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MORE ABOUT ZEROS
Why do we try to prove things anyway? I think because we want to
understand them. We also want a sense of certainty. Mathematics is a very
deep field. Its results are stacked very high, and they depend on each other a
lot. You build a tower of blocks but if one block is a bit wobbly, you can’t
build the tower very high before it will fall over. So I think mathematicians
are concerned about rigor, which gives us certainty. But I also think proofs are
so that we can understand. I guess I like explanations better than step-by-step
rigorous demonstrations.
William P. Thurston
The problem that
infected me with such
virulence . . . concerned
solving cubic equations and
the answer had been
known since Cardano
published it in 1545. What I
did not know was how to
derive it. The sages who
had designed the
mathematics curricula . . .
had stopped at solving
quadratic equations.
Questions by curious
students about cubic and
higher-order equations
were deflected with
answers such as “This is too
advanced for you” or “You
will learn this when you
study higher mathematics,”
thereby creating a
forbidden-fruit aura about
the subject.
Mark Kac
We now have considerable experience with graphs of polynomial functions and a
feeling for the nature of their zeros. When we have completely factored a polynomial, we obviously know its zeros, but how do we find the zeros (or equivalently,
the factors) of a nonfactored polynomial?
Exact form answers are not always available, even theoretically (see the Historical Note, “There Is No Quintic Formula”). There are useful theorems about the
nature of zeros, and we now have technological tools undreamed of by earlier
mathematicians. This section adds to our arsenal of theorems about the nature of
polynomial zeros, and then gives an informal introduction to Newton’s Method, an
important tool for approximating zeros with great precision.
Number of Zeros of Polynomial Functions
How many zeros does a polynomial function of degree n have? Returning again to
quadratic functions (degree 2, parabola as the graph), there can be two real zeros
(when the parabola crosses the x-axis twice), a repeated real zero (when the parabola is tangent to the x-axis), or two nonreal conjugate zeros (when the parabola
does not touch the x-axis). Such behavior is typical of polynomial functions in
general. We explore some cubics in the following two examples.
cEXAMPLE 1 Zeros of a family of cubics Consider the family ^ of cubic
curves given by f ~x! 5 x 3 2 cx 1 2. Graph members of ^ for c 5 0, 1, 2, 3, 4.
Decide what values of c give a cubic whose graph crosses the x-axis (a) just once,
(b) exactly three times. (c) For what value of c does f have a repeated zero? Check
by factoring.
Solution
(a) and (b) When c 5 0, the graph is a vertical shift of the cubic y 5 x 3, with
only one real zero. When c 5 1 or 2, the graph has two turning points but crosses
the x-axis only once, so there is only one real zero. At c 5 3, the graph appears to
just touch the x-axis at the point (1, 0). See Figure 16. For values of c larger than
3, there are exactly three real zeros. (c) When c 5 3, it appears that there is a
repeated zero at 1, and the other crossing looks like ~22, 0!. To check, we want to
show that a factored form of x 3 2 3x 1 2 is ~x 2 1!2~x 1 2!.
~x 2 1!2~x 1 2! 5 ~x 2 2 2x 1 1!~x 1 2! 5 x 3 2 3x 1 2. b
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(0, 2)
(0, 2)
[– 2.5, 2] by [– 2.5, 4.5]
(a) c = 0
[– 2.5, 2] by [– 2.5, 4.5]
(b) c = 1
(0, 2)
(0, 2)
[– 2.5, 2] by [– 2.5, 4.5]
(d) c = 3
[– 2.5, 2] by [– 2.5, 4.5]
(c) c = 2
FIGURE 16
y 5 x 3 2 cx 1 2
cEXAMPLE 2 Zeros of a family of cubics, continued
^ of Example 1, f ~x! 5 x 3 2 cx 1 2.
Consider the family
(a) Find a value c for which 2 is a zero of f , and find the other zeros of f in exact
form.
