000 – 2= 3000 – 2xx SubstitutingSubstituting this value of this value of yy into the expression for into the expression for AA gives: gives:

A A = = xx(3000 – 2(3000 – 2xx) = 3000) = 3000xx – 2 – 2xx22

Finally,Finally, x x and and yy represent represent distancesdistances, so they must be , so they must be nonnegativenonnegative, so , so xx 0 0 and and yy = 3000 – 2= 3000 – 2xx 0 0 (or (or xx 1500 1500). ).

Thus, the Thus, the required functionrequired function is: is: ff((xx) =) = 30003000xx – 2 – 2xx22 (0 (0 xx 1500) 1500)


Applied Example:Applied Example: Charter-Flight Revenue Charter-Flight Revenue

If exactly If exactly 200200 people sign up for a people sign up for a charter flightcharter flight, Leasure , Leasure World Travel Agency charges World Travel Agency charges $300$300 per person. per person.

However, if However, if moremore than than 200200 people sign up for the flight people sign up for the flight (assume this is the case), then each fare is (assume this is the case), then each fare is reducedreduced by by $1$1 for each additional personfor each additional person..

Letting Letting xx denote the denote the number of passengers abovenumber of passengers above 200200, find , find a a functionfunction giving the giving the revenuerevenue realized by the company. realized by the company.


Applied Example:Applied Example: Charter-Flight Revenue Charter-Flight Revenue

SolutionSolution This information was given.This information was given.

✦ If there are If there are xx passengers above passengers above 200200, then the number of , then the number of passengers signing uppassengers signing up for the flight is for the flight is 200 + 200 + xx..

✦ The The farefare will be will be (300 – (300 – xx)) dollars dollars per passengerper passenger.. The The revenuerevenue will be will be

RR = (200 + = (200 + xx)(300 – )(300 – xx))

= –= – xx22 + 100 + 100xx + 60,000 + 60,000 The quantities The quantities must be positivemust be positive, so , so xx 0 0 and and 300 – 300 – xx 0 0

(or (or xx 300 300). ). So the So the required functionrequired function is: is:

ff((xx) =) = –– xx22 + 100 + 100xx + 60,000 (0 + 60,000 (0 xx 300) 300)


2.42.4LimitsLimits

( )f x400

300

200

100

x

y

10 20 30 40 50 60

Introduction to CalculusIntroduction to Calculus

Historically, the development of calculus by Historically, the development of calculus by Isaac NewtonIsaac Newton and and Gottfried W. LeibnizGottfried W. Leibniz resulted from the investigation resulted from the investigation of the following problems:of the following problems:

1.1. Finding the Finding the tangent line to a curvetangent line to a curve at a given point on the at a given point on the curve:curve:

t

yT


Historically, the development of calculus by Historically, the development of calculus by Isaac NewtonIsaac Newton and and Gottfried W. LeibnizGottfried W. Leibniz resulted from the investigation resulted from the investigation of the following problems:of the following problems:

2.2. Finding the Finding the area of planar regionarea of planar region bounded by an bounded by an arbitrary curve.arbitrary curve.

t

y

R


The study of the The study of the tangent-line problemtangent-line problem led to the led to the creation of creation of differential calculusdifferential calculus, which relies on the , which relies on the concept of the concept of the derivativederivative of a function. of a function.

The study of the The study of the area problemarea problem led to the creation of led to the creation of integral calculusintegral calculus, which relies on the concept of the , which relies on the concept of the anti-derivativeanti-derivative, or , or integralintegral, of a function., of a function.

Example:Example: A Speeding Maglev A Speeding Maglev

From data obtained in a From data obtained in a test runtest run conducted on a conducted on a prototype of maglevprototype of maglev, which moves along a straight , which moves along a straight monorail track, engineers have determined that the monorail track, engineers have determined that the positionposition of the maglev (in feet) of the maglev (in feet) from the originfrom the origin at at timetime tt is given byis given by

ss = = ff((tt) = 4) = 4tt22 (0 (0 ≤≤ t t ≤≤ 30) 30)

Where Where ff is called the is called the position functionposition function of the maglev. of the maglev. The The positionposition of the maglev at time of the maglev at time tt = 0, 1, 2, 3, … , 10= 0, 1, 2, 3, … , 10 is is

ff(0) = 0 (0) = 0 ff(1) = 4 (1) = 4 ff(2) = 16 (2) = 16 ff(3) = 36 … (3) = 36 … ff(10) = 400(10) = 400

But what if we want to find the But what if we want to find the velocityvelocity of the maglev at of the maglev at any given any given point in timepoint in time??


Say we want to find the Say we want to find the velocityvelocity of the maglev at of the maglev at t t = 2= 2.. We may compute the We may compute the average velocityaverage velocity of the maglev of the maglev

over an over an interval of timeinterval of time, such as , such as [2, 4][2, 4] as follows: as follows:

or or 24 feet/second24 feet/second.. This is This is notnot the velocity of the maglev at the velocity of the maglev at exactlyexactly t t = 2= 2, ,

but it is a useful but it is a useful approximationapproximation..

2 2

Distance covered (4) (2)

Time elapsed 4 2

4(4 ) 4(2 )

264 16

224

f f

2 2

Distance covered (4) (2)

Time elapsed 4 2

4(4 ) 4(2 )

264 16

224

f f


We can find a We can find a better approximationbetter approximation by choosing a by choosing a smaller intervalsmaller interval to compute the speed, say to compute the speed, say [2, 3][2, 3]..

MoreMore generally generally, let , let tt > 2 > 2. Then, the . Then, the average velocityaverage velocity of of the maglev over the the maglev over the time intervaltime interval [2, [2, tt]] is given by is given by

2 2

2

Distance covered ( ) (2)

Time elapsed 2

4( ) 4(2 )

2

4( 4)

2

f t f

t

t

t

t

t

2 2

2

Distance covered ( ) (2)

Time elapsed 2

4( ) 4(2 )

2

4( 4)

2

f t f

t

t

t

t

t


By choosing the values of By choosing the values of tt closer and closercloser and closer to to 22, we , we obtain obtain average velocitiesaverage velocities of the maglev over of the maglev over smaller and smaller and smaller time intervalssmaller time intervals..

The The smallersmaller the the time intervaltime interval, the , the closercloser the the average velocity becomes to the becomes to the instantaneous velocityinstantaneous velocity of the train of the train at at tt = 2 = 2, as the table below demonstrates:, as the table below demonstrates:

The The closercloser tt gets to gets to 22, the , the closercloser the the average velocityaverage velocity gets gets to to 16 feet/second16 feet/second. .

Thus, the Thus, the instantaneous velocityinstantaneous velocity at at tt = 2 = 2 seems to be seems to be 16 feet/second16 feet/second..

