11.1 POWER FUNCTIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011,...

11.1

POWER FUNCTIONS

Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Proportionality and Power Functions

• Example 1 The area, A, of a circle is proportional to the square of its radius, r: A = π r2.

• Example 2 The weight, w, of an object is inversely proportional to the square of the object’s distance, d, from the earth’s center: w =k/d2 = kd−2.



A quantity y is (directly) proportional to a power of x if y = k xn, k and n are constants.

A quantity y is inversely proportional to xn if y =k/xn , k and n are constants.

A power function is a function of the form f(x) = k xp, where k and p are constants.

Proportionality and Power Functions

• Example 3 Which of the following functions are power functions? For each power function, state the value of the constants k and p in the formula y = k xp.

(a) f(x) =

(b) g(x) = 2 (x + 5)3

(c) u(x) =

(d) v(x) = 6 ・ 3x

• Solution: The functions f (k=13, p=1/3) and u (k=5, p=-3/2) are power functions; the functions g and v are not power functions.


313 x

3/25 x

The Effect of the Power p• Graphs of the Special Cases y = x0 and y = x1

• The power functions corresponding to p = 0 and p = 1 are both linear. The function y = x0 = 1, except at x = 0. Its graph is a horizontal line with a hole at (0,1). The graph of y = x1 = x is a line through the origin with slope +1. Both graphs contain the point (1,1).


1 1 2 3 4 5x

1

1

2

3

4

5

y

y = x1

y = x0(1,1)

●

The Effect of the Power pPositive Integer Powers


4 2 0 2 4x

2

4

6

8

10

12y

(1,1)

y = x4y = x2

(-1,1)● ●

2 1 1 2x

4

2

2

4

y y = x3 y = x5

(1,1)

(-1,-1)

●

●

Graphs of positive even powers of x are U-shaped

Graphs of positive odd powers of x are “chair”-shaped

3 2 1 0 1 2 3x

1

2

3

4

5

6y

3 2 1 1 2 3x

4

2

2

4

y

The Effect of the Power pNegative Integer Powers


(1,1)

(-1,-1)

y = x-2

(1,1)(-1,1)

Both graphs have a vertical asymptote of x = 0

Both graphs have a horizontal asymptote of y = 0

y = x-1

x 0.1 0.05 0.01 0.001 0.0001 0 1/x 10 20 100 1000 10000 undefined 1/x2 100 400 10,000 1,000,000 100,000,000 undefined

x 0 10 20 30 40 50 1/x undefined 0.1 0.05 0.033333 0.025 0.02 1/x2 undefined 0.01 0.0025 0.001111 0.000625 0.0004

4 2 2 4x

2

1

1

2y

4 2 2 4x

2

1

1

2y

The Effect of the Power pGraphs of Positive Fractional Powers


(1,1) y = x1/4

y = x1/2y = x1/3

y = x1/5(1,1)

(-1,-1)

Graphs of even roots of x are not defined for x < 0

Graphs of odd roots of x are defined for all values of x

11.2

POLYNOMIAL FUNCTIONS


A General Formula for the Family of Polynomial Functions

The general formula for the family of polynomial functions can be written as

p(x) = an xn + an−1 xn−1 + . . . + a1 x + a0, where n is a positive integer called the degree of p and where an ≠ 0.

• Each power function aixi in this sum is called a term.

• The constants an, an−1, . . . , a0 are called coefficients.

• The term a0 is called the constant term. The term with the

highest power, anxn, is called the leading term.

• To write a polynomial in standard form, we arrange its

terms from highest power to lowest power, going from left

to right.Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally


Like the power functions from which they are built, polynomials are defined for all values of x. Except for polynomials of degree zero (whose graphs are horizontal lines), the graphs of polynomials do not have horizontal or vertical asymptotes; they are smooth and unbroken. The shape of the graph depends on its degree; typical graphs are shown below.

Quadratic Cubic Quartic Quintic n = 2 n = 3 n = 4 n=5

The Long-Run Behavior of Polynomial Functions

When viewed on a large enough scale, the graph of the polynomial p(x) = anxn + an−1xn−1 + ・・・ + a1x + a0 looks like the graph of the power function y = anxn. This behavior is called the long-run behavior of the polynomial. Using limit notation, we write

• •


nn

xxxaxp

lim)(lim

nn

xxxaxp

lim)(lim

Zeros of PolynomialsExample 3

Given the polynomial q(x) = 3x6 − 2x5 + 4x2 − 1,where q(0) = −1, is there a reason to expect a solution to the equation q(x) = 0? Ifnot, explain why not. If so, how do you know?

