Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 43 ppsx

404 CHAPTER 11 A Portrait of Polynomials and Rational FunctionsExploratory Problems for Chapter 11 Functions and Their Graphs: Tinkering with Polynomials and Rational Functions 1.Find a

9 Find a possible equation to fit each polynomial graph below Notice that a is a negative number

y

x

d

c

(a)

y

x

(b)

(–1, 3)

10 Find a polynomial to fit the graph below

y

x

1

1

–3

11 Each of the graphs on the following page is the graph of a polynomial P (x) For each graph do the following

(a) Determine whether the degree of P (x) is even or odd

(b) Despite the fact that you have just categorized each of the polynomials as being

of either odd or even degree, none of the polynomials graphed are even functions

and none are odd functions Explain

(c) Determine whether the leading coefficient is positive or negative

(d) Determine a good lower bound for the degree of the polynomial Explain your reasoning (For example, the last graph on the right has one turning point, so it must be of degree 2 or more It is not a parabola since it has a point of inflection; therefore we know the degree is higher than 2 It cannot be a polynomial of degree

3 because for |x| large enough, P (x) is positive Therefore, it must be a polynomial

of degree 4 or more.)

402 CHAPTER 11 A Portrait of Polynomials and Rational Functions

P

x

(i)

P

x

(ii)

P

x

(iii)

12 (a) Suppose P (x) is a polynomial of degree 5 Which of the statements that follow must necessarily be true? If a statement is not necessarily true, provide a coun-terexample (an example for which the statement is false)

i P (x) has at least one zero

ii P (x) has no more than four zeros

iii The graph of P (x) has at least one turning point

iv The graph of P (x) has at most four turning points

(b) Suppose P (x) is a polynomial of degree 5 with its natural domain (−∞, ∞) If

P(π ) = 0 and P(π ) = 5, then which one of the following statements is true?

Explain your answer

i P has a local minimum at x = π but this local minimum is not an absolute minimum

ii P has a local minimum at x = π and this local minimum may be an absolute

minimum

iii P has a local maximum at x = π but this local maximum is not an absolute maximum

iv P has a local maximum at x = π and this local maximum may be an absolute

maximum

13 (a) Suppose P (x) is a polynomial of degree 6 Which of the statements that follow must necessarily be true? If a statement is not necessarily true, provide a coun-terexample (an example for which the statement is false)

i P (x) has at least one zero

ii P (x) has no more than five zeros

iii The graph of P (x) has at least one turning point

iv The graph of P (x) has at most five turning points

(b) Suppose P (x) is a polynomial of degree 6 with its natural domain (−∞, ∞) If

P(2) = 0 and P(2) = −1, then which one of the following statements is true?

Explain your answer

i P has a local minimum at x = 2 but this local minimum is not an absolute minimum

ii P has a local minimum at x = 2 and this local minimum may be an absolute

minimum

iii P has a local maximum at x = 2 but this local maximum is not an absolute maximum

iv P has a local maximum at x = 2 and this local maximum may be an absolute

maximum

14 For each of the graphs below, all vertical and horizontal asymptotes are indicated with dotted lines If there are no dotted lines there are no asymptotes

(a) Which of the following could possibly be the graph of a polynomial function? If

the graph could be the graph of a polynomial, what can you say about the degree

of the polynomial? Can you determine whether the degree is even or odd? Can you determine an n such that the degree of the polynomial is at least n?

(b) Which could possibly be the graph of a function of the form f (x) = Cbx

+ D, where C, b, and D are constants?

(c) For each of the remaining graphs (graphs not listed as answers to the previous two questions), what characteristic of the graph made you rule it out?

y

x

y

x

y

x

y

x

y

x

y

x

15 The functions that follow in this exercise are not polynomials We ask you about their range, domain, and graphs with the goal of having you appreciate how nicely polynomial functions behave For each of the following functions:

(a) Determine the domain

(b) Determine the range

(c) Sketch a graph of the function Do this using your knowledge of flipping, stretch-ing, shrinkstretch-ing, shiftstretch-ing, and of graphingf (x)1 ; check your graph with your graphing calculator

Your answers to parts (a) and (b) ought to agree with your answer to part (c) You can use your answers to parts (a) and (b) to select an appropriate viewing window in your calculator

is y = 1/x.)

is y =√x iii h(x) =√ 1

reciprocal.)

the reciprocal.)

