Download Chapter 1 - Resequencing Calculus

Chapter 1
Functions
1.1
1.2
1.3
1.4
1.5
1.6
Functions and Their Graphs
Algebra of Functions
Library of Functions
Implicit Functions and Conic Sections
Polar Functions
Parametric Functions
2
CHAPTER 1. FUNCTIONS
1.1 Functions and Their Graphs
It is impossible to overstate the significance of functions—both as a foundational
mathematical concept and as a tool for describing relationships between real-world
variables. The language of functions permeates every branch of the physical and
social sciences: IQ scores are a function of nature and nurture, pressure is a function
of temperature and volume, consumer demand is a function of price, population is a
function of food supply, and computational power is a function of processor speed.
Generally speaking, functions arise whenever the value of one or more variables
determines the value of another.
Just as it is impossible to overstate the significance of functions, it is impossible
to overstate the importance of calculus as a tool for analyzing functions and for extending our ability to use functions to solve real-world problems. However, before
we begin the development of calculus, we’ll spend some time in this chapter reviewing the essential terminology and concepts that form the algebraic foundations for
the study of functions.
Definition of Function
At an intuitive level, a function is nothing more than a correspondence between two
quantities with the special property that each value of one quantity determines a
unique value of another. Table 1.1 gives several examples of correspondences with
this special property.
Table 1.1: Correspondences defining functions
The cost to overnight a package via
FedEx from Los Angeles to New
York. . .
The area of a circle. . .
The velocity of an object dropped
from the top of a tall building. . .
is uniquely determined by
...
the weight of the package.
is uniquely determined by
...
is uniquely determined by
...
the radius of the circle.
the number of seconds
since it was dropped.
The mathematical definition of a function simply formalizes the notion that one
value uniquely determines another.
Function Terminology
A function f from a set A to a set B is a correspondence or rule that assigns to
each element x of A exactly one element y of B. The set A is called the domain
of the function and each element x of A is called an input value. An element
y of B that corresponds to an input value x is called an output value, and y is
usually denoted by f (x). The set of all possible output values is called the range
of the function.
Functions can be expressed in a number of different ways. In Table 1.1 we saw
several examples of functions described verbally. A symbolic or algebraic description
of a function involves the use of an equation. For example, the equation A = πr2
describes how the area of a circle is uniquely determined by the radius. In this
case, we say that r is the independent variable and that A is the dependent
3
1.1. FUNCTIONS AND THEIR GRAPHS
variable. We may also choose to use function notation and write f (r) = πr2 .
Graphs of various types can also be used to represent functions. The graph in
Figure 1.1 describes the average annual global temperature as a function of the
year. Finally, a table of data values can be used to describe a function. Table 1.2
suggests how annual income may be a function of the number of years of education
beyond high school.
Table 1.2: Mean income as a function
of years of education
Years of Education
Beyond High School
0
2
4
6
10
Mean Income
$55,729
$72,097
$99,070
$119,559
$147,529
(Data source: U.S. Census Bureau)
Figure 1.1
There is often much interplay between the various ways of describing functions.
If a function is described symbolically, we may wish to study it by using a table
of values or a graph. We’ll consider graphs of functions shortly. If a function is
defined using a graph or a table, we may wish to construct a function in equation
form that “fits” the data given in the graph or table. We’ll consider this topic in
Section 1.3.
Recall from the definition given earlier that the domain of a function consists
of the set of input values, while the range is the set of possible output values. If
the domain is not explicitly stated, it is understood to be the set of all input values
that produce valid output values. In other words, the domain is the set of values
that “work” in the function.
Example 1 Finding domain and range √
Find the domain and range of h(x) = x − 2.
Solution Since we are only considering values of x for which h(x) is real, the valid
input values of x are those for which x − 2 is not negative—that is, for which
x − 2 ≥ 0. Solving this linear inequality for x yields x ≥ 2. Thus, the domain of
h is [2, ∞). Since the principal square root of any number is positive, the range is
the set [0, ∞).