24( 4)Average velocity

2

t

t

24( 4)Average velocity

2

t

t

tt 2.52.5 2.12.1 2.012.01 2.0012.001 2.00012.0001

Average VelocityAverage Velocity 1818 16.416.4 16.0416.04 16.00416.004 16.000416.0004

Intuitive Definition of a LimitIntuitive Definition of a Limit Consider the function Consider the function g,g, which gives the which gives the average velocityaverage velocity

of the maglev:of the maglev:

Suppose we want to find the value that Suppose we want to find the value that gg((tt)) approachesapproaches as as t t approachesapproaches 22..✦ We take values of We take values of tt approachingapproaching 22 from the rightfrom the right

(as we did before), and we find that (as we did before), and we find that gg((tt)) approachesapproaches 1616::

✦ Similarly, we take values of Similarly, we take values of t t approachingapproaching 22 from the leftfrom the left, , and we find that and we find that gg((tt)) also also approachesapproaches 1616::

24( 4)( )

2

tg t

t

24( 4)( )

2

tg t

t

tt 2.52.5 2.12.1 2.012.01 2.0012.001 2.00012.0001

gg((tt)) 1818 16.416.4 16.0416.04 16.00416.004 16.000416.0004

tt 1.51.5 1.91.9 1.991.99 1.9991.999 1.99991.9999

gg((tt)) 1414 15.615.6 15.9615.96 15.99615.996 15.999615.9996

Intuitive Definition of a LimitIntuitive Definition of a Limit

We have found that as We have found that as tt approachesapproaches 22 from either sidefrom either side, , gg((tt)) approachesapproaches 1616..

In this situation, we say that In this situation, we say that the limitthe limit of of gg((tt)) as as tt approachesapproaches 22 is is 1616..

This is written asThis is written as

Observe that Observe that tt = 2 = 2 is is notnot in the domain in the domain of of gg((tt)) . . But this But this does not matterdoes not matter, since , since tt = 2 = 2 does not play any does not play any

role in computing this limit.role in computing this limit.

2

2 2

4( 4) ( ) 16

2lim limt t

tg t

t

2

2 2

4( 4) ( ) 16

2lim limt t

tg t

t

Limit of a FunctionLimit of a Function

The The functionfunction f f has a has a limitlimit LL as as xx approachesapproaches aa, written, written

If the value of If the value of ff((xx)) can be made as can be made as closeclose to the number to the number LL as we please by taking as we please by taking x x valuesvalues sufficiently closesufficiently close to to (but (but not equal tonot equal to) ) aa..

( )limx a

f x L

( )limx a

f x L

ExamplesExamples

Let Let ff((xx) = ) = xx33. Evaluate . Evaluate

SolutionSolution You can see in the graph You can see in the graph

that that ff((xx) ) can becan be as as closeclose to to 8 8 as we please by taking as we please by taking xx sufficiently closesufficiently close to to 22..

Therefore, Therefore,

2( )lim .

xf x

2( )lim .

xf x

3

28lim

xx

3

28lim

xx

– – 2 2 – – 11 11 22 33

88

66

44

22

–– 22

xx

yy ff((xx)) = = xx33

Example 1, page 101

ExamplesExamples

Let Let Evaluate Evaluate

SolutionSolution1

( )lim .x

g x1

( )lim .x

g x

– – 2 2 – – 11 11 22 33

55

33

11

xx

yy gg((xx))

2 1( )

1 1

if

if

x xg x

x

2 1( )

1 1

if

if

x xg x

x

You can see in the graph You can see in the graph that that gg((xx) ) can becan be as as closeclose to to 3 3 as we please by as we please by taking taking xx sufficiently sufficiently closeclose to to 11..

Therefore, Therefore, 1

( ) 3lim x

g x

1

( ) 3lim x

g x

Example 2, page 101

ExamplesExamples

Let Let Evaluate Evaluate

SolutionSolution0

( )lim .x

f x0

( )lim .x

f x

– – 2 2 – – 11 11 22

55

xx

yy

2

1( )f x

x 2

1( )f x

x

The graph shows us that as The graph shows us that as xx approachesapproaches 00 from from either either sideside, , ff((xx)) increases without increases without boundbound and thus does and thus does notnot approach any specific real approach any specific real number.number.

Thus, the limit of Thus, the limit of ff((xx)) does does notnot exist exist as as xx approachesapproaches 00..

2

1( )f x

x 2

1( )f x

x

Example 3b, page 101

Theorem 1Theorem 1Properties of LimitsProperties of Limits

SupposeSuppose andand

Then,Then,

1.1. rr, a real number, a real number

2.2. cc, a real number, a real number

3.3.

4.4.

5.5. Provided thatProvided that MM ≠ ≠

00

SupposeSuppose andand

Then,Then,

1.1. rr, a real number, a real number

2.2. cc, a real number, a real number

3.3.

4.4.

5.5. Provided thatProvided that MM ≠ ≠

00

( )lim x a

g x M

( )lim x a

g x M

( )limx a

f x L

( )limx a

f x L

( ) ( )lim limr

r r

x a x af x f x L

( ) ( )lim lim

rr r

x a x af x f x L

( ) ( )lim limx a x a

cf x c f x cL

( ) ( )lim limx a x a

cf x c f x cL

( ) ( ) ( ) ( )lim lim limx a x a x a

f x g x f x g x L M


f x g x f x g x L M


f x g x f x g x LM


f x g x f x g x LM

( )( )

( ) ( )

limlim

limx a

x ax a

f xf x L

g x g x M

( )( )

( ) ( )

limlim

limx a

x ax a

f xf x L

g x g x M

ExamplesExamples

Use Use theorem 1theorem 1 to evaluate the following limits: to evaluate the following limits:

33

22 8lim

xx

33

22 8lim

xx

3

2 lim

xx

3

2 lim

xx

3/23/2

45 5(4) 40lim

xx

3/23/2

45 5(4) 40lim

xx

3/2

4 5lim

xx

3/2

4 5lim

xx

44

1 15 2 5(1) 2 3lim lim

x xx

44

1 15 2 5(1) 2 3lim lim

x xx

4

1 (5 2)lim

xx

4

1 (5 2)lim

xx

Example 4, page 102

ExamplesExamples

Use Use theorem 1theorem 1 to evaluate the following limits: to evaluate the following limits:

32 3 2

3 32 7 2(3) (3) 7 216lim lim

x xx x

32 3 2

3 32 7 2(3) (3) 7 216lim lim

x xx x

3 2

3 2 7lim

xx x

3 2

3 2 7lim

xx x

22

2

2

(2 1) 2(2) 1 93

( 1) 2 1 3

lim

lim

x

x

x

x

22

2

2

(2 1) 2(2) 1 93

( 1) 2 1 3

lim

lim

x

x

x

x

2

2

2 1

1limx

x

x

2

2

2 1

1limx

x

x

Example 4, page 102

Indeterminate FormsIndeterminate Forms

Let’s considerLet’s consider

which we evaluated earlier for the which we evaluated earlier for the maglev examplemaglev example by by looking at values for looking at values for xx nearnear xx = 2 = 2..