(b) Find all zeros in exact form when c 5 21.
Solution
y
(a) To find a value of c for which f ~2! 5 0, we substitute 2 for x and solve for c:
23 2 c · 2 1 2 5 0
2c 5 10, or c 5 5.
6
4
(0, 2)
2
(2, 0)
x
–4
–2
2
4
Thus, the cubic in ^ that has 2 as a zero is f ~x! 5 x 3 2 5x 1 2. We know then
that ~x 2 2! is a factor, and we can use long division (or synthetic division, or
simple factoring) to find the other factor:
x 3 2 5x 1 2 5 ~x 2 2!~x 2 1 2x 2 1!.
f(x) = x3 – 5x + 2
FIGURE 17
The remaining zeros of f are the roots of x 2 1 2x 2 1 5 0, which the
quadratic formula gives as 21 6 Ï2 (Check.) See Figure 17.
(b) For c 5 21, f ~x! 5 x 3 1 x 1 2 has one real zero and no turning points, and
the x-intercept point appears to be (21, 0), which we can verify by substituting
21 for x: f ~21! 5 ~21!3 1 ~21! 1 2 5 0. Therefore, ~x 1 1! is a factor.
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f ~x! 5 x 3 1 x 1 2 5 ~x 1 1!~x 2 2 x 1 2!
As in part (a), the remaining zeros are the roots of x 2 2 x 1 2 5 0, which the
quadratic formula gives as 1 62Ï7i . b
Examples 1 and 2 remind us again of the utility of the factor and remainder
theorems of Section 3.2, and they reinforce our ideas about the nature of zeros of
cubic polynomials. A cubic always has three zeros, at least one real zero. There can
be a repeated zero, and there is a possibility of two nonreal conjugate zeros.
Fundamental Theorem of Algebra
The general situation is summed up in the fundamental theorem of algebra and
some of its consequences, called corollaries. These are stated in terms of the
complex-number system and proofs necessarily involve complex numbers as well.
The fundamental theorem was first proved by one of the greatest mathematicians
of all times. See the Historical Note, “Carl Friedrich Gauss.”
Fundamental theorem of algebra
Suppose p is a polynomial function of degree n, n $ 1. There is at least one
number c where p~c! 5 0; that is, p has at least one zero (which may be a
nonreal complex number).
Corollary 1: In the complex number system, p has exactly n zeros (counting
multiplicities).
Corollary 2: If the coefficients of p~x! are real numbers, then the graph of p
can cross (or touch) the x-axis in at most n points.
The fundamental theorem of algebra is what mathematicians call an existence
theorem. For any given polynomial function of positive degree, the theorem states
that zeros exist, but it provides no help for finding any particular zero. For linear
and quadratic functions we can find the zeros exactly; for higher-degree polynomial functions, the situation becomes more difficult.
For most of the problems that arise in applications we can only approximate
zeros. Virtually all numerical techniques to do this are rooted in the locator
theorem, our most fundamental tool. Mathematicians have found a number of
theorems, however, that can help in the search for certain kinds of zeros.
Gauss’ proof of the fundamental theorem applies to polynomial functions with
both imaginary and real coefficients. We emphasize, however, that in this chapter
we discuss only polynomial functions with real number coefficients.
Nature of Zeros of Polynomial Functions
The next theorem generalizes what we already know about quadratic polynomials.
In certain situations zeros of quadratic functions come in pairs. For example,
if f ~x! 5 x 2 2 4x 1 1, then the zeros of f are 2 1 Ï3 and 2 2 Ï3;
if g~x! 5 x 2 2 2x 1 2, then the zeros of g are 1 1 i and 1 2 i.
The numbers a 1 bi and a 2 bi are called complex conjugates. Certain kinds of
zeros must occur in pairs in higher-degree polynomials, as well.
