Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 41 pptx

EXERCISE 11.4 Construct a polynomial equation with the specification given.. a a fifth degree polynomial with no real zeros b a fourth degree polynomial with no real zeros Answer is prov

In this course we will mainly be working with the real number system as opposed to the complex number system, so unless otherwise stated, when we ask about the number of roots

it is understood that we mean real roots

EXERCISE 11.4 Construct a polynomial equation with the specification given (The answers to this exercise

are not unique!)

(a) a third degree polynomial equation with roots at x = −2, x = 3, and x = 0 (b) a second degree polynomial equation with a double root at x = −1 (c) a second degree polynomial equation with no roots

(d) a fifth degree polynomial equation with roots only at x = −2, x = 3, and x = 0

Answers are provided at the end of the section.

EXERCISE 11.5 Identify which of the following two polynomials is possible to construct, and construct it

(a) a fifth degree polynomial with no real zeros (b) a fourth degree polynomial with no real zeros

Answer is provided at the end of the section.

Finding the Zeros of a Polynomial

Finding the zero of a linear function is simple Finding the zeros of a quadratic is no problem when you use the quadratic formula It is considerably harder to find the roots of a general cubic equation A systematic procedure for finding a cubic’s roots does exist, but it is much more complicated than the quadratic formula and in practice is not often used The story

of the discovery of the cubic formula in Italy in the mid-1500s is one worth reading about

in a math history book It involves alleged lies about the discovery, a public competition to solve cubic equations being won by Nicolo Tartaglia and his “secret” method being passed

on to Girolamo Cardano in confidence but then published anyway, and a final dispute from which one of the men is said to have been lucky to have escaped alive.1There is an algorithm for finding the zeros of fourth degree polynomials, but in the early 1800s the Norwegian mathematician Niels Abel (1802–1829) proved that a formula for finding the zeros of a fifth

degree polynomial does not exist.

For a general nth degree polynomial it can be difficult to find the zeros, and as is clear from the last paragraph, we cannot give a recipe for pinning them down exactly We can, however, take advantage of the continuity of polynomials; if f (a) > 0 and f (b) < 0, then there is a c somewhere between a and b such that f (c) = 0

EXAMPLE 11.3 Find the zeros of f (x) = x3− 2x2− 8x

SOLUTION We can factor out an x from each term

x3− 2x2− 8x = x(x2− 2x − 8) Now we have a quadratic that we can factor

= x(x − 4)(x + 2)

So, the zeros are x = 0, x = 4, and x = −2

1

SOLUTION In this case, it is not clear how to factor f (x) Graph f on a graphing calculator.

10

y

x

Figure 11.7

We see that there is a zero near x = 1 By zooming in, it looks like the zero might be exactly x = 1 To be sure, evaluate f at x = 1 f (1) = (1)3+ 3(1)2− 3(1) − 1 = 0, so x = 1

is a zero There are two other zeros By zooming in, we estimate that they are at x ≈ −3.73 and x ≈ −0.27

Let’s try to find the exact values x = 1 is a zero of f (x); therefore, (x − 1) is a factor

of f (x) In other words, we can write x3+ 3x2− 3x − 1 = (x − 1)p(x), where p(x) is a new polynomial, a polynomial of degree 2

To find p(x), we do long division of polynomials Long division of polynomials is analogous to long division for numbers Let’s review long division

Find 7324 Write denominator ) numerator

4 goes into 7 one time Write the 1 on the top and 1 • 4 under the 7.

Subtract 4 from 7 Bring down the 3 from 732 to get 33.

4 goes into 33 eight times Write the 8 on the top and 8 • 4 under the 33.

Subtract 32 from 33 Bring down the 2 from 732 to get 12.

4 goes into 12 three times Write the 3 on the top and 3 • 4 under the 12.

Subtract 12 from 12 This gives zero We've accounted for 732: 7 • 100 + 3 • 10 + 2,

so we're done.