If we attempt to evaluate this expression by applying If we attempt to evaluate this expression by applying Property 5Property 5 of limits, we get of limits, we get

In this case we say that the limit of the In this case we say that the limit of the quotientquotient ff((xx)/)/gg((xx)) as as xx approachesapproaches 22 has the has the indeterminate form indeterminate form 0/00/0..

This expression does This expression does notnot provide us with a provide us with a solutionsolution to our to our problem.problem.

2

2

4( 4)

2limx

x

x

2

2

4( 4)

2limx

x

x

222

22

4( 4) 4( 4) 0

2 2 0

limlim

limx

xx

xx

x x

222

22

4( 4) 4( 4) 0

2 2 0

limlim

limx

xx

xx

x x

Strategy for Evaluating Indeterminate FormsStrategy for Evaluating Indeterminate Forms

1.1. ReplaceReplace the given function with an appropriate the given function with an appropriate one that takes on the one that takes on the same valuessame values as the as the original function original function everywhereeverywhere exceptexcept at at x x == a a..

2.2. EvaluateEvaluate the limit of this function as the limit of this function as xx approachesapproaches aa..

ExamplesExamples

EvaluateEvaluate

SolutionSolution As we’ve seen, here we have an As we’ve seen, here we have an indeterminate formindeterminate form 0/00/0.. We can rewrite We can rewrite

xx ≠ 2≠ 2

Thus, we can say thatThus, we can say that

Note that Note that 1616 is the same value we obtained for the is the same value we obtained for the maglev maglev exampleexample through through approximationapproximation..

2

2

4( 4)

2limx

x

x

2

2

4( 4)

2limx

x

x

24( 4) 4( 2)( 2)4( 2)

2 2

x x xx

x x

24( 4) 4( 2)( 2)4( 2)

2 2

x x xx

x x

2

2 2

4( 4)4( 2) 16

2lim limx x

xx

x

2

2 2

4( 4)4( 2) 16

2lim limx x

xx

x

Example 5, page 104

ExamplesExamples

EvaluateEvaluate

SolutionSolution Notice in the graphs below that Notice in the graphs below that the two functions yield the the two functions yield the

same graphssame graphs, , except for the valueexcept for the value xx = 2 = 2::

2

2

4( 4)

2limx

x

x

2

2

4( 4)

2limx

x

x

( ) 4( 2)g x x 24( 4)

( )2

xf x

x

2020

1616

1212

88

44

xx

yy

– – 3 3 –– 2 2 – – 11 11 22 33

2020

1616

1212

88

44

xx

yy

– – 3 3 –– 2 2 – – 11 11 22 33Example 5, page 104

ExamplesExamples

EvaluateEvaluate

SolutionSolution As we’ve seen, here we have an As we’ve seen, here we have an indeterminate formindeterminate form 0/00/0.. We can rewrite (with the We can rewrite (with the constraintconstraint that that hh ≠≠ 0 0):):

Thus, we can say thatThus, we can say that

0

1 1limh

h

h

0

1 1limh

h

h

1 1 1 1 1

( 1 1) 1

1 1

1 1 1

h h h

h h hh h h

h

1 1 1 1 1

( 1 1) 1

1 1

1 1 1

h h h

h h hh h h

h

0 0

1 1 1 1 1

21 1 1 1lim limh h

h

h h

0 0

1 1 1 1 1

21 1 1 1lim limh h

h

h h

Example 6, page 105

Limits at InfinityLimits at Infinity

There are occasions when we want to know whether There are occasions when we want to know whether ff((xx)) approachesapproaches a unique numbera unique number as as xx increases increases without without boundbound..

In the graph below, as In the graph below, as xx increases increases without boundwithout bound, , ff((xx)) approaches approaches the number the number 400400..

We call the line We call the line yy = 400 = 400 a a horizontal asymptotehorizontal asymptote..

In this case, we can say In this case, we can say thatthat

and we call this and we call this a limita limitof a function at infinityof a function at infinity..

( ) 400limx

f x

( ) 400limx

f x

( )f x( )f x400

300

200

100

x

y

10 20 30 40 50 60

ExampleExample

Consider the functionConsider the function

Determine what happens to Determine what happens to ff((xx)) as as xx gets gets larger and largerlarger and larger..

SolutionSolution We can pick a We can pick a sequence of valuessequence of values of of xx and and substitute them substitute them

in the functionin the function to obtain the following values: to obtain the following values:

As As xx gets gets larger and largerlarger and larger, , ff((xx)) gets gets closer and closercloser and closer to to 22.. Thus, we can say thatThus, we can say that

2

2

22

1limx

x

x

2

2

22

1limx

x

x

2

2

2( )

1

xf x

x

2

2

2( )

1

xf x

x

xx 11 22 55 1010 100100 10001000

ff((xx)) 11 1.61.6 1.921.92 1.981.98 1.99981.9998 1.9999981.999998

Limit of a Function at InfinityLimit of a Function at Infinity

The function The function ff has the has the limitlimit LL as as xx increases without increases without boundbound (as (as x x approaches infinityapproaches infinity), written), written

if if ff((xx)) can be made arbitrarily can be made arbitrarily close toclose to LL by taking by taking xx large enough.large enough.

Similarly, the function Similarly, the function ff has the has the limitlimit MM as as xx decreases decreases without boundwithout bound (as (as x x approaches negative infinityapproaches negative infinity), ), writtenwritten

if if ff((xx)) can be made arbitrarily can be made arbitrarily close toclose to MM by taking by taking xx large enough in absolute value.large enough in absolute value.

( )limx

f x L

( )limx

f x L

( )limx

f x M

( )limx

f x M

ExamplesExamples

Let Let

EvaluateEvaluate andand

SolutionSolution Graphing Graphing ff((xx)) reveals that reveals that

( )limx

f x

( )limx

f x

1 if 0( )

1 if 0

xf x

x

1 if 0( )

1 if 0

xf x

x

( ) 1limx

f x

( ) 1limx

f x

( ) 1limx

f x

( ) 1limx

f x

( )f x( )f x11

–– 1 1

xx

yy

– – 3 3 33

Example 7, page 107

ExamplesExamples

Let Let

EvaluateEvaluate andand

SolutionSolution Graphing Graphing gg((xx)) reveals that reveals that

( )limx

g x

( )limx

g x

( )limx

g x

( )limx

g x

2

1( )g x

x 2

1( )g x

x

( ) 0limx

g x

( ) 0limx

g x

( ) 0limx

g x

( ) 0limx

g x

2

1( )g x

x 2

1( )g x

x

xx

yy

– – 3 3 – – 2 2 – – 1 1 11 22 33

Example 7, page 107


For all For all nn > 0 > 0,, andand

provided that is defined.provided that is defined.