Solution


1 .0 0 .5 0 .5 1 .0x

2

4

6

8

10

12

q

Plot of q(x)Since q(0) is negative and since the leading term, 3x6 , goes to positive infinity as x increases in either the positive or negative direction, we know that the graph must cross the x-axis for at least one positive and one negative value of x.

11.3

THE SHORT-RUN BEHAVIOR OF POLYNOMIALS


Visualizing Short-Run and Long-Run Behaviors of a Polynomial

Example 1Compare the graphs of the polynomials f, g, and h given by

f(x) = x4 − 4x3 + 16x − 16, g(x) = x4 − 4x3 − 4x2 + 16x, h(x) = x4 + x3 − 8x2 − 12x.


4 2 2 4x

30

20

10

10

20

30

y

10 5 5 10x

10 00

20 00

30 00

40 00y

A close-up look near zeros and turns A larger scale look resembling x4

Factored Form, Zeros, and the Short-Run Behavior of a Polynomial

Example 2Investigate the short-run behavior of the third degree polynomial u(x) = x3 − x2 − 6x.(a) Rewrite u(x) as a product of linear factors.(b) Find the zeros of u(x).

Solution(a) u(x) = x (x2 – x – 6) = x (x + 2) (x – 3)(b) u(x) = 0 if and only if one of its linear factors is zero, so

the zeros of u(x) occur at x = 0, x = -2, and x = 3.Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Example 2 (continued)Investigate the short-run behavior of the third degree polynomial u(x) = x3 − x2 − 6x.(c) Describe the graph of u(x). Where does it cross the x-

axis? the y-axis? Where is u(x) positive? Negative?


3 2 1 1 2 3 4x

20

10

10

20

y y = x3 y =u(x)

(-2,0)(3,0)

(0,0)

u negative → ↖ u negative

← u positive

u positive ↘

The graph of u(x) has zeros at x = -2, 0, and 3. The function changes sign at each of these zeros. Its long-run behavior resembles y = x3.

Factors and Zeros of Polynomials

Suppose p is a polynomial. • If the formula for p has a linear factor, that is,

a factor of the form (x − k), then p has a zero at x = k.

• Conversely, if p has a zero at x = k, then p has a linear factor of the form (x − k).


The Number of Factors, Zeros, and Bumps

The graph of an nth degree polynomial has at most n zeros and turns at most (n − 1) times.


4 2 2 4x

30

20

10

10

20

30

y

Consider the three 4th degree polynomials in Example 1

The red graph has two zeros.

The blue graph has three zeros.

The green graph has four zeros.

Multiple Zeros

If p is a polynomial with a repeated linear factor, then p has a multiple zero. • If the factor (x − k) occurs an even number

of times, the graph of y = p(x) does not cross the x-axis at x = k, but “bounces” off the x-axis at x = k.

• If the factor (x − k) occurs an odd number of times, the graph of y = p(x) crosses the x-axis at x = k, but it looks flattened there.


Example 3Describe in words the zeros of the 4th-degree polynomials f(x), g(x), and h(x), in the graphs below


2 1 1 2 3x

3 0

2 0

1 0

1 0

y

2 1 1 2 3 4x

30

20

10

10

y

3 2 1 1 2 3x

30

20

10

10

y

The graph suggests that f has a single zero at x = −2. The flattened appearance near x = 2 suggests that f has a multiple zero there. Since the graph crosses the x-axis at x = 2 (instead of bouncing off it), this zero must occur an odd number of times. Since f is 4th degree, f has at most 4 factors, so there must be a triple zero at x = 2.

The graph of g has four single zeros.