Occasionally, it is desirable to use two or more formulas to define a function.
Consider, for example, the function f defined by
(
−x
for x ≤ −1
f (x) =
2x + 2 for x > −1
This piecewise definition indicates that for an input value x less than or equal to
−1, the output value is −x. For an input value x greater than −1, the output value
is 2x + 2. Thus, f (−4) = −(−4) = 4, whereas f (1) = 2(1) + 2 = 4.
4
CHAPTER 1. FUNCTIONS
Graphs of Functions
The graph of a function f is the set of points (x, y) such that y = f (x). In other
words, the graph of a function f is simply the graph of the equation y = f (x). Thus,
the techniques you are familiar with for graphing equations can also be applied to
graphing functions.
y = f (x)
f (x)
(x, f (x))
x
The graph of a function depicts the correspondence between input values in the
domain and output values in the range. Indeed, if (x, y) is a point on the graph of
a function f , then y is the value of the function at x, as shown in Figure 1.2.
Since a function has the property that for each x in the domain there is exactly
one y, no vertical line can intersect the graph of a function more than once. Conversely, a graph that does intersect a vertical line more than once cannot be the
graph of a function of the form y = f (x). Thus, the vertical line test provides a
convenient method for determining if an equation defines a function.
Figure 1.2
The Vertical Line Test
An equation in x and y defines a function of the form y = f (x) if and only if no
vertical line crosses the graph of the equation more than once.
Applying the vertical line test, we see in Figure 1.3 that the equation x2 +y 2 = 4
does not define y as a function of x, whereas we can conclude from Figure 1.4 that
x3 + y 3 = 1 does define y as a function of x.
4
3
2
4
x2 + y 2 = 4
1
-4 -3 -2 -1
-1
3
2
1
1
2
3
4
-4 -3 -2 -1
-1
-2
-2
-3
-3
-4
-4
Figure 1.3
x3 + y 3 = 1
1
2
3
4
Figure 1.4
Example 2 Graphing a function with a restricted domain
Sketch the graph of the function f (x) = x1 .
Solution We first note that 0 is not in the domain of f . So there is no point on the
graph with x-coordinate 0. Next, we find and plot some points satisfying y = x1 , as
shown in Figure 1.5.
5
1.1. FUNCTIONS AND THEIR GRAPHS
x
1
2
4
3
f (x) =
1
x
y=
1
1
2
1
3
3
−1
−2
−1
− 21
−3
− 31
1
x
1
x
x
y=
1
2
1
3
2
4
3
3
− 21
−2
− 31
−3
2
1
-4
-1
-1
1
2
3
4
-4
2
1
-4
-1
-1
1
2
3
4
Figure 1.5
Notice that as the x-values get larger (in magnitude), the corresponding y-values
get smaller, and so the points get closer and closer to the x-axis. Moreover, as the
x-values get smaller, the y-values get larger, shooting up or down along the y-axis.
We use these observations to complete the graph, as shown in Figure 1.6.
-4
Figure 1.6
As we noted in Example 2, there is no point on the graph with x-coordinate 0.
As a consequence, the graph does not cross the line x = 0 but rather approaches it
arbitrarily closely. More precisely, the graph rises (or falls) without bound as x gets
close to 0. For this reason, the line x = 0 (the y-axis) is called a vertical asymptote.
The line y = 0 (the x-axis), on the other hand, is a horizontal asymptote since the
graph approaches it arbitrarily closely as x goes toward ∞ and −∞. We will study
vertical asymptotes in detail in Section 2.2 and horizontal asymptotes in Section
2.4.