1 0lim

nx x

1 0lim

nx x 1

0limnx x

1

0limnx x

1nx

1nx

All properties of limits listed in Theorem 1 are valid when a is replaced by or –.

In addition, we have the following properties for limits to infinity:

ExamplesExamples

Evaluate Evaluate

SolutionSolution The The limitslimits of both the of both the numeratornumerator and and denominatordenominator do not do not

existexist as as x x approaches infinity, so approaches infinity, so property 5property 5 is is notnot applicable.applicable.

We can find the solution instead by We can find the solution instead by dividingdividing numerator numerator andand denominator denominator byby xx33::

2

3

3

2 1limx

x x

x

2

3

3

2 1limx

x x

x

2 3 2 3

3 3

3

1 1 3( 3) / 0 0 0 0

01(2 1) / 2 0 22

lim limx x

x x x x x xx x

x

2 3 2 3

3 3

3

1 1 3( 3) / 0 0 0 0

01(2 1) / 2 0 22

lim limx x

x x x x x xx x

x

Example 8, page 108

ExamplesExamples

Evaluate Evaluate

SolutionSolution Again, we see that Again, we see that property 5property 5 does does notnot apply. apply. So we So we dividedivide numerator numerator andand denominator denominator byby xx22::

2

2

3 8 4 2 4 5

limx

x x

x x

2

2

3 8 4 2 4 5

limx

x x

x x

2 2 2

2 2

2

8 43(3 8 4) / 3 0 0 3

4 5(2 4 5) / 2 0 0 22lim limx x

x x x x xx x x

x x

2 2 2

2 2

2

8 43(3 8 4) / 3 0 0 3

4 5(2 4 5) / 2 0 0 22lim limx x

x x x x xx x x

x x

Example 9, page 108

ExamplesExamples

Evaluate Evaluate

SolutionSolution Again, we see that Again, we see that property 5property 5 does does notnot apply. apply. But But dividingdividing numerator numerator andand denominator denominator byby xx22 does does notnot

help in this case:help in this case:

In other words, In other words, the limit does not existthe limit does not exist.. We indicate this by writingWe indicate this by writing

3 2

2

2 3 1

2 4limx

x x

x x

3 2

2

2 3 1

2 4limx

x x

x x

3 2 2 2

2 2

2

12 3(2 3 1) /

2 4( 2 4) / 1lim limx x

xx x x xx x x

x x

3 2 2 2

2 2

2

12 3(2 3 1) /

2 4( 2 4) / 1lim limx x

xx x x xx x x

x x

3 2

2

2 3 1

2 4limx

x x

x x

3 2

2

2 3 1

2 4limx

x x

x x

Example 10, page 109

2.52.5One-Sided Limits and ContinuityOne-Sided Limits and Continuity

ax

y

b c d

One-Sided LimitsOne-Sided Limits


Its graph shows that Its graph shows that ff doesdoes notnot have a have a limitlimit as as xx approaches approaches zerozero, because , because approaching from each approaching from each side results in side results in different different valuesvalues..

1 if 0( )

1 if 0

x xf x

x x

1 if 0( )

1 if 0

x xf x

x x

( )f x( )f x

11

–– 1 1

xx

yy

– – 1 1 11



If we If we restrictrestrict xx to be to be greatergreater than zerothan zero (to the right of zero), (to the right of zero), we see that we see that ff((xx)) approachesapproaches 11 as as close to as we please as close to as we please as xx approachesapproaches 0 0..

In this case we say that the In this case we say that the right-hand limitright-hand limit of of ff as as xx approachesapproaches 00 is is 11, written, written

1 if 0( )

1 if 0

x xf x

x x

1 if 0( )

1 if 0

x xf x

x x

11

–– 1 1

xx

yy

– – 1 1 11

0( ) 1lim

xf x

0( ) 1lim

xf x

( )f x( )f x



Similarly, if we Similarly, if we restrictrestrict xx to be to be lessless than zero than zero (to the left of zero), (to the left of zero), we see that we see that ff((xx)) approachesapproaches –1–1 as close to as we please as as close to as we please as xx approachesapproaches 0 0..

In this case we say that the In this case we say that the left-hand limitleft-hand limit of of ff as as xx approachesapproaches 00 is is –– 11, written, written

1 if 0( )

1 if 0

x xf x

x x

1 if 0( )

1 if 0

x xf x

x x

11

–– 1 1

xx

yy

– – 1 1 11

0( ) 1lim

xf x

0( ) 1lim

xf x

( )f x( )f x


The function The function ff has the has the right-hand limitright-hand limit LL as as xx approaches from the rightapproaches from the right, written, written

If the values of If the values of ff((xx)) can be made as close to can be made as close to LL as we as we please by taking please by taking xx sufficiently close to (but not equal sufficiently close to (but not equal to) to) aa and and to the rightto the right of of aa..

Similarly, the function Similarly, the function ff has the has the left-hand limitleft-hand limit LL as as xx approaches from the leftapproaches from the left, written, written

If the values of If the values of ff((xx)) can be made as close to can be made as close to LL as we as we please by taking please by taking xx sufficiently close to (but not equal sufficiently close to (but not equal to) to) aa and and to the leftto the left of of aa..

( )limx a

f x L

( )limx a

f x L

( )limx a

f x L

( )limx a

f x L


Let Let ff be a function that is defined for all values of be a function that is defined for all values of xx close to close to x x == a a with the possible exception of with the possible exception of aa itself. itself. ThenThen

( ) ( ) ( )lim lim lim if and only if x a x a x a

f x L f x f x L

( ) ( ) ( )lim lim lim if and only if x a x a x a

f x L f x f x L

The connection between one-side limits and the two-sided limit defined earlier is given by the following theorem.

– – 22 – – 1 1 11 22

ExamplesExamples

Show that Show that exists by studying the exists by studying the one-sidedone-sided

limitslimits of of ff as as xx approachesapproaches 00: :

SolutionSolution For For xx > 0 > 0, we find, we find

And for And for xx ≤≤ 0 0, we find, we find

Thus, Thus,

if 0( )

if 0

xxf x

xx

if 0( )

if 0

xxf x

xx

0( )lim

xf x

0( )lim

xf x

0( ) 0lim

xf x

0( ) 0lim

xf x

0( ) 0lim

xf x

22

11

xx

yy

( )f x( )f x

Example 1, page 118

Show that Show that does not exist. does not exist.

SolutionSolution For For xx << 0 0, we find, we find

And for And for xx 0 0, we find, we find

Thus, Thus, does does not exist.not exist.