404 CHAPTER 11 A Portrait of Polynomials and Rational Functions

Exploratory Problems for Chapter 11

Functions and Their Graphs: Tinkering with Polynomials and Rational Functions

1.Find a polynomial function P (x) that fits the graph drawn below

The x-intercepts should be at x = −2 and x = 0 and the function should have a global minimum of −6 It is not clear exactly where this minimum is attained

y

x

–2

–6

f (x)

Figure 11.19

The goal of this problem is to encourage you to tinker with the equation using what you know about polynomials and their derivatives The first few questions below are designed to steer you in the right direction

(a) Is P (x) a polynomial of even degree, or of odd degree?

(b) Give a lower bound for the degree of P (x) Explain

(c) What are the roots of P (x)? Take a first guess at the equation for P (x) in factored form

(d) What can you do (or what have you done) to make the graph

of P (x) flatten out at x = 0?

(e) Adjust your formula to assure that the minimum value of P (x) will be −6

Do you want to stretch vertically or do you want to shift vertically? You don’t want to uproot the x-intercepts you have

so carefully nailed into place.

(f ) Write a formula for P (x)

(g) Given your answer to the last question, determine where P (x)

takes on its minimum value

Your answer should come from your function and be deter-mined by analyzing its derivative; don’t simply guess by read-ing off the graph.

The next set of problems asks you to think about rational functions, the topic of the next section of this chapter

2.Graph each function f and under it, graph its reciprocal, f (x)1

Then answer the following questions in as much generality as you

can

(a) How is the sign off (x)1 related to the sign of f ?

(b) How is the magnitude (the absolute value) of f (x)1 related to

the magnitude of f ?

(c) What characteristic(s) of f determines the location and type

of vertical asymptote of f (x)1 ?

3.This problem should be done with the aid of a graphing calculator

or computer Give a very rough sketch of the graph of each of the

following If you like, a group can get together and split up the

work, each person graphing a couple of these functions on his or

her calculator The important thing is for you to think about the

relationships between the equations and their graphs

(d) y =x2 (x−1)1 (e) y =x(x−1)1 2 (f ) y =x2 (x−1)1 2

Think about the relationships between these equations and their

graphs, the effect of the factors in the denominators, and the effect

of squaring certain factors Present as many observations as you

can come up with

406 CHAPTER 11 A Portrait of Polynomials and Rational Functions

An Introduction to Rational Functions

Rational functions are functions of the form f (x) =polynomial in xpolynomial in x They are a class of functions that includes the polynomials9(a well-behaved family) as well as some more unruly relatives Rational functions can exhibit much wilder and more varied behavior than polynomials They may be undefined for certain values of x, and therefore may be discontinuous Not only may they be discontinuous, but the magnitude of f may blow

up around a point of discontinuity In other words, it is possible that limx→c|f (x)| = ∞ for some finite number c In this case, the rational function has a vertical asymptote It

is possible that limx→∞f (x) = k for some finite constant k, in which case f (x) has a horizontal asymptote Once you become accustomed to the behavior of rational functions and learn the relationship between the function and the behavior of its graph, you may very well find that rational functions are fun to work with That alone could constitute a reason

to get to know these functions But there are more practical reasons as well

An economist interested in the average cost per pound of producing q pounds of a good will divide the total cost function, C(q), by the number of pounds produced If C(q)

is modeled by a polynomial, then

average cost per item =C(q)

q

is a rational function

Any time two variables are inversely proportional to one another, there is a functional relationship of the form f (x) = k/x, a form with which we have longstanding familiarity Scientists observing naturally occurring phenomena have found such relationships ubiqui-tous For instance, chemists use the combined gas laws relating the pressure, P , temperature,

T, and volume, V , of a gas:

Physicists have found that the gravitational attraction between two objects is inversely proportional to the square of the distance between them For example, a rocket on a journey

in space will be subject to the gravitational force of the earth The acceleration due to the gravitational attraction of the earth is given by

r2 , where G is the universal gravitational constant, mE is the mass of the earth, and r is the distance from the rocket to the center of the earth If the rocket is journeying from the earth

to the moon, then the primary forces acting on it are the gravitational forces of the earth and the moon The acceleration due to the gravitational attraction of the moon is given by

R2 ,

9

where G is as above, mMis the mass of the moon, and R is the distance from the rocket to the moon’s center