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CARL FRIEDRICH GAUSS (1777–1855)
Called the “prince of
17 sides when he was almost 17
mathematicians,” Gauss is clearly
himself. His construction was
among the greatest mathematicians
published before he turned 19.
of all time. He contributed to all
Gauss was the first to use i for
areas of mathematics, as well as to
Ï21 and thoroughly understood
astronomy and physics, and we are
the importance of complex
still building directly on
numbers in the solution of
foundations he laid.
equations. Although he left college
Most of us could construct an
before receiving his doctorate, he
equilateral triangle or a square with
submitted his dissertation and
a compass and ruler. The Greeks
attained the degree by the age of
Mathematician, astronomer,
also constructed a pentagon. Angle
21. His thesis established the
and physicist Carl Friedrich
Gauss.
bisection allows us to double the
fundamental theorem of algebra.
number of sides, so it is theoretically
This theorem so fascinated him that
possible to construct regular polygons of n sides if
he gave three different proofs during his lifetime.
n is a power of 2 or n 5 2k · 3 or n 5 2k · 5,
Returning to constructions, he proved in 1826
where k is any nonnegative integer. Gauss made
that an n-gon is constructible for an odd prime n
k
the first significant progress in 2000 years when he
only when n has the form 22 1 1 (for instance, 3,
5, and 17).
discovered how to construct a regular polygon of
Conjugate zeros theorem
Let p be any polynomial function with real number coefficients.
If the nonreal complex number a 1 bi is a zero of p, then a 2 bi is also
a zero. Here b is not zero.
If, in addition, p has integer coefficients and a 1 Ïb is a zero of p, then
a 2 Ïb is also a zero. Here b is not a perfect square.
Strategy: Since the
coefficients of p are real
numbers, the conjugate zeros theorem applies; if i is a
zero, then 2i must also be a
zero.
The conjugate zeros theorem is illustrated in the next example.
cEXAMPLE 3 Conjugate zeros Given that p~i! 5 0, find all zeros of the
polynomial p~x! 5 2x 4 2 2x 3 1 x 2 2 2x 2 1.
y
Solution
As the strategy suggests, both i and 2i are zeros, so both x 2 i and x 1 i are factors
of p~x!. Thus p~x! has x 2 1 1 as a factor since
3
2
1
~x 2 i!~x 1 i! 5 x 2 1 1.
–1
x
–3 –2
(0, – 1)
1
–1
2
3
Dividing p~x! by x 2 1 1, we find that the other factor is 2x 2 2 2x 2 1 and so
p~x! 5 ~x 2 1 1!~2x 2 2 2x 2 1!.
–2
–3
FIGURE 18
p~x! 5 2x 4 2 2x 3 1
x 2 2 2x 2 1
The remaining zeros of p are the roots of 2x 2 2 2x 2 1 5 0, which the quadratic
formula gives as 1 62Ï3 . Therefore, p has two pairs of conjugate zeros: i and 2i, and
1 12 Ï3, 12 2 12 Ï3. The graph in Figure 18 shows the two real zeros and suggests,
as we now know, that there are no others. b
1
2
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Approximating Real Zeros: Newton’s Method
Without some initial information it may be effectively impossible to find zeros of
a polynomial function. In Example 3, we are given the fact that i is a zero, which,
with the conjugate zeros theorem, is enough for us to find all zeros in exact form.
We know how to zoom in on zeros from a graph, but the process is slow, and it is
difficult to get much precision.
Many graphing calculators will immediately give approximations for all zeros
of a polynomial function, to full calculator accuracy, at the press of a single key.
How do they do it? A comprehensive answer is beyond the scope of this course, but
most SOLVE routines essentially build on an iterative technique for approximation
called Newton’s (or the Newton-Raphson) method. The idea behind the process has
Tangent
a simple graphical interpretation that we can describe with the aid of a few pictures.
line
(x0 , f(x 0)) For the pictures we need to speak of the “slope” or “direction” of a curve at a given
point, another idea that is made precise in calculus is the derivative of a function.