11

–– 1 1

ExamplesExamples

1 if 0( )

1 if 0

xg x

x

1 if 0( )

1 if 0

xg x

x

0( )lim

xg x

0( )lim

xg x

0( ) 1lim

xg x

0( ) 1lim

xg x

0( ) 1lim

xg x

0( ) 1lim

xg x

0( )lim

xg x

0( )lim

xg x

xx

yy

( )g x( )g x

Example 1, page 118

Continuous FunctionsContinuous Functions

Loosely speaking, Loosely speaking, a function is continuousa function is continuous at a given point if at a given point if its graph at that point has its graph at that point has nono holesholes, , gapsgaps, , jumpsjumps, or , or breaksbreaks..

Consider, for example, the graph of Consider, for example, the graph of ff

This function is This function is discontinuousdiscontinuous at the following points: at the following points:✦ At At x x == a a,, f f is is not definednot defined ( (x x == a a is is not in the domainnot in the domain of of f f ).).

aaxx

yy




This function is This function is discontinuousdiscontinuous at the following points: at the following points:✦ At At x x == b b,, f f((bb)) is is not equal to the limitnot equal to the limit ofof f f((xx)) asas x x approachesapproaches bb..

aaxx

yy

bb




This function is This function is discontinuousdiscontinuous at the following points: at the following points:✦ At At x x == c c,, the function the function does not have a limitdoes not have a limit, since the , since the left-handleft-hand

and and right-hand limitsright-hand limits are are not equalnot equal..

xx

yy

bb cc




This function is This function is discontinuousdiscontinuous at the following points: at the following points:✦ At At x x == d d,, the the limit of the function does not existlimit of the function does not exist, resulting in a , resulting in a

break in the graph.break in the graph.

xx

yy

cc dd

Continuity of a Function at a NumberContinuity of a Function at a Number

A function A function ff is is continuouscontinuous at a number at a number x = ax = a if the following if the following conditionsconditions are satisfied: are satisfied:

1.1. ff((aa)) is defined. is defined.

2.2.

3.3.

( )lim exists.x a

f x

( )lim exists.x a

f x

( ) ( )limx a

f x f a

( ) ( )limx a

f x f a

If If ff is not continuous at is not continuous at x = ax = a, then , then ff is said to is said to be be discontinuous atdiscontinuous at x = ax = a..

Also, Also, ff is is continuous on an intervalcontinuous on an interval if if ff is is continuous at every number in the interval.continuous at every number in the interval.

ExamplesExamples

Find the values of Find the values of xx for which the function is for which the function is continuouscontinuous::

SolutionSolution The function The function ff is is continuous everywherecontinuous everywhere because the three because the three

conditions for continuity are satisfied for all values of conditions for continuity are satisfied for all values of xx..

( ) 2f x x ( ) 2f x x

– – 22 – – 1 1 11 22

55

44

33

22

11

xx

yy

( ) 2f x x ( ) 2f x x

Example 2, page 120

ExamplesExamples


SolutionSolution The function The function gg is is discontinuous at discontinuous at x x = 2= 2 because because gg is is not not

defineddefined at that number. It is at that number. It is continuous continuous everywhere else. everywhere else.

2 4( )

2

xg x

x

2 4( )

2

xg x

x

2 4( )

2

xg x

x

2 4( )

2

xg x

x

– – 22 – – 1 1 11 22

55

44

33

22

11

xx

yy

Example 2, page 120

ExamplesExamples


SolutionSolution The function The function hh is is continuouscontinuous everywhere everywhere exceptexcept at at x x = 2= 2

where it is where it is discontinuous discontinuous becausebecause

2 2( )

1 2

i f

if

x xh x

x

2 2( )

1 2

i f

if

x xh x

x

( )y h x ( )y h x

– – 22 – – 1 1 11 22

55

44

33

22

11

xx

yy

2(2) 1 ( ) 4lim

xh h x

2(2) 1 ( ) 4lim

xh h x

Example 2, page 120

ExamplesExamples


SolutionSolution The function The function FF is is discontinuous discontinuous atat x x = 0= 0 because the limit because the limit

of of FF fails to existfails to exist as as xx approachesapproaches 00. It is . It is continuous continuous everywhere else. everywhere else.

1 0( )

1 0

if

i

f

xF x

x

1 0( )

1 0

if

i

f

xF x

x

( )y F x ( )y F x11

–– 1 1

xx

yy

Example 2, page 120

ExamplesExamples


SolutionSolution The function The function GG is is discontinuous discontinuous atat x x = 0= 0 because the limit because the limit

of of GG fails to existfails to exist as as xx approachesapproaches 00. It is . It is continuous continuous everywhere else. everywhere else.

10

( )1 0

if

if

xG x x

x

10

( )1 0

if

if

xG x x

x

( )y G x ( )y G x

– 1

x

y

Example 2, page 120

Properties of Continuous FunctionsProperties of Continuous Functions

1.1. The The constantconstant function function f f((xx) = ) = cc is is continuous everywherecontinuous everywhere..

2.2. The The identityidentity function function ff((xx) = ) = xx is is continuous everywherecontinuous everywhere..

IfIf f f andand g g areare continuous continuous atat x = ax = a, then, then

3.3. [[ff((xx)])]nn, where , where nn is a real number, is is a real number, is continuouscontinuous at at x = ax = a whenever it is defined at that number. whenever it is defined at that number.

4.4. f ± gf ± g is is continuouscontinuous at at x = ax = a..

5.5. fgfg is is continuouscontinuous at at x = ax = a..

6.6. ff /g/g is is continuouscontinuous at at gg((aa) ) ≠ ≠ 00..

Properties of Continuous FunctionsProperties of Continuous Functions

Using these properties, we can obtain the following Using these properties, we can obtain the following additional propertiesadditional properties..

1.1. A A polynomial functionpolynomial function y = Py = P((xx)) is continuous at every is continuous at every value of value of xx..

2.2. A A rational functionrational function RR((xx)) = = pp((xx)/)/qq((xx)) is continuous at is continuous at every value of every value of xx where where qq((xx)) ≠ 0≠ 0..

ExamplesExamples

Find the Find the values ofvalues of x x for which the function is for which the function is continuouscontinuous..

SolutionSolution The function The function ff is a is a polynomial function of degree 3polynomial function of degree 3, so , so ff((xx))

is is continuous for all values ofcontinuous for all values of xx..

3 2( ) 3 2 10f x x x x

Example 3, page 121

ExamplesExamples


SolutionSolution The function The function gg is a is a rational functionrational function. . Observe that the Observe that the denominatordenominator of of gg is is never equal to zeronever equal to zero.. Therefore, we conclude that Therefore, we conclude that gg((xx)) is is continuous for all continuous for all

values ofvalues of xx..

10 2

2

8 4 1( )

1

x xg x

x

Example 3, page 121

ExamplesExamples


SolutionSolution The function The function hh is a is a rational functionrational function. . In this case, however, the In this case, however, the denominatordenominator of of hh is is equal to zeroequal to zero

at at x x = 1= 1 and and x x = 2= 2, which we can see by factoring., which we can see by factoring. Therefore, we conclude that Therefore, we conclude that hh((xx)) is is continuous everywhere continuous everywhere

except except at at x x = 1= 1 and and x x = 2= 2..