Let G be the acceleration of the rocket due to the combined gravitational forces of the earth and moon

A =
acceleration due tothe earth’s gravity



−
acceleration due tothe moon’s gravity

 The two terms have opposite signs because, from the perspective of the rocket, the forces act in opposite directions The distance between the center of the earth and the center of the moon is roughly 240,000 miles We’ll call this distance D Then, when the rocket is a distance x from the center of the earth, its distance from the center of the moon is D − x

Earth

Figure 11.20

(D − x)2, or A(x) = G



mE(D − x)2− mMx2

x2(D − x)2



Ais a rational function of the rocket’s distance from the center of the earth

Removable and Nonremovable Discontinuities and Asymptotes

The main ways in which rational functions can deviate from the behavior of polynomials are that they can have discontinuities (removable or nonremovable) and can have vertical and horizontal asymptotes.10

Points of Discontinuity

A rational function f (x) will be undefined (and hence discontinuous) wherever the denom-inator is zero

rational function are both zero at x = b, then the numerator and denominator have a common factor of x − b If these factors occur with the same multiplicity,11 then the graph has a pinhole at x = b A pinhole, i.e., a situation in which limx→c +f (x) = limx→c −f (x) = L where L is finite, but f (c) = L, is referred to as a removable discontinuity

10 The nonremovable discontinuities show up as vertical asymptotes.

11 What is actually required is that the multiplicity in the numerator is greater than or equal to that in the denominator For example, if f (x) =(x2+1)xx 2, then f (x) =

 (x 2 + 1)x for x = 0 undefined for x = 0 and the graph of f has a pinhole at x = 0.

If g(x) = (x 2 +1)x

x 2 , then, g(x) =

 x 2

+1

x for x = 0 undefined at x = 0 The graph of g(x) has a vertical asymptote at x = 0.

408 CHAPTER 11 A Portrait of Polynomials and Rational Functions

y

x

y =(x2 + 1)x

x pinhole at x = 0

Figure 11.21

at x = b and the numerator is nonzero, then the graph has a vertical asymptote at x = b.12

We need a bird’s-eye view because as x approaches b, f (x) will either increase without bound or decrease without bound If f has a vertical asymptote at x = b, then near x = b the graph of f will look like one of the graphs shown in Figure 11.22

x = b

For instance,

f (x) = 1

(x–b)2

x = b

For instance,

f (x) = 1

x –b

x = b

For instance,

f (x) = –1

(x–b)2

x = b

For instance,

f (x) = –1

x –b

Figure 11.22

Simple sign information will distinguish between the four options The base idea behind this

is just what we discussed in Chapter 7 when looking at limx→0+ 1

x= ∞ and limx→0 − 1

x=

−∞ The graph will never cross its vertical asymptotes because the function is undefined there

Horizontal Asymptotes: A bird’s eye-view of rational functions.13 A horizontal asymp-tote supplies information about f as the magnitude of x increases without bound; it indicates the behavior of the function toward the extremities of the graph in the case where these ex-tremities look like horizontal lines If limx→∞f (x) = K for some (finite) constant K, we say f has a horizontal asymptote at K, and similarly if limx→−∞f (x) = K Recall that polynomials never have horizontal asymptotes, and exponential functions have one-sided

horizontal asymptotes For any rational function f , if limx→∞f (x) = K where K is finite, then limx→−∞f (x) = K as well; the horizontal asymptotes are two-sided In order to in-vestigate whether or not the graph of a rational function f (x) has a horizontal asymptote,

we must look at the behavior of f as x → ∞ and x → −∞ For x very large in magnitude, any polynomial is dominated by its term of highest degree; therefore, we will break down our investigations into cases in which we are concerned with the relative degrees of the

12What is actually required is that b is a zero of the denominator, and if it is also a zero of the numerator, the multiplicity of

the root in the numerator is less than that in the denominator.