4 732 4

0 33 183

32 12 12

Find x3+3x2−3x−1

x−1 Write denominator ) numerator

x times x2 gives x3 Write the x2 on the top and x2 • (x –1) under the x3 + 3x2

Subtract x3 – x2 from x3 + 3x2 Bring down the –3x from above to get 4x2 –3x.

x times 4x gives 4x2 Write the 4x on the top and 4x • (x –1) under the 4x2 –3x Subtract 4x2 –4x from 4x2 –3x Bring down the –1 from above to get x –1.

x times 1 gives x Write the 1 on the top and 1 • (x –1) under the x –1.

Subtract x –1 from x –1 This gives zero We've accounted for x3 + 3x2 –3x –1,

so we're done.

x –1 x3 + 3x2 –3x –1

x2 + 4x +1

4x2 – 3x 4x2 – 4x

0

By doing long division as shown, we see that p(x) = x2+ 4x + 1, so f (x) = x3+ 3x2− 3x − 1 = (x − 1)(x2+ 4x + 1) We use the quadratic formula to find the roots of

x2+ 4x + 1 = 0 and find that

x = −4 ±

√

16 − 4

√ 12

√ 3

Verify that the values of these two roots are approximately equal to our previous estimates from the graph

The zeros of f (x) = x3+ 3x2− 3x − 1 are x = 1, x = −2 +√3, and x = −2 −√3 Notice that if we had not known an exact zero (x = 1 in this case), then we could not have used long division and would have needed to approximate all three zeros graphically without learning their exact values.2

EXERCISE 11.6 Find all the roots of f (x) = x3− 1 = 0 (This is the cubic discussed in Section 7.3.)

Characteristics of Polynomials and Differentiation of Polynomials

Let’s return for a moment to the building blocks of polynomials, functions of the form

y = xkfor k = 0, 1, 2, 3, , and look at them more closely They are graphed below on the same set of axes Two scales are used, one allowing us to compare the graphs for x near zero, the other facilitating comparison for large values of x

2 Another alternative is to use what is called “Newton’s Method” for approximating roots See the Appendix on Newton’s

(1, 1)

–1

1

–1

x

x x x

10 100

–100 –10

Figure 11.8 Behavior for small x (a bug’s-eye view)

The larger k is, the flatter the graph of xkis on the interval −1 < x < 1

In other words, for |x| < 1, the higher the power to which x is raised, the smaller the magnitude of xk.3For instance,125is smaller than122

Behavior for large x (a bird’s-eye view)

For |x| > 1, the magnitude of xkincreases dramatically as k increases For example, consider the values of y corresponding to x = 10 in each of the polynomials above 105

is much greater than 102

By summing up building block functions of the form xk and weighting them with coeffi-cients, we obtain a general nth degree polynomial P (x) = a0+ a1x + a2x2+ · · · + anxn Characteristics of such a polynomial are discussed next

A Bird’s-Eye View of Polynomials

For x large enough in magnitude (zoom out a lot on your graphing calculator) the graph of

an nth degree polynomial, P (x) = a0+ a1x + a2x2+ a3x3+ · · · + anxn, is dominated by

anxnand can be characterized as follows:

For n even, if anis positive, then the graph reaches upward; .

if anis negative, then the graph reaches downward . For n odd, if anis positive, then the graph reaches down on the left

if anis negative, the graph reaches up on the left

More precisely, let P (x) = a0+ a1x + a2x2+ a3x3+ · · · + anxn be a polynomial of degree n

3 By “magnitude of x k ” we mean |x k | We are referring to the size, without regard to the sign.

Suppose n is even: If an>0, then limx→∞P (x) = +∞ and limx→−∞P (x) = +∞.

If an<0, then limx→∞P (x) = −∞ and limx→−∞P (x) = −∞ Suppose n is odd: If an>0, then limx→∞P (x) = +∞ and limx→−∞P (x) = −∞

If an<0, then limx→∞P (x) = −∞ and limx→−∞P (x) = +∞ This bird’s-eye view tells us that a polynomial of odd degree will have neither an absolute maximum value nor an absolute minimum value Its range is (−∞, ∞) On the other hand,

a polynomial of even degree will have a global maximum but no global minimum if the leading coefficient is negative and vice versa if the leading coefficient is positive

A Bug’s-Eye View of Polynomials

If we zoom in enough anywhere on the graph of any polynomial, the graph will eventually look like a straight line Polynomial functions have the delightful property that they are locally linear everywhere Another way to say this is that polynomials are differentiable everywhere Their graphs are smooth; there are no sharp corners