The graph of h has two single zeros (at x = 0 and x = 3) and a double zero at x = −2. The multiplicity of the zero at x = −2 is not higher than two because h is of degree n = 4.

f(x) g(x) h(x)

Finding the Formula for a Polynomial from its Graph

Example 4Find a possible formula for the polynomial function f graphed to the right.SolutionBased on its long-run behavior, f is of odd degree greater than or equal to 3. The polynomial has zeros at x = −1 and x = 3. We see that x = 3 is a multiple zero of even power, because the graph bounces off the x-axis here instead of crossing it. Therefore, we try the formula f(x) = k (x + 1) (x – 3)2, where k represents a stretch factor. The shape of the graph shows that k must be negative. To find k, we use the fact that f(0) = kˑ1ˑ(-3)2= −3, so k= – 1/3 and f(x) = – 1/3 (x + 1) (x – 3)2 is a possible formula for this polynomial.


2 2 4x

1 0

5

5

1 0

y

y = f(x)

11.4

RATIONAL FUNCTIONS


The Average Cost of Producing a Therapeutic Drug

A pharmaceutical company wants to begin production of a new drug. The total cost C, in dollars, of making q grams of the drug is given by the linear function

C(q) = 2,500,000 + 2000q.The fact that C(0) = 2,500,000 tells us that the company spends $2,500,000 before it starts making the drug. This quantity is known as the fixed cost because it does not depend on how much of the drug is made. It represents the cost for research, testing, and equipment. In addition, the slope of C tells us that each gram of the drug costs an extra $2000 to make. This quantity is known as the variable cost per unit. It represents the additional cost, in labor and materials, to make an additional gram of the drug.


The Average Cost of Producing a Therapeutic Drug – continued

We define the average cost, a(q), as the cost per gram to produce q grams (q > 0) of the drug:


q

qqqC

qa2000000,500,2)(

)(grams ofNumber

cost Total

0 5 0 0 0 1 0 0 0 0 1 5 0 0 0 2 0 0 0 0q , n u m b e r o f g r a m s

1 0 0 0

2 0 0 0

3 0 0 0

4 0 0 0

5 0 0 0

6 0 0 0

y , a v e r a g e c o s t d o l l a r s p e r g r a m y = a(q)

y= 2000: horizontal asymptote

The graph of y = a(q), a rational function, has a horizontal asymptote at y = 2000 and a vertical asymptote at q = 0

What is a Rational Function?

If r can be written as the ratio of polynomial functions p(x) and q(x), that is, if

then r is called a rational function. We assume that q(x) is not the constant polynomial q(x) = 0.


,)(

)()(

xq

xpxr

The Long-Run Behavior of Rational Functions

For x of large enough magnitude (either positive or negative), the graph of the rational function r looks like the graph of a power function. If r(x) = p(x)/q(x), then the long-run behavior ofy = r(x) is given by

Using limits, we write


q

py

of termLeading

of termLeading

q

p

xq

xpxx of termLeading

of termLeadinglim

)(

)(lim


Example 1For positive x, describe the positive long-run behavior of the rational function SolutionThe leading term in the numerator is x and the leading term in the denominator is x. Thus for large enough values of x,


.2

3)(

x

xxr

.11lim)(lim and 1)(xx

xrx

xxr


Example 1 - continuedFor positive x, describe the positive long-run behavior of the rational function Solution


.2

3)(

x

xxr

0 5 1 0 1 5 2 0 2 5x

0 .5

1 .0

1 .5

y

y = 1: horizontal asymptote

2

3

x

xy For large enough values of x, the

graph of y = r(x) looks like the line y = 1, its horizontal asymptote.However, for x > 0, the graph of r is above the line since the numerator is larger than the denominator.


Example 2For positive x, describe the positive long-run behavior of the rational function SolutionThe leading term in the numerator is 3x and the leading term in the denominator is x2. Thus for large enough values of x,


.2

13)(

2

xx

xxg

.03

lim)(lim and 33

)(xx2

xxg

xx

xxg


Example 2 - continuedFor positive x, describe the positive long-run behavior of the rational function SolutionPlotting both functions for values of x > 1:


.2

13)(

2

xx

xxg

2 4 6 8 10x

0 .5

1 .0

1 .5

2 .0

2 .5

y

y = 3/x

2

132

xx

xy

What Causes Asymptotes?

Examples 1 and 2 – Horizontal AsymptotesWe have seen that horizontal asymptotes, shown as red dashed lines, can occur as x approaches positive or negative infinity and these can be determined using the leading terms.