Tangent
For our purposes, we will explain how to work with derivatives of polynomials of
line
degrees 3 and 4, but we could just as well use the numerical derivative in the form
(c, 0)
x already built-in in several graphing calculators.
x2 x1
x0
y = f(x)
First we look at a picture of a typical function near a zero we want to approximate. See Figure 19. This could be almost any function, under any degree of
FIGURE 19
magnification. If we know that the number x 0 is near the desired zero (which we
have indicated as c in Figure 19), then we want a mechanical procedure for getting
a better approximation, a new number nearer to c than x0. The idea is that if we go
to the point on the curve ~x0 , f ~x0!! and essentially “take aim” along the curve in
the direction of what is called the tangent line to the curve, we should hit the x-axis
at a point x 1 nearer c than x0 . Repeating the process from x 1 should take us nearer
still, after which we could move to a point x 2 still nearer, and so forth.
Fortunately, the general process (derived in calculus) is given in a simple
formula that can be implemented easily on our calculators. We take the following
information from calculus. Associated with every polynomial function f is a related
function called the derivative of f , denoted by f 9. We give formulas for the derivatives of cubic and quartic functions, leaving explanations for later courses. (It may
also be possible to use a built-in program of your calculator; see the Technology Tip
following Example 4.)
Function
Derivative
f ~x! 5 ax 1 bx 1 cx 1 d
Example: f ~x! 5 x 3 2 4x 2 1 2x 2 1
f ~x! 5 Ax 4 1 Bx 3 1 Cx 2 1 Dx 1 E
Example: f ~x! 5 2x 4 1 5x 3 2 x 1 17
f 9~x! 5 3ax 2 1 2bx 1 c
f 9~x! 5 3x 2 2 8x 1 2
f 9~x! 5 4Ax 3 1 3Bx 2 1 2Cx 1 D
f 9~x! 5 8x 3 1 15x 2 2 1
3
2
Having a way to write the derivative, we can describe the approximation
process of Newton’s method. If x 0 is an approximation to the zero c, then the new
approximation is given by
x 1 5 x 0 2 f ~x 0!yf 9~x0!,
where f 9~x 0! is the slope of the tangent line, the number we get when the derivative
function (or the numerical derivative) of f is evaluated at x 0 .
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The process can be programmed very simply with almost any calculator, but
we outline the steps as an algorithm that can be performed on your home screen.
Then we illustrate with an example. Follow Example 4 with your calculator.
Algorithm for Newton’s method
1. Make an initial guess, a reasonable approximation to the desired zero.
Note: the better the first guess, the more efficient the method.
2. Store your guess in memory register x. (For example, if your initial guess
is 1, 1 A x, ENTER.)
3. Evaluate the next approximation, using x 1 5 x 0 2 f ~x 0!yf 9~x0!, store it in
memory register x and ENTER. This displays the new number and stores it
for use in the next step.
4. Now press ENTER again and again, getting a new approximation with each
repetition, until the display doesn’t change, indicating that we have the
best approximation the calculator can deliver.
cEXAMPLE 4 Approximating zeros The polynomial p~x!5x 3 2 5x 1 2
has a real zero between 0 and 1. Use Newton’s method, beginning with an initial
guess of 0.5, to approximate the zero to ten decimal-place accuracy.
Solution
We would normally get our initial guess from a graph, but following directions, we
will let x 0 5 0.5, and store: .5 A x ENT. For p~x! 5 x 3 2 5x 1 2, we have
p9 5 3x 2 2 5, so we enter
X 2 ~X3 2 5X 1 2!y~3X2 2 5! A X
ENTER
.
The display (ten decimals) reads .4117647059, and as we repeatedly
sequence of displayed numbers is as follows.