3 2

2

4 3 1( )

3 2

x xh x

x x

Example 3, page 121

Intermediate Value TheoremIntermediate Value Theorem

Let’s look again at the Let’s look again at the maglevmaglev example. example. The train The train cannot vanishcannot vanish at any instant of time and at any instant of time and cannot cannot

skip portions of trackskip portions of track and reappear elsewhere. and reappear elsewhere.

Intermediate Value TheoremIntermediate Value Theorem

MathematicallyMathematically, recall that the position of the maglev is a , recall that the position of the maglev is a function of time given by function of time given by ff((tt) = 4) = 4tt22 forfor 0 0 tt 30 30::

24y t 24y t

tt11

ss22

ss33

ss11tt

yy

Suppose the Suppose the positionposition of the maglev is of the maglev is ss11 at some at some timetime tt11 and and its its positionposition is is ss22 at some at some timetime tt22..

Then, if Then, if ss33 is any number betweenis any number between ss11 and and ss22, , there must be there must be at least oneat least one tt33 between between tt11 and and tt22 giving the time at which the giving the time at which the maglev is at maglev is at ss33 ( (ff((tt33) = ) = ss33).).

tt33 tt22

Theorem 4Theorem 4 Intermediate Value TheoremIntermediate Value Theorem

The Maglev example carries the gist of the The Maglev example carries the gist of the intermediate value theoremintermediate value theorem::

If If ff is a is a continuous functioncontinuous function on a on a closed intervalclosed interval [[aa, , bb]] and and MM is any number between is any number between ff((aa) ) and and ff((bb)),, then then there there is at least one numberis at least one number cc in in [[aa, , bb]] such that such that ff((cc) ) = M= M..

( )y f x ( )y f x

aa

ff((bb))

MM

ff((aa))

xx

yy

cc bb

( )y f x ( )y f x

xx

yy

cc11 cc22 cc33bbaa

ff((bb))

MM

ff((aa))

Theorem 5 Theorem 5 Existence of Zeros of a Continuous FunctionExistence of Zeros of a Continuous Function

A A special casespecial case of this theorem is when a continuous of this theorem is when a continuous functionfunction crosses the crosses the xx axisaxis..

If If ff is a is a continuous functioncontinuous function on a on a closed intervalclosed interval [[aa, , bb]], , and if and if ff((aa) ) and and ff((bb)) have opposite signs, then there is have opposite signs, then there is at at least one solutionleast one solution of the equation of the equation ff((xx) ) = = 0 0 in the in the interval interval ((aa, , bb))..

( )y f x ( )y f xff((bb))

ff((aa))

xx

yy

cc bb

( )y f x ( )y f x

xx

yy

cc22 cc33bbaa

ff((bb))

ff((aa))

aa cc11

ExampleExample

Let Let ff((xx) = ) = xx33 + + x x + 1+ 1..

a.a. Show that Show that ff is is continuouscontinuous for all values of for all values of xx..

b.b. Compute Compute ff(–1)(–1) and and ff(1)(1) and use the results to deduce that and use the results to deduce that there must be at least one number there must be at least one number xx = = cc, where , where cc lies in the lies in the interval interval (–1, 1)(–1, 1) and and ff((cc) = 0) = 0..

SolutionSolution

a.a. The function The function ff is a is a polynomial function of degree 3polynomial function of degree 3 and is and is therefore therefore continuous everywherecontinuous everywhere..

b.b. f f (–1) = (–1)(–1) = (–1)33 + (–1) + (–1) + 1 = –1 + 1 = –1 andand f f (1) = (1)(1) = (1)33 + (1) + (1) + 1 = 3 + 1 = 3

Since Since f f (–1) (–1) andand f f (1) (1) have have opposite signsopposite signs, , Theorem 5Theorem 5 tells tells us that there must be at least one number us that there must be at least one number x = cx = c with with –1 < c < 1–1 < c < 1 such that such that ff((cc) = 0) = 0..

Example 5, page 124

2.62.6The DerivativeThe Derivative

x

y

f(x + h) – f(x)

hA

f(x + h)

f(x)

x x + h

An Intuitive ExampleAn Intuitive Example

Consider the Consider the maglevmaglev example from Section 2.4. example from Section 2.4. The The positionposition of the maglev is a function of of the maglev is a function of timetime given by given by

ss == ff((tt) = 4) = 4tt22 (0 (0 tt 30) 30)

where where ss is measured in is measured in feetfeet and and tt in in secondsseconds.. Its graph is:Its graph is:

( )s f t ( )s f ts s (ft)(ft)

11 22 33 44

6060

4040

2020

t t (sec)(sec)

An Intuitive ExampleAn Intuitive Example

The graph rises The graph rises slowlyslowly at firstat first but but more rapidlymore rapidly over timeover time.. This suggests the This suggests the steepnesssteepness of of ff((tt)) is related to the is related to the speedspeed of of

the maglev, which the maglev, which also increases also increases over time.over time. If so, we might be able to If so, we might be able to findfind the the speedspeed of the maglev at of the maglev at

any given time by finding the any given time by finding the steepnesssteepness of of f f at that time.at that time. But how do we find the steepness of a point in a curve?But how do we find the steepness of a point in a curve?

( )s f t ( )s f ts s (ft)(ft)

11 22 33 44

6060

4040

2020

t t (sec)(sec)

AA

Slopes of Lines and of CurvesSlopes of Lines and of Curves

The The slopeslope at a point of a at a point of a curvecurve is given by the is given by the slope of theslope of the tangenttangent to the curve at that point: to the curve at that point:

yy

Suppose we want to find the Suppose we want to find the slopeslope at point at point AA..

The The tangent linetangent line has the has the same slopesame slope as the as the curvecurve does does at point at point AA..

xx

1.8Slope 1.8

1

y

x

1.8

Slope 1.81

y

x

The The slopeslope of a point in a of a point in a curvecurve is given by the is given by the slope of theslope of the tangenttangent to the curve at that point: to the curve at that point:

y y = 1.8= 1.8

x x = 1= 1AA


The The slopeslope of the of the tangent tangent in in this case is this case is 1.81.8::

Slope = 1.8Slope = 1.8

xx

yy

To calculate To calculate accuratelyaccurately the slope of a the slope of a tangenttangent to a to a curvecurve, we , we must make the must make the change inchange in xx as as smallsmall as possible:as possible:

y y = 3= 3

x x = 4= 4

3Slope 0.75

4

y

x

As we let As we let xx get get smallersmaller, the , the slopeslope of theof the secantsecant becomes becomes

more and more similarmore and more similar to the to the slopeslope of theof the tangenttangent to the to the

curve at that point.curve at that point.