13

numerator and the denominator of the rational function We will use the fact that for any positive integer n and any constant k, limx→∞ xkn = 0

Case I Degree of Numerator Less Than Degree of Denominator

EXAMPLE 11.11 f (x) = x−1

2x 2

−x−3 Calculate limx→∞2x2x−1

−x−3

SOLUTION Both the numerator and the denominator of this fraction are growing without bound, but

the denominator grows much more rapidly than the numerator (Try this out numerically on your calculator for very large x.) For x very large in magnitude the term of highest degree dominates any polynomial, so the numerator “looks like” x and the denominator like 2x2 Therefore, from a bird’s-eye view, for x very large in magnitude f (x) looks like2xx2 =2x1 limx→∞ 2x1 = 0

More formally, we can get a handle on this limit by dividing the numerator and denominator by the highest power of x occurring in the denominator:

lim

x→∞

x − 1 2x2− x − 3= limx→∞

x

x 2 −x12 2x 2

x 2 −xx2 −x32

= lim

x→∞

1

x−x12

2 −1x−x32

2 = 0

This argument can be generalized to show that if the degree of the numerator is less than the degree of the denominator, then the rational function has a horizontal asymptote at

y = 0, the x-axis

Case II Degree of Numerator Equal to Degree of Denominator

EXAMPLE 11.12 f (x) = x2−1

2x 2 −x−3 Calculate limx→∞ x2−1

2x 2 −x−3

SOLUTION Both the numerator and the denominator of this fraction are growing without bound, but

the denominator is growing about twice as rapidly as is the numerator From a bird’s-eye view, for x very large in magnitude the numerator “looks like” x2and the denominator like 2x2 Therefore, for x very large in magnitude f (x) looks like x2

2x 2 =12 Again, you can test this out numerically on your calculator for very large x

More formally, we can get a handle on this limit by dividing the numerator and denominator by the highest power of x occurring in the denominator:

lim

x→∞

x2− 1 2x2− x − 3= limx→∞

x2

x 2 −x12 2x 2

x 2 −xx2 −x32

= lim

x→∞

1 − x12

2 −1x−x32

2. Similarly, we can show limx→−∞f (x) =21

This argument can be generalized to show that if the degree of the numerator is equal

to the degree of the denominator, then the rational function has a horizontal asymptote at

y = leading coefficient of numerator leading coefficient of denominator.

Case III Degree of Numerator Greater Than Degree of Denominator

EXAMPLE 11.13 f (x) = x3−1

2x 2 −x−3 Calculate limx→∞ x3−1

2x 2 −x−3

410 CHAPTER 11 A Portrait of Polynomials and Rational Functions

SOLUTION Both the numerator and the denominator of this fraction are growing without bound, but

the numerator is growing much more rapidly than the denominator From a bird’s-eye view, for x very large in magnitude the numerator looks like x3and the denominator like 2x2 Therefore, for x very large in magnitude f (x) looks like x3

2x 2 =x2 As x → ∞, x2→ ∞ More formally, we can get a handle on this limit by dividing the numerator and denominator by the highest power of x occurring in the denominator:

lim

x→∞

x3− 1 2x2− x − 3= limx→∞

x3

x 2 −x12 2x 2

x 2 −xx2 −x32

= lim

x→∞

x −x12

2 −x1−x32

= lim

This argument can be generalized to show that if the degree of the numerator is greater than the degree of the denominator, then the rational function has no horizontal asymptote

Graphs from Equations/Equations from Graphs

A rational function may be discontinuous At a point of discontinuity the function can change sign without passing through zero If a function is discontinous its derivative will

be discontinuous as well (Recall that differentiability guarantees continuity, and therefore discontinuity guarantees a lack of differentiability.) Therefore the derivative can change sign without passing through zero and the function can change from increasing to decreasing or vice versa without having a horizontal tangent line

Graphing a Rational Function f (x) =polynomial in xpolynomial in x

some information is easiest to get when the expression is factored If there is a common factor in the numerator and denominator, cancel with care If the common factor is (x − c),

then the function is undefined at c, so its graph has a pinhole or a vertical asymptote at c.

f (x) =(x+1)(x+2)(x+1)(x+3)=(x+2)(x+3)for x = −1 f (x)is undefined when x = −1

The graph has a pinhole at x = −1

f (x) =(x+1)(x+1)2 =x+11 for x = −1 f (x)is undefined when x = −1

The graph has a vertical asymptote at x = −1

f (x) =(x(x22+1)(x+2)+1)(x+3)=(x+2)(x+3) The graph has no pinholes (x2+ 1) = 0

1 Identify all vertical asymptotes.Where is the denominator of the simplified expres-sion zero?

f (x)will be undefined at a vertical asymptote and nearby |f (x)| → ∞