ENTER,
the
.4142119097
.4142135624
.4142135624
This last number does not change when we continue to press ENTER. Thus, to ten
decimal-place accuracy, the desired zero is .4142135624. This is the same polynomial function we considered in Example 2a, in which we found an exact form for
the same zero, 21 1 Ï2. b
TECHNOLOGY TIP r Numerical derivatives
If your calculator has a built-in function for calculating a numerical
derivative, as the TI-82 (MATH 8), TI-85 (2nd CALC F2), HP–38 (▪ CHARS, Choose ­,
OK; you want ­X~F1~X!), or Casio fx 9700 (SHIFT dydx), you need not write out the
derivative. Enter the function in your Y1 register (or on your SHIFT MEM list as f1).
Then follow the first two steps of the algorithm. In place of having to write
out both the function and its derivative in Step 3, simply enter X 2 Y1/nDER(Y1) A
X and iterate.
While this may not seem a significant advantage for polynomials, using
X 2 Y1/nDER(Y1) A X will work with an appropriate initial guess for any function
that has a reasonably smooth graph.
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Polynomials in Application
In many applied problems, we are not interested in finding polynomial zeros in
exact form; we may not care about the number of zeros. The nature of the problem
may dictate that only certain zeros have physical meaning, as the next example
illustrates.
cEXAMPLE 5 Application leading to a polynomial Twin radio towers are
to be erected on opposite sides of a 10-foot roadway, where 50-foot guy wires reach
from the top of each tower to the base of the other. The wires must cross high
enough to leave a 12-foot clearance above the road. How tall can the towers be, and
how close to the side of the roadway can they be built?
50' 50'
h
h
24~x 1 5!
12
h
.
, or h 5
5
x
x
2x 1 10
12'
x
x
10'
Roadway
Using the Pythagorean theorem with nACB,
h 2 1 ~2x 1 10!2 5 50 2.
(a)
B
50
h
D
A
x
12
10
E
Solution
We sketch a diagram to help us visualize the situation. See Figure 20. Because of
the symmetry, all of the pertinent information is contained in the second part of the
diagram, where x is the distance from the edge of the road to the tower of height h.
From similar right triangles, nACB and nAED, we have
x
(b)
FIGURE 20
C
242~x 1 5!2
1 4~x 1 5!2 5 2500, then expand
x2
and simplify to get a polynomial equation,
We substitute 24~x 1 5!yx for h,
x 4 1 10x 3 2 456x 2 1 1440x 1 3600 5 0.
Clearly, only positive values for x have any meaning, and from the picture, we
estimate that x is probably between 5 and 15, so we look for zeros in that range.
In a @5, 15# 3 @25, 50# window, we see just two vertical lines, but there appear to
be two zeros. To get a picture that looks a little more like the polynomials we are
familiar with, we need a much larger y-range. With a y-range of @28000, 7000#,
we see the curve in Figure 21. We still see two zeros, one near 14 and one near 5.7.
The corresponding heights are given by h < 32.6 and h < 45.2.
How do we interpret two different answers? When we look more closely, it
turns out that we can get at least a 12-foot clearance over the road if the distance
5.7
45.2'
14
32.6'
[5, 15] by [– 8000, 7000]
y = x 4 + 10x3 – 456x2 + 1440x + 3600
FIGURE 21
12'
10'
5.7
14'
FIGURE 22
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from the roadway is anything between 5.7 feet and 14 feet. The maximum tower
height decreases as we move away from the road. See Figure 22. For maximum
height, we can erect 45-foot towers 5.7 feet on each side of the roadway. b
EXERCISES 3.3
Check Your Understanding
Exercises 1–7 True or False. Draw graphs when helpful.
1. The function f ~x! 5 x 3 2 8x 2 1 5x 1 4 has three
real zeros.
2. The graph of f ~x! 5 x 4 1 2x 2 1 1 crosses the x-axis.
3. The equation x 3 2 7x 1 5 5 0 has two negative and
one positive root.