Polynomials are continuous, locally linear, and easy to evaluate This makes polynomial functions very user-friendly In fact, polynomials are so delightful to work with that people use them to approximate local behavior of functions that are more difficult to handle.4 Another perk is that polynomial functions are a pleasure to differentiate We can compute the derivative of xnfor n any positive integer, so we can differentiate any polynomial

d

dx[xn] = nxn−1

We can conclude that the derivative of an nth degree polynomial is a polynomial of degree

n − 1

P (x) = a0 + a1x + a2x2 + a3x3 + · · · + anxn

P(x) = a1 + 2a2x2 + 3a3x3 + · · · + nanxn−1

Critical Points of a Polynomial

The natural domain of polynomial functions is (−∞, ∞) and polynomials are differentiable everywhere; therefore, the critical points of a polynomial are simply the zeros of its derivative The derivative of an nth degree polynomial is a polynomial equation of degree

(n − 1) A polynomial of degree (n − 1) can have at most (n − 1) real roots We conclude

that

a polynomial of degree n can have at most (n − 1) turning points.

This makes sense graphically If a polynomial of degree n has n zeros (the maximum possible), then between each of those zeros the polynomial must have a turning point so its graph can return to the x-axis This means there would have to be (n − 1) turning points

Notice also that a polynomial of degree n may have fewer than n − 1 turning points Consider

the graphs of f (x) = xkin Figure 11.6; each of them had either zero or one turning point

4 You might, for instance, try to approximate e x near x = 0 by a third degree polynomial One of the treats awaiting you is the amazing discovery that you can actually express functions such as e x (and some of the trigonometric functions) as polynomials of

view) we can argue that

a polynomial of odd degree must have at least one zero and

a polynomial of even degree must have at least one turning point

Complete this argument as an exercise

EXERCISE 11.7 Justify the assertion made above

Symmetry. Recall that a function is said to be

evenif f (−x) = f (x) for all x in its domain (the graph is symmetric about the y-axis)

oddif f (−x) = −f (x) for all x in its domain (the graph is symmetric about the origin)

y

x

f (x) = x2

y

x

f (x) = x3

Figure 11.9

Looking back at the graphs of xkin Figure 11.6, we can see why the names “even” and

“odd” were chosen for these types of symmetry Functions of the form xk, where k is even, have even symmetry, and functions of the form xk, where k is odd, have odd symmetry Any random polynomial may be even, odd, or neither The following example is meant

to illustrate criteria for a function to be even and to be odd

EXAMPLE 11.5 For each of the following, determine whether the function is even, odd, or neither even nor

odd

(a) g(x) = −5x4+ 2x2+ 1 (b) h(x) = 2x3− x

(c) f (x) = 2x3− x + 1

SOLUTION Our strategy is to look at g(−x) and see if it is equal to g(x), −g(x), or neither

(a) g(−x) = − 5(−x)4+ 2(−x)2+ 8

= − 5x4+ 2x2+ 8

= g(x)

So g(x) is even

(b) h(−x) = 2(−x)3− (−x)

= − 2x3+ x

= − (2x3− x)

= − h(x)

So h(x) is odd

(c) f (−x) = 2(−x)3− (−x) + 1

= − 2x3+ x + 1

This is neither f (x) nor −f (x), so f is neither even nor odd

We can deduce the following:

A polynomial with the variable raised to even powers only is even

A polynomial with no constant term and the variable raised to odd powers only, with

no constant terms, is odd

Terminology:The polynomial f (x) = x2+ x is a polynomial of even degree, but it is not

an even function Notice that f (x) = x(x + 1), so the zeros of the polynomial are at 0 and

−1; the graph of f is not symmetric with respect to the y-axis Similarly, the polynomial g(x) = x3+ 1 is a polynomial of odd degree, but it is not an odd function The graph of g

has a y-intercept of 1, so it is not symmetric about the origin

Answers to Selected Exercises

Answers to Exercise 11.4

(a) y = x(x + 2)(x − 3) (b) y = (x + 1)2

Answer to Exercise 11.5

(b) y = x4+ 3

P R O B L E M S F O R S E C T I O N 1 1 2

1 Determine whether or not the expression given is a polynomial

(a) √1

2x +√33x2+1911

(b) 2x2+ 3x−1+ 5x3

(c) 2x + x1/2+ 5x5

(d) 2x+2x3 + 1

(e) 5−1/2x + 3−1x2+π21

−2 + 2 (f ) (x2+ 1)−1

characteristics Answers to these problems are not unique.