5 5x

15

10

5

5

10

15y

4 2 2 4x

5

5

y

.2

3)(

x

xxr .

2

13)(

2

xx

xxg

x=-2 x=-2 x=1

y = 1y = 0

What Causes Asymptotes?

Examples 1 and 2 – Vertical AsymptotesVertical asymptotes occur whenever a denominator goes to zero, while the numerator does not. The functions given in Examples 1 and 2 both possess vertical asymptotes (blue vertical lines) as well as horizontal asymptotes (red dashed lines).


5 5x

15

10

5

5

10

15y

4 2 2 4x

5

5

y

.2

3)(

x

xxr .

2

13)(

2

xx

xxg

x=-2 x=-2 x=1

y = 0y = 1

11.5

THE SHORT-RUN BEHAVIOR OF RATIONAL FUNCTIONS


The Zeros and Vertical Asymptotes of a Rational Function

Example 2Find the zeros and vertical asymptotes of the rational function r(x). This was explored in the previous section.SolutionNoting that the numerator is zero when x = -3, we will expect the graph to cross the x-axis at x = -3. Noting that the denominator is zero when x = -2, we will expect a vertical asymptote when x = -2


.2

3)(

x

xxr

4 2 2 4x

4

2

2

4

6

8

y

2

3)(

x

xxr The horizontal asymptote, y = 1,

was determined in the previous section.

The Graph of a Rational Function

If r is a rational function given by

where p and q are polynomials with different zeros, then:• The long-run behavior and horizontal asymptote (if any)

of r are given by the ratio of the leading terms of p and q.• The zeros of r are the same as the zeros of the numerator,

p.• The graph of r has a vertical asymptote at each of the

zeros of the denominator, q.


)(

)()(

xq

xpxr

Can a Graph Cross an Asymptote?The graph of a rational function never crosses a vertical asymptote. However, the graphs of some rational functions cross their horizontal asymptotes. The difference is that a vertical asymptote occurs where the function is undefined, so there can be no y-value there, whereas a horizontal asymptote represents the limiting value of the function as x →± ∞. Example:


2

2 32)(

x

xxxr

Horizontal asymptote at y = 1 and r(3/2) = 1

6 4 2 2 4 6x

3

2

1

1

y

y = 1y=r(x)

Rational Functions as Transformations of Power Functions

• The average cost function from Section 11.4 can be written as

• Thus, the graph of a is the graph of the power function y = 2,500,000q−1 shifted up 2000 units.

• Many rational functions can be viewed as translations of power functions.


2000000,500,22000000,500,2

)( 1

qq

qqa

6 4 2 2 4 6x

3

2

1

1

y

Finding a Formula for a Rational Function from its Graph

The graph of a rational function can give a good idea of its formula. Zeros of the function correspond to factors in the numerator and vertical asymptotes correspond to factors in the denominator.Exercise 34Find a possible formula for the function whose graph is given.

Solution


It is important to note that the horizontal asymptote at y =1 implies that the polynomials in the numerator and denominator have the same degree.

y = 1

2)2(

)1)(3()(

x

xxxr

(-3,0) (1,0)

(0,-3/4)

Possible rational function:

When Numerator and Denominator Have the Same Zeros: Holes

The rational function

is undefined at x = 1 because the denominator equalszero at x = 1. However, the graph of h does not have avertical asymptote at x = 1 because the numerator of halso equals zero at x = 1; h(1) = 0/0 which is undefined.

In fact, h(x) = x + 2 everywhere except at x = 1, so the graph of y = h(x) looks just like that for y = x + 2, except that there is a hole (not big enough to see) at the point (1,3) in the graph of y = h(x), since h(1) is not defined.


1

2)(

2

x

xxxh

11.6

COMPARING POWER, EXPONENTIAL, AND LOG FUNCTIONS


Comparing Power Functions

Example 1Let f(x) = 100x3 and g(x) = x4 for x > 0. Compare the long-term behavior of these two functions using graphs.Solution


When comparing power functions with positive coefficients, higher powers dominate.