4
3
y = −x,
x ≤ −1
2
y = 2x + 2,
x > −1
Example 3 Graphing a piecewise-defined function
Sketch the graph of the function
(
−x
for x ≤ −1
f (x) =
2x + 2 for x > −1
1
-4 -3 -2 -1
-1
1
2
3
4
-2
-3
Solution The graph of f is obtained by plotting the line y = −x for x ≤ −1 and
the line y = 2x + 2 for x > −1, as shown in Figure 1.7. Note that the point (−1, 1)
is on the graph but (−1, 0) is not.
-4
Figure 1.7
A graphing utility (like a graphing calculator or other graphing software) can be
useful for determining the domain and range of a function. Since the domain is the
set of inputs (x-values) to a function and the range is the set of outputs (y-values),
the domain and range of a function can be found graphically. In particular, the
domain is the set of all x-values for points on the graph, and the range is the set of
all y-values.
10
Example 4 Finding domain and range graphically
√
Find the domain and range of h(x) = x2 − 4 − 4.
-10
10
-10
Figure 1.8
Solution Graphing h(x) with a graphing utility set to the standard viewing window
gives the graph shown in Figure 1.8. For points (x, y) on the graph of h, it appears
that x ≥ 2 or x ≤ −2. Thus, the domain appears to consist of the intervals [2, ∞)
and (−∞, −2]. All y-values on the graph appear to be greater than or equal to −4,
so the range is [−4, ∞).
6
CHAPTER 1. FUNCTIONS
Functions are generally the method of choice for expressing relationships between
variables in the real world, and even the not-so-real world, as the following example
illustrates.
Example 5 Computing the power output of a cyclist
For a 170-pound cyclist on a 30-pound bicycle pedaling at v miles per hour, the
power output in watts is given by P (v) = 0.0178678v 3 + 2.01168v.
a. Find the power output if the cyclist is traveling at 20 miles per hour.
b. A small power plant will produce around 500,000,000 watts of power. Find the
speed required for a single cyclist to generate enough power to replace a small
power plant. Answer to the nearest mile per hour.
Solution
a. Evaluating P (20), we have
P (20) = 0.0178678(20)3 + 2.01168(20) ≈ 183.176 watts
b. We could set P (v) = 500, 000, 000 and then attempt to solve for v algebraically.
Unfortunately, there are no convenient algebraic methods for solving the cubic
equation 0.0178678v 3 + 2.01168v = 500, 000, 000. Instead, we just approximate
a solution by plotting the graph of P with a graphing utility and searching for
a point with y-coordinate 500, 000, 000. After some experimentation, we settle
on the view shown in Figure 1.9. Next, we zoom in until we can estimate the
x-coordinate of the point to the nearest mile per hour, as shown in Figure 1.10.
Thus, a cyclist capable of traveling at 3036 miles per hour would generate enough
power to replace a small power plant.
600000000
0
3500
X=3035.9865
Y=500006337
0
Figure 1.9
Figure 1.10
f (x) = x3 − 3x2
4
3
Increasing and Decreasing Functions
2
1
-4 -3 -2 -1
-1
Rising
1
2
-2
4
Rising
-3
-4
3
(2, −4)
Falling
Figure 1.11
The graph of a function can be used to unveil many important characteristics.
Consider, for example, the function f (x) = x3 − 3x2 graphed in Figure 1.11. Notice
that as we move from left to right, the graph rises until x = 0, falls in the interval
from x = 0 to x = 2, and then rises to the right of x = 2. We say that the
function f is increasing in the interval (−∞, 0), decreasing in the interval (0, 2),
and increasing in the interval (2, ∞). The points (0, 0) and (2, −4), where the
graph of f changes direction, are called turning points. More generally, we make
the following definitions.
7
1.1. FUNCTIONS AND THEIR GRAPHS
Increasing, Decreasing, and Turning Points
A function f is said to be increasing or decreasing on an interval according to
the following properties.
Type of behavior
Increasing
Graphical Property
As we move from left to
right in the interval, the
graph rises.
Decreasing
As we move from left to
right in the interval, the
graph falls.
Algebraic Property
For any x1 and x2 in
interval with x1 <
f (x1 ) < f (x2 ).