2 A second degree polynomial with zeros at x = 1 and x = −3

3 A third degree polynomial with zeros at x = −1, 0, and 5

4 A fourth degree polynomial with no zeros

5 A fourth degree polynomial whose only zero is at x =√2

6 A fifth degree polynomial with a zero of multiplicity two at x = 9 and zeros at x = 0,

3, and −e

7 A third degree polynomial whose only zero is at x = π + 1, and whose y-intercept

is 1

8 A fourth degree polynomial whose only turning point is at (−3, 2)

In Problems 9 through 14, construct a polynomial P (x) with the specified character-istics Determine whether or not the answer to the problem is unique Explain and/or illustrate your answer.

9 A fourth degree polynomial with zeros of multiplicity two at x = 2 and x = −3, and a

y-intercept of −2

10 A fifth degree polynomial with zeros of multiplicity two at x = 0 and x = π, and a zero

at x = −2; limx→∞P (x) = ∞

11 A third degree polynomial whose only zero is at x = −1 and such that limx→∞P (x) = ∞

12 The graph is a parabola with a vertex at (π , 2) and a y-intercept of 0

13 A fifth degree polynomial with a zero of multiplicity 3 at x = 0 and zeros at x = 1 and

x = −2, and passing through the point (−1, 2)

14 A third degree polynomial with zeros at x = 1 and x = 2, a turning point at x = 1, and

a y-intercept of√

e

In Problems 15 through 21, find the (real) zeros of the polynomial given.

15 (a) f (x) = 2x3+ 2x2− 12x (b) g(x) = 2x3+ 2x2+ 12x

16 (a) f (x) = −x3− x2− 5x (b) g(x) = 0.5x4− 0.5

17 (a) P (x) = x3− x2− 4x + 4

(b) Q(x) = x3− x2+ 4x − 4

Hint:Either guess a zero by observation or use a graphing calculator to guess a root; then use long division

18 f (x) = −x5+ 16x4

19 P (x) = x4− 2x3− 6x2+ 12x (Hint: At a certain point, guess a root.)

20 g(x) = 3x3+ 3

21 h(x) = x3− 8

In Problems 22 through 24, P (x) is a polynomial with the characteristics specified For each statement following the characteristic, determine whether the statement is definitely true; possibly true, but not necessarily true; or definitely false Explain.

22 The graph of P (x) is symmetric about the origin

(a) The degree of P (x) is even

(b) The degree of P (x) is odd

(c) P (x) has no zeros

(d) P (x) has at least one zero

(e) The number of turning points of P (x) is of the form 2n where n is a nonnegative integer

23 P (x) is a polynomial whose degree is even and nonzero P (0) = 0

(a) The graph of P (x) is symmetric about the origin

(b) The graph of P (x) is symmetric about the y-axis

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

(d) The graph of P (x) has an even number of turning points

(e) The graph of P (x) has an odd number of turning points

24 P (x) is a fifth degree polynomial limx→∞P (x) = ∞

(a) P (x) has five distinct real roots

(b) P (x) has no more than five roots

(c) P (x) has five turning points

(d) P (x) has four turning points

(e) P (x) has no more than four turning points

(f ) P (x) has at least one real root

(g) limx→−∞P (x) = ∞

P (x)

x

(a) The degree of P (x) is at least (b) What is the sign of the leading coefficient of P (x)?

26 Below are graphs of polynomials, each with a zero at x = a

(a) For each graph determine whether the zero is a simple zero, a zero of even multiplicity, or a zero of odd multiplicity

(b) For each graph, determine whether the derivative at x = a is positive, negative, or zero

y

x a

x a (ii)

y

x a

x a (v)

y

x a

(iii)