4. Since Ï3 is a zero of f ~x! 5 x 3 2 2x 2 Ï3, then
2Ï3 must also be a zero.
5. All zeros of f ~x! 5 x 3 2 8x 2 1 5x 1 8 lie between
21 and 8.
6. If f ~x! 5 2x 4 1 x 3 1 7x 2 2 x 2 6, then f ~x! , 15
for every x.
7. Based on what can be seen from the graph of
f ~x! 5 x 3 2 40x 2 2 400x 1 1600 using @224, 50# 3
@25100, 17,000#, we can conclude that f has one negative and two positive zeros.
Exercises 8–10 Fill in the blank so that the resulting
statement is true.
8. For the family of functions f ~x! 5 x 3 1 cx 2 5, the
value of c for which 21 is a zero is
.
4
9. The number of real zeros of f ~x! 5 x 1 2x 3 2
.
2x 2 2 4x is
10. In applying Newton’s method, if f ~x! 5 x 3 2 2x 1 3,
then f 9~x! 5
.
Develop Mastery
Exercises 1–4 Zeros from Graph (a) Graph y 5 p (x)
and locate each zero between two consecutive integers.
(b) Find an approximation (one decimal place) for the
largest zero.
1. p~x! 5 x 3 2 3x 2 2 x 1 2
2. p~x! 5 x 3 2 3x 2 2 3
3. p~x! 5 x 3 2 2x 2 2 x 1 3
4. p~x! 5 x 3 1 2x 2 2 3x 2 2
Exercises 5–8 Locate Intersection Graph the two equations on the same screen and find the coordinates of the
point of intersection (one decimal place).
6. y 5 x 2 x 3
5. y 5 x 3 2 3x
y 5 23
y51
8. y 5 1 2 x 3
7. y 5 x 3
y5x11
y 5 x2 2 2
Exercises 9–12 Zeros to Function Find a polynomial
function of lowest degree with integer coefficients, a leading
coefficient of 1, and the given numbers as zeros. Give the
result in standard (expanded) form. Use the conjugate zeros
theorem, if needed.
9. 1 1 i, 1 2 Ï2
10. 1, Ï2
11. 2, 21 1 Ï3
12. Ï2, 1 2 i
Exercises 13–16 Conjugate Zeros Theorem The given
number is a zero of f. Find the remaining zeros. (Hint: Use
the conjugate zeros theorem and long division.)
13. f ~x! 5 x 3 2 4x 2 1 3x 1 2; 1 2 Ï2
14. f ~x! 5 x 3 2 2x 2 2 9x 2 2; 2 1 Ï5
15. f ~x! 5 2x 3 2 9x 2 1 2x 1 1; 2 2 Ï5
16. f ~x! 5 2x 3 2 3x 2 2 4x 2 1; 1 1 Ï2
Exercises 17–24 Find All Zeros Assume that the domain of the variable is the set of complex numbers. Find all
zeros in exact form.
17. f ~x! 5 x 3 2 4x 2 1 2x 2 8
18. f ~x! 5 4x 3 2 4x 2 2 19x 1 10
19. f ~x! 5 x 3 2 2.5x 2 2 7x 2 1.5
20. f ~x! 5 3x 3 2 1.5x 2 1 x 2 0.5
21. f ~x! 5 6x 4 2 13x 3 1 2x 2 2 4x 1 15
22. f ~x! 5 2x 4 1 3x 3 1 2x 2 2 1
23. f ~x! 5 4x 4 2 4x 3 2 7x 2 1 4x 1 3
24. f ~x! 5 4x 4 1 8x 3 1 9x 2 1 5x 1 1
Exercises 25–32 Exact Form Roots
exact form.
25. 6x 3 2 2x 2 2 9x 1 3 5 0
26. 6x 3 2 x 2 2 13x 1 8 5 0
27. x 4 2 x 3 2 3x 2 1 x 1 2 5 0
28. x 4 2 x 3 2 8x 1 8 5 0
29. x 3 2 3x 1 2 5 0
30. 18x 3 1 27x 2 1 13x 1 2 5 0
~14x 2 1 17x!