For any x1 and x2 in
interval with x1 <
f (x1 ) > f (x2 ).
the
x2 ,
the
x2 ,
The points where f changes from increasing to decreasing or decreasing to increasing are called turning points.
Example 6 Using a graphing utility to find turning points
Use a graphing utility to approximate the locations of the turning points for the
function f (x) = 18 x3 − x2 + 2. Also, indicate the intervals on which f is increasing
and the intervals on which it is decreasing.
Solution The graph of y = 81 x3 − x2 + 2, shown in Figure 1.12, has two turning
points. One occurs where x = 0, and the other is in the fourth quadrant. Since
f (0) = 2, the first turning point is (0, 2). By tracing the graph, as shown in Figure
1.13, we find that the second turning point is approximately (5.37, −7.48).
Since the graph of f is rising for x less than 0 and also for values of x greater
than 5.37 (approximately), f is increasing on (−∞, 0) ∪ (5.37, ∞). Since the graph
of f is falling between x = 0 and x ≈ 5.37, f is decreasing on (0, 5.37).
10
-10
10
10
-10
10
X=5.3684211
-10
Y=-7.480245
-10
Figure 1.13
Figure 1.12
Sequence Notation
A sequence is nothing more than a list of numbers, such as
1 1
1, , , . . .
2 3
To be more precise, we define sequences as functions with integer domains, as
follows.
8
CHAPTER 1. FUNCTIONS
Definition of Sequence
A sequence is a function a whose domain is a set of integers. The function
values a(n) are called terms and are denoted an . Unless indicated otherwise, an
is assumed to be defined for n = 1, 2, . . ..
Example 7 Finding terms of a sequence
List the first five terms of the sequence defined by an =
1
n2 .
Solution We simply substitute values for n, starting with n = 1.
1
12
1
= 2
2
1
= 2
3
1
= 2
4
1
= 2
5
a1 =
=
a2
=
a3
a4
a5
=
=
=
1
1
1
4
1
9
1
16
1
25
Example 8 Defining a sequence
Find an expression for the nth term of a sequence bn whose first few terms are
1, 2, 4, 8, 16, . . ..
Solution The given numbers are successive powers of 2, beginning with 20 . Thus,
bn = 2n for n = 0, 1, 2, . . ..
1.1 Exercises
Exercises 1–4 Answer true or false.
5. y = −x2 + 1
4
3
2 y = −x2 + 1
1
1. The domain of a function is the set of all output
values.
-4 -3 -2 -1
-2
-3
-4
2. No circle is the graph of y = f (x), where f is a
function.
3. If there is a vertical line that crosses the graph of
an equation once, then the graph is the graph of a
function.
4. If the graph of a function f lies entirely in the third
quadrant, then 2 is in neither the domain nor the
range of f .
Exercises 5–10 Use the graph to determine whether the
given equation defines y as a function of x.
6.
x2
y2
+
=1
16
9
4
3
2
1
-4 -3 -2 -1
-2
-3
-4
1 2 3 4
x2
16
+
y2
9
1 2 3 4
=1
9
1.1. FUNCTIONS AND THEIR GRAPHS
7. x = y 3 − 4y
4
3 x = y 3 − 4y
2
1
-4 -3 -2 -1
-2
-3
-4
1 2 3 4
16.
8. x = y 3
4
3
2
1
-4 -3 -2 -1
-2
-3
-4
17.
x = y3
1 2 3 4
18.
9. xy = 4
4
3
2
1
-4 -3 -2 -1
-2
-3
-4
xy = 4
19.
1 2 3 4
10. |x| − |y| = 0
4
3
2
1
-4 -3 -2 -1
-2
-3
-4
20.
|x| − |y| = 0
1 2 3 4
Exercises 11–20 Evaluate the given function as indicated.