AA

xx

yy


y y = 2.4= 2.4

x x = 3= 3AA



2.4Slope 0.8

3

y

x

As we let As we let xx get get smallersmaller, the , the slope slope of theof the secantsecant becomes becomes

more and more similarmore and more similar to the to the slope slope of theof the tangenttangent to the to the


xx

yy


y y = 2.2= 2.2

x x = 2= 2AA



2.2Slope 1.1

2

y

x




xx

yy

y y = 1.5= 1.5

x x = 1= 1


AA



1.5Slope 1.5

1

y

x




xx

yy


y y = 0.00179= 0.00179x x = 0.001= 0.001

AA



0.00179Slope 1.8

0.001

y

x




xx

yy

In general, we can express the In general, we can express the slope slope of theof the secant secant as follows: as follows:


x

y

f(x + h) – f(x)

hA

f(x + h)

f(x)

x x + h

( ) ( ) ( ) ( )Slope

( )

y f x h f x f x h f x

x x h x h

( ) ( ) ( ) ( )

Slope( )

y f x h f x f x h f x

x x h x h

Thus, as Thus, as hh approaches zeroapproaches zero, the , the slopeslope of theof the secantsecant approachesapproaches the the slopeslope of theof the tangenttangent to the curve at that point: to the curve at that point:

x

y

f(x + h) – f(x)

hA

f(x + h)

f(x)

x x + h



x

y

f(x + h) – f(x)

hA

f(x + h)

f(x)

x x + h



x

y

f(x + h) – f(x)

hA

f(x + h)

f(x)

x x + h



x

y

f(x + h) – f(x)

hA

f(x + h)

f(x)

x x + h



x

y

f(x + h) – f(x)Af(x + h)f(x)

xx + h

h


Thus, as Thus, as hh approaches zeroapproaches zero, the , the slopeslope of theof the secantsecant approachesapproaches the the slopeslope of theof the tangenttangent to the curve at that point. to the curve at that point.

Expressed in Expressed in limits notationlimits notation::

TheThe slope slope of the tangentof the tangent line to the graph line to the graph of of ff at the point at the point PP((x, fx, f((xx)))) is given by is given by

if it exists.if it exists.

0

( ) ( )limh

f x h f x

h

0

( ) ( )limh

f x h f x

h


Average Rates of ChangeAverage Rates of Change

We can see that measuring theWe can see that measuring the slopeslope of theof the tangent linetangent line to to a graph is a graph is mathematically equivalentmathematically equivalent to finding the to finding the rate of changerate of change of of ff at at xx..

The number The number ff((x x + + hh) – ) – ff((xx)) measures the measures the changechange in in yy that that corresponds to a corresponds to a changechange hh in in xx..

Then the Then the differencedifference quotient quotient

measures the measures the averageaverage rate of change rate of change of of yy with respect to with respect to xx over the interval over the interval [[xx, , x x + + hh]]..

In the In the maglev examplemaglev example, if , if yy measures the measures the positionposition the train the train at time at time xx, then the , then the quotientquotient give the give the average velocityaverage velocity of the of the train over the time interval train over the time interval [[xx, , x x + + hh]]..

( ) ( )f x h f x

h

Average Rates of ChangeAverage Rates of Change

The The average rate of changeaverage rate of change of of ff over the over the intervalinterval [[xx, , x x + + hh]] or or slopeslope of the of the secant linesecant line to the graph of to the graph of ff through the points through the points ((xx, , ff((xx)))) and and ((x x + + hh, , ff((x + hx + h)))) is is

( ) ( )f x h f x

h

( ) ( )f x h f x

h

Instantaneous Rates of ChangeInstantaneous Rates of Change

By taking theBy taking the limit limit of theof the difference quotient difference quotient as as hh goes to goes to zero, evaluatingzero, evaluating

we obtain the we obtain the rate of changerate of change of of f f at at xx.. This is known as the This is known as the instantaneousinstantaneous rate of change rate of change of of f f at at xx

(as opposed to the (as opposed to the averageaverage rate of changerate of change).). In the maglev example, if In the maglev example, if y y measures the measures the positionposition of a of a

train at time train at time xx, then the, then the limit limit gives the gives the velocityvelocity of the train of the train at time at time xx..

0

( ) ( )limh

f x h f x

h

Instantaneous Rates of ChangeInstantaneous Rates of Change

The The instantaneousinstantaneous rate of change rate of change of of f f at at xx or or slopeslope of theof the tangent line tangent line to the graph of to the graph of ff at at ((xx, , ff((xx)))) is is

This limit is called the This limit is called the derivativederivative of of f f at at xx . .

0

( ) ( )limh

f x h f x

h

0

( ) ( )limh

f x h f x

h

The Derivative of a FunctionThe Derivative of a Function

The The derivative derivative of a function of a function f with respect to with respect to x is the is the function function f′′′ (read “(read “f prime”). prime”).

The The domaindomain of of f ′′ is the set of is the set of allall xx where the where the limit existslimit exists..

Thus, the Thus, the derivativederivative of function of function ff is a is a functionfunction f f ′′ that that gives the gives the slopeslope of the tangentof the tangent to the line to the graph of to the line to the graph of ff at any point at any point ((xx, , ff((xx)))) and also the and also the rate of changerate of change of of ff at at xx..

0

( ) ( )( ) lim

h

f x h f xf x

h

The Derivative of a FunctionThe Derivative of a Function

Four Step Process for Finding Four Step Process for Finding ff ′′((xx))

1.1. Compute Compute ff((x + hx + h))..

2.2. Form the difference Form the difference ff((x + hx + h) –) – ff((xx)). .

3.3. Form the quotientForm the quotient

4.4. Compute Compute 0

( ) ( )( ) lim .

h

f x h f xf x

h

0

( ) ( )( ) lim .

h

f x h f xf x

h

( ) ( )f x h f x

h

.

( ) ( )f x h f x

h

.

ExamplesExamples

Find the Find the slope of the tangent lineslope of the tangent line to the graph to the graph ff((xx) = 3) = 3xx + 5 + 5 at any point at any point ((xx, , ff((xx))))..

SolutionSolution The required slope is given by the The required slope is given by the derivativederivative of of ff at at xx.. To find the derivative, we use the To find the derivative, we use the four-step processfour-step process::

Step 1.Step 1. ff((x + hx + h) = 3() = 3(x + hx + h) + 5 = 3) + 5 = 3xx + 3 + 3hh + 5 + 5..

Step 2.Step 2. ff((x + hx + h) – ) – ff((xx) = 3) = 3xx + 3 + 3hh + 5 – (3 + 5 – (3xx + 5) = 3 + 5) = 3hh..

Step 3.Step 3.

Step 4.Step 4. 0 0

( ) ( )( ) 3 3lim lim .

h h

f x h f xf x

h

( ) ( ) 33.

f x h f x h

h h

Example 2, page 138

ExamplesExamples

Find the Find the slope of the tangent lineslope of the tangent line to the graph to the graph ff((xx) = ) = xx22 at at any point any point ((xx, , ff((xx))))..