31. x 3 1 2 5 2
3
32. x 4 1 4x 3 2 5x 2 5 36x 1 36
Find all roots in
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Exercises 33–34 Solution Set, Exact Form Find the solution set for (a) f ~x! $ 0 (b) f ~x 2 1! $ 0. Give answers
in exact form.
33. f ~x! 5 x 3 1 x 2 2 11x 2 15
34. f ~x! 5 x 3 2 6x 2 1 4x 1 16
Exercises 43–46 Explore The equation defines a family
of polynomial functions of degree 3. Experiment with several integer values of c and for each draw a graph. Describe
the role that c appears to play. Include information about
local extrema, number of real zeros, and any other graphical features you observe. Use complete sentences.
43. f ~x! 5 x 3 1 cx 2 2 4x 1 3
44. f ~x! 5 x 3 1 2x 2 1 cx 1 3
45. f ~x! 5 x 3 1 2x 2 2 4x 1 c
46. f ~x! 5 cx 3 1 2x 2 2 4x 1 3
Exercises 51–53 Explore Composition with Absolute
Value (See Section 2.6)
51. Try several polynomial functions f of degree 3 and in
each case draw a graph of f and then a graph of g~x! 5
f ~_ x _ !. For what functions f will the composition function g have
(a) no zeros
(b) two zeros
(c) four zeros
(d) six zeros
52. Draw a graph of f ~x! 5 _ x 3 1 x 2 2 2x _ . Explain why
the graph of f cannot be the graph of a polynomial
function. (Hint: Read “Smoothness and End Behavior”
in Section 3.1.)
53. (a) Take any polynomial function f of degree 3 and
draw the graph of g~x! 5 _ f ~x! _ . Use the graph to
explain why g cannot be a polynomial function.
(b) Is there a polynomial function f of degree 4 for
which g~x! 5 _ f ~x! _ is also a polynomial function?
Explain.
54. For each real number c, the graph of f ~x! 5 0.1~x 1
c!3 1 0.3~x 1 c!2 2 ~x 1 c! 2 8 is a horizontal
translation of the graph of g~x! 5 0.1x 3 1 0.3x 2 2
x 2 8. For what integer values of c will f have a negative zero?
55. Explore The graph of f ~x! 5 x 3 2 6x 2 1 9x 2 6
has one real zero and local extrema at ~1, 22! and
~3, 26!. Draw a graph in a shifted decimal window to
support this claim. Consider the family of functions
g~x! 5 f ~x! 1 c.
(a) For what integer values of c will g have three distinct real zeros?
(b) For what integer values of c will g have a repeated
zero? Justify algebraically.
(c) For what integer values of c will g have a negative
zero?
56. Explore Consider the family of functions f ~x! 5
0.3x 3 2 4x 1 c. Let P be the local maximum point
and Q be the local minimum point. For what integer
values of c will (a) P and Q be below the x-axis?
(b) P be above and Q be below the x-axis? (c) P and
Q be above the x-axis?
Exercises 47–50 Explore For each real number k, f is a
polynomial function of degree 3. Experiment with several
integer values of k and determine the values of k for which
f will have (a) three real zeros, (b) exactly one real zero, (c)
one negative and no positive zeros.
48. f ~x! 5 x 3 1 kx 1 3
47. f ~x! 5 x 3 1 kx 2 2
3
50. f ~x! 5 x 3 1 kx 2 5
49. f ~x! 5 x 1 kx 1 8
Exercises 57–60 Newton’s method Use Newton’s method
to find the largest zero of f (nine decimal places). See Example 4.