2
11. f (x) = x + 3x + 2
a) f (2)
b) f (−3)
c) f (0)
12. g(t) = t3 − 4t
a) g(1)
b) g(−2)
c) g(3)
3x + 2
13. f (x) =
4x − 5
a) f (4)
b) f (5)
c) f (−t)
14. h(y) =
Exercises 21–24 Express the given function using function
notation.
21. h assigns to every real number the absolute value of
the number.
22. g assigns to every real number twice the cube of the
number.
23. f assigns to every real number in its domain the
reciprocal of 3 more than the number.
24. p assigns to each real number in its domain the reciprocal of the square root of the number.
Exercises 25–36 Find the domain of the function.
25. f (x) = x2
4
26. g(x) =
x
a) h(2)
b) h(−6)
c) h(x)
15. x(t) = 3t2
3
x−4
r(s) = |s|
5t2 + 3t + 1
r(t) =
t2 − 9
x+1
p(x) = 2
x − 5x + 6
√
f (x) = x + 1
√
g(t) = t − 3
27. h(x) =
28.
2
y −3
y+7
a) x(t2 )
b) x(t − 1)
c) x 5t
2x + 2
f (x) =
2x − 3
a) f (t)
b) f (x − 1)
c) f x2
g(x) = (x + 1)2
a) g(x − 1)
b) g ( )
c) g 1−a
a
√
p (x) = 9 − x2
a) p ( )
b) p (x + 3)
c) p (x − 3)
1
f (x) =
, g (x) = x2
x−1
a) f (g (3))
b) f (g (t))
c) g (f (x + 1))
f (x) = x2 , g(y) = 2y + 1
a) f (g(2))
b) f (g(t))
c) g(f (n2 ))
29.
30.
31.
32.
10
CHAPTER 1. FUNCTIONS
√
x+2
Exercises 45–46 Use a graphing utility to sketch the graph
of the function and then use the graph to help identify or
x2 − 4x + 3
2
approximate the domain and range of the function.
34. g(y) = √
√
2y − 6
45. a) f (x) = 9 − x2
2
√
35. f (x) =
b) f (x) = x + 9 − x2
|x| − 3
√
c) f (x) = x 9 − x2
x
36. g(x) =
x
x
d) f (x) = √
9 − x2
Exercises 37–40 Complete the table and use the resulting
√
46. a) f (x) = x
points to sketch the graph of the function.
√
b) f (x) = x + 4
√
37. x f (x) = −2x + 5
c) f (x) = x + 4
0
1
d) f (x) = √
2
x+4
4
Exercises 47–54 Use a graphing utility to approximate the
38.
x f (x) = 32 x − 2
intervals on which the given function is increasing, the
−3
intervals on which it is decreasing, and the coordinates of
0
any turning points.
3
47. f (x) = x2 − 4x + 5
39.
2
x
f (x) = (x + 1) − 3
48. g(x) = −x2 − 6x − 5
−3
49. h(x) = 2x3 − 3x2 − 6x + 5
−2
50. p(x) = x3 − 4x2 + x + 1
−1
5
0
51. q(x) = 2
x
−
4x
+5
1
−3
40.
52. r(x) = 2
x − 6x + 10
x f (x) = −(x − 2)2 + 4
53. f (x) = x4 − 3x2 + x
0
1
54. g(x) = −2x4 + 5x2 + 7
2
Exercises 55–70 List the first four terms of the sequence
3
with the given nth term.
4
33. f (x) =
Exercises 41–44 Evaluate the piecewise-defined function as
indicated and then sketch its graph.