SolutionSolution The required slope is given by the The required slope is given by the derivativederivative of of ff at at xx.. To find the derivative, we use the To find the derivative, we use the four-step processfour-step process::

Step 1.Step 1. ff((x + hx + h) = () = (x + hx + h))22 = = xx22 + 2 + 2xhxh + + hh22..

Step 2.Step 2. ff((x + hx + h) – ) – ff((xx) = ) = xx22 + 2 + 2xhxh + + hh22 – – xx22 = = hh(2(2xx + + hh))..

Step 3.Step 3.

Step 4.Step 4. 0 0

( ) ( )( ) (2 ) 2lim lim .

h h

f x h f xf x x h x

h

( ) ( ) (2 )2 .

f x h f x h x hx h

h h

Example 3, page 138

ExamplesExamples

Find the Find the slope of the tangent lineslope of the tangent line to the graph to the graph ff((xx) = ) = xx22 at at any point any point ((xx, , ff((xx))))..

The slope of the tangent line is given byThe slope of the tangent line is given by f ′′(x) = 2x.. Now, Now, find and interpretfind and interpret f ′′(2)..

SolutionSolution

f ′′(2) = 2(2) = 4.. This means that, This means that, at the pointat the point

(2, 4)(2, 4)……

… … the slopethe slope of the tangent of the tangent line to the graph is line to the graph is 44..

–2 –1 0 1 2x

5

4

3

2

1

(2, 4)(2, 4)

1

4

y

Example 3, page 138

Applied Example:Applied Example: Demand for Tires Demand for Tires

The management of Titan Tire Company has determined The management of Titan Tire Company has determined that the that the weekly demandweekly demand function of their Super Titan tires function of their Super Titan tires is given byis given by

where where pp is measured in is measured in dollarsdollars and and x x is measured in is measured in thousands of tiresthousands of tires..

Find the Find the average rate of changeaverage rate of change in the in the unit priceunit price of a tire if of a tire if the quantity demanded is between the quantity demanded is between 50005000 and and 60006000 tires; tires; between between 50005000 and and 51005100 tires; and between tires; and between 50005000 and and 50105010 tires.tires.

What is the What is the instantaneous rate of changeinstantaneous rate of change of the unit price of the unit price when the quantity demanded is when the quantity demanded is 50005000 tires? tires?

2( ) 144p f x x



SolutionSolution The The average rate of changeaverage rate of change of the unit price of a tire if the of the unit price of a tire if the

quantity demanded is between quantity demanded is between xx and and x x ++ h h is is

2 2( ) ( ) [144 ( ) ] (144 )f x h f x x h x

h h

2 2 2144 2 144x xh h x

h

22xh h

h

( 2 )h x h

h

2x h



SolutionSolution The The average rate of changeaverage rate of change is given by is given by –2–2x x – – hh.. To find the To find the average rate of changeaverage rate of change of the unit price of a of the unit price of a

tire when the tire when the quantity demandedquantity demanded is between is between 50005000 and and 60006000 tires tires [5, 6][5, 6], we take , we take xx = 5 = 5 and and h h = 1= 1, obtaining, obtaining

or or –$11–$11 per per 10001000 tires. tires. Similarly, with Similarly, with xx = 5 = 5, and , and hh = 0.1 = 0.1, we obtain , we obtain

or or –$10.10–$10.10 per per 10001000 tires. tires. Finally, with Finally, with xx = 5 = 5, and , and hh = 0.01 = 0.01, we get , we get

or or –$10.01–$10.01 per per 10001000 tires. tires.

2(5) 0.1 10.1

2(5) 1 11

2(5) 0.01 10.01



SolutionSolution The The instantaneous rate of changeinstantaneous rate of change of the unit price of a tire of the unit price of a tire

when the quantity demanded is when the quantity demanded is xx tires is given by tires is given by

In particularIn particular, the instantaneous rate of change of the price , the instantaneous rate of change of the price per tire per tire when quantity demandedwhen quantity demanded is is 50005000 is given by is given by –2(5)–2(5), , or or –$10–$10 per tire. per tire.

0 0

( ) ( )( ) ( 2 ) 2lim lim

h h

f x h f xf x x h x

h


Differentiability and ContinuityDifferentiability and Continuity

Sometimes, one Sometimes, one encounters continuousencounters continuous functions that functions that fail fail to be differentiableto be differentiable at certain values in the at certain values in the domaindomain of the of the function function ff..

For exampleFor example, consider the continuous function , consider the continuous function ff below: below:✦ It It fails to be differentiablefails to be differentiable at at x = ax = a , because the graph , because the graph

makes an makes an abrupt changeabrupt change (a corner) at that point. (a corner) at that point.(It is not clear what the(It is not clear what the slopeslope is at that point)is at that point)

ax

y

Differentiability and ContinuityDifferentiability and Continuity

Sometimes, one Sometimes, one encounters continuousencounters continuous functions that functions that fail fail to be differentiableto be differentiable at certain values in the at certain values in the domaindomain of the of the function function ff..

For exampleFor example, consider the continuous function , consider the continuous function ff below: below:✦ It also It also fails to be differentiablefails to be differentiable and and x = bx = b because the because the

slope is not definedslope is not defined at that point. at that point.

ax

y

b

Applied Example:Applied Example: Wages Wages

Mary works at the B&O department store, where, on a Mary works at the B&O department store, where, on a weekday, she is weekday, she is paidpaid $8$8 an hour for the an hour for the firstfirst 8 8 hours and hours and $12$12 an hour of an hour of overtimeovertime. .

The functionThe function

gives Mary’s gives Mary’s earningsearnings on a weekday in which she worked on a weekday in which she worked xx hourshours..

Sketch the graphSketch the graph of the function of the function f f and and explain why it is not explain why it is not differentiabledifferentiable at at xx = 8 = 8..

8 0 8( )

12 32 8

x xf x

x x

if

if

8 0 8( )

12 32 8

x xf x

x x

if

if


Applied Example:Applied Example: Wages Wages

SolutionSolution

The graph of The graph of ff has a has a cornercorner at at x = x = 88 and so is and so is not not differentiabledifferentiable at that point. at that point.

8 0 8( )

12 32 8

x xf x

x x

if

if

8 0 8( )

12 32 8

x xf x

x x

if

if

yy130130

110110

9090

7070

5050

3030

1010

22 44 66 88 1010 1212xx

(8, 64)(8, 64)


End of End of Chapter Chapter

2 Functions and Their Graphs The Algebra of Functions Functions and Mathematical Models Limits One-Sided Limits and Continuity The Derivative.

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2 Functions and Their Graphs The Algebra of Functions Functions and Mathematical Models Limits One-Sided Limits and Continuity The Derivative.