57. f ~x! 5 x 3 2 4x 1 2
58. f ~x! 5 x 3 1 4x 2 2 8
59. f ~x! 5 x 4 2 5x 2 1 3x 2 1
60. f ~x! 5 x 4 2 9x 2 1 8x
Exercises 35–38 Evaluating Inverse (a) Graph the
function to support the claim that f is either an increasing or
decreasing function (tell which), so that f has an inverse, f 21.
(b) Find f 21~3! (1 decimal place). (Hint: In x 5 f ~y! replace
x by 3 and solve for y.)
35. f ~x! 5 2x 3 1 3x 2 4
36. f ~x! 5 x 3 2 3x 2 1 4x 1 5
37. f ~x! 5 2 2 x 1 x 2 2 2x 3
38. f ~x! 5 4 2 3x 2 2x 3
39. For f ~x! 5 x 3 2 3x 2 2 x 1 3 and g~x! 5 _ x _ , how
many zeros does the function f 8 g have? Draw a graph.
Does your answer contradict the corollaries to the fundamental theorem of algebra? Explain.
40. Repeat Exercise 39 for f ~x! 5 x 3 2 3x 2 2 4x 2 4.
41. Explore For f ~x! 5 x 3 1 cx 1 4, choose several integer values for c (positive and negative) and draw a graph
of the corresponding function. What values of c give graphs
with (a) turning points, (b) more than one x- intercept
point?
42. Repeat Exercise 41 for f ~x! 5 x 3 1 cx 2 4.
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61. A box with an open top is constructed from a rectangular piece of tin that measures 9 inches by 12 inches by
cutting out from each corner a square of side x, and then
folding up the sides as shown in the diagram.
(a) If V denotes the volume of the box, find a formula
for V as a function of x. What is the domain of the
function?
x
x x
xx
9
x
12
(b) What size corners should be cut out so that the box
will have a capacity of 81 cubic inches? (Hint:
There are two answers, both between 1 and 2.)
62. A storage tank consists of a right circular cylinder
mounted on top of a hemisphere, as shown in the diagram. If the height of the cylindrical portion is 12 feet
and the tank is to have a capacity of 1250p cubic feet,
find the radius r of the cylinder to 3 significant digits.
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65. A storage tank has the shape of a cube. If one of the
dimensions is increased by 2 and another by 3, while
the third is decreased by 4, then the resulting rectangular tank will have a volume of 600 cubic units. What is
the length of an edge of the original cube to 3 significant
digits?
66. A rectangular storage container is 3 by 4 by 5 feet.
(a) What is its capacity (volume)?
(b) If we increase each of the dimensions by x feet in
order to get a container with a capacity five times as
large as the original, how large must x be, rounded
off to 3 significant digits?
67. Two vertical poles, AB and CD, are connected by guy
wires of lengths 30 ft and 40 ft, intersecting at point F,
12 feet from the ground. See the diagram. Find the
heights u and v, of the poles and the distance d between
the poles.
D
B
30 ft 40 ft
F
u
v
12 ft
A
12
r
E
C
68. In Example 5 we found that the two towers can be
placed anywhere from 5.7 ft to 14 ft from the road.
How far from the road should they be located if we want
maximum clearance above the road? (Hint: Let u be the
desired distance and v be the distance, _DE _ , from the
right edge of the road to the guy wire AB.) Write v as a
function of u. See diagram.
y
63. An isosceles triangle has the dimensions shown in the
diagram. If the area is equal to 10 square units, find the
length of the altitude h (to the nearest tenth).
B(0, h)
Tower
x+1
h
50 ft
D
C
12
v
u 10 u
E
x+1
h
A (2u + 10, 0)
x
Roadway
2x
64. What are the dimensions of a rectangle of area 7 square
inches that has a diagonal 1 inch longer than the length
of one of its sides? Give the result rounded off to 3
significant digits.
69. Solve the problem in Example 5 if the guy wires are 48
feet long.
70. In Exercise 69, how far from the road should the two
towers be located so that there is maximum clearance
above the road? See Exercise 68.