0 2 4 6 8 1 0x

2 0 0 0 0

4 0 0 0 0

6 0 0 0 0

8 0 0 0 0

1 0 0 0 0 0y

Close-up view

0 100 200 300 400 500x

1 10 10

2 10 10

3 10 10

4 10 10

5 10 10y

Far-away view

0 20 40 60 80 10 0x

2 .0 10 7

4 .0 10 7

6 .0 10 7

8 .0 10 7

1 .0 10 8

1 .2 10 8

1 .4 10 8

y

Point of intersection: (100,1004)

f(x)

f(x)g(x)

g(x)

BlueWins

Comparing Exponential Functions and Power Functions

Example Compare the functions f(x) = x4 and g(x) = 2x for x > 0. Solution


Any positive increasing exponential function eventually grows faster than any power function.

0 2 4 6 8 1 0x

2 0 0 0

4 0 0 0

6 0 0 0

8 0 0 0

1 0 0 0 0y

0 20 40 60 80 100x

2 107

4 107

6 107

8 107

1 108y

Close-up view

Far-away view

0 5 10 15 20x

50 000

100 000

150 000

200 000y

Point of intersection: (16, 65,536)

f(x)

f(x)

g(x)

g(x)

BlueWins

Decreasing Exponential Functions and Decreasing Power Functions

Example Compare the functions f(x) = x-2 and g(x) = 1.1-x for x > 0. Solution


Any positive decreasing exponential function eventually approaches the horizontal axis faster than any positive decreasing power function.

0 20 40 60 80x

0 .002

0 .004

0 .006

0 .008

0 .010y

0 50 10 0 15 0 20 0 25 0x

0 .000 2

0 .000 4

0 .000 6

0 .000 8

0 .001 0yClose-up view

Far-away view

f(x)

f(x)

g(x)g(x)

BlueWins

Comparing Log and Power Functions

Example Compare the functions f(x) = x1/2 , g(x) = log x, and h(x) = ln x for x > 0. Solution


Any positive increasing power function eventually grows more rapidly than y = log x and y = ln x.

0 5 10 15 20x

1

2

3

4

5y

f(x)

g(x)

h(x)

Red Wins

11.7

FITTING EXPONENTIALS AND POLYNOMIALS TO DATA


The Spread of AIDSVisualizing the DataThe data in the table gives the total number of deaths in the US from AIDS from 1981 to 1996. The graph suggests that a linear function may not give the best possible fit for these data.


t N1 1592 6223 21304 56355 126076 247177 411298 622489 90039

10 12157711 15819312 19928713 24392314 29258615 34095716 375904

0 2 4 6 8 10 12 14 16 180

50

100

150

200

250

300

350

400

US Deaths from AIDS (1981 - 1996)

t (years since 1980)

N (i

n th

ousa

nds)

The Spread of AIDSFitting Functions to the Data


0 2 4 6 8 10 12 14 160

50

100

150

200

250

300

350

400



N (i

n th

ousa

nds)

0 2 4 6 8 10 12 14 160

50

100

150

200

250

300

350

400



N (i

n th

ousa

nds)

Exponential Fit FunctionN = 630e0.47t

Power Fit FunctionN = 107t3.005

Linear Fit FunctionN = −97311 + 25946t

0 2 4 6 8 10 12 14 160

50

100

150

200

250

300

350

400



N (i

n th

ousa

nds)

Which Function Best Fits the Data?

Despite the fact that all three functions fit the data reasonably well up to 1996, it is important to realize that they give wildly different predictions for the future. If we use each model to estimate the total number of AIDS deaths by the year 2010 (when t = 30), • the exponential model gives N = 630e(0.47)30 ≈

837,322,467, about triple the current US population;• the power model gives N = 107(30)3.005 ≈ 2,938,550, or

about 1% of the current population;• and the linear model gives N = −97311 + 25946ˑ30 =

681,069, or about 0.22% of the current population.Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Analyzing the Results with Newer Data

Long-term Predictions Are Difficult


0 5 10 15 20 25 300

100

200

300

400

500

600

700

U S Deaths from AIDS1981–2007


N (i

n th

ousa

nds)

When data from the entire period from 1981 to 2007 are plotted together, we see that the rate of increase of AIDS deaths reaches a peak sometime around 1995 and then begins to taper off. Since none of the three types of functions we have used to model AIDS deaths exhibit this type of behavior, some other type of function is needed to describe the number of AIDS deaths accurately over the entire 26-year period.

11.1 POWER FUNCTIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011,...

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