(
−x − 4 for x ≤ 0
41. f (x) =
2x − 4 for x > 0
f (−2), f (0), f (3)
(
3x + 2 for x < 0
42. g(x) =
−x + 2 for x ≥ 0
g(−4), g(0), g(1)


for x < −1
x
43. f (x) = 1
for − 1 ≤ x ≤ 1


x + 2 for x > 1
f (−3), f (−1), f (4)


−x − 1 for x < −2
44. h(x) = x + 3
for − 2 ≤ x ≤ 0


3
for x > 0
h(−3), h(0), h(6)
55. an = 3n + 1
56. un = 4 − n
57. zn = (−2)n
58. cn = − 21n
59. rn =
60. bn =
1
n2
n−1
n+1
61. cn = 1 + (−1)n
62. xn = (−1)n n2
63. vn =
(−1)n−1
n
64. rn =
(−1)n+1
3n
p
n(n + 1)
√
√
66. yn = n + 1 − n
n
67. bn = 1 + n1
65. an =
68. rn =
n2
2n
11
1.1. FUNCTIONS AND THEIR GRAPHS
69. sn = 1 + 2 + · · · + n
70. an = 1 · 2 · · · · · n
Exercises 71–78 Find an expression for the nth term of a
sequence whose first few terms are given.
71. 2, 4, 6, 8, . . .
72. 3, 9, 27, 81, . . .
1
73. − 12 , 14 , − 18 , 16
,...
74. 3, 7, 11, 15, . . .
75.
1 2 3 4
2, 3, 4, 5, . . .
76. 2 · 3, 3 · 4, 4 · 5, 5 · 6, . . .
77. 1 − 11 , 1 − 12 , 1 − 13 , 1 − 14 , . . .
78. 1 + 12 , 1 − 14 , 1 + 81 , 1 −
1
16 , . . .
Applications
79. Volume of a Sphere The volume of a sphere with radius r is given by V = 43 πr3 . Express the volume of
a sphere as a function of its diameter.
80. Area of a Circle The area of a circle with radius r is
given by A = πr2 . Express the area of a circle as a
function of its diameter.
81. Area of a Square A square is inscribed in a circle.
Express the area of the square as a function of the
radius of the circle.
82. Volume of a Cube A cube is inscribed in a sphere.
Express the volume of the cube as a function of the
radius of the sphere.
83. Training Heart Rate The maximum heart rate m for
an adult male aged a years can be approximated by
m = 220 − a
The training heart rate for a fit man should be about
85% of his maximum heart rate.
a) Find a function t(a) that gives the target training heart rate for a man aged a years.
b) What is the feasible domain of this function,
that is, for what values of a does this formula
apply?
84. Logs to Lumber The volume in board feet V that
can be obtained from a log L feet long and with a
diameter of D inches can be approximated (using
the Scribner log rule) by
V = (0.79D2 − 2D − 4)L
For a certain collection of logs, a mill worker observes
that the diameter of a log in inches is typically equal
to its length in feet minus 2.
a) Express the mill worker’s observation as an algebraic equation.
b) Find a function f (D) that gives the volume for
a log of diameter D inches, assuming the mill
worker’s observation is true.
c) Find a function g(L) that gives the volume for
a log L feet long, assuming the mill worker’s
observation is true.
85. Sphere of Influence Electromagnetic waves—like
those in radio and television transmissions, for
example—travel at the speed of light, often in all
directions. Although our radios pick up a signal almost instantaneously, the signal continues to travel
through space at the speed of light. Thus, for example, Dr. Martin Luther King’s famous 1963 “I Have
A Dream” speech is even now reaching new parts
of the galaxy. The radio waves from the broadcast
began traveling through space at the speed of light
in all directions from the point of origin. Since light
travels at a rate of approximately 5.9 × 1012 miles
per year, the distance the waves will have traveled
after t years is given by r = (5.9 × 1012 )t. Moreover,
when the waves are a distance r from the point of
origin, the volume of space (ignoring the fact that
some of the signal will be blocked by the Earth itself) through which they will have traveled is given
by V = 43 πr3 . Express the volume V as a function
of t and compute the volume of space through which
the speech had traveled by the year 2000.
86. Radius of an Oil Spill An oil tanker develops a
leak and begins spilling oil at a rate of 10, 000, 000
cubic centimeters per minute. The volume of oil
that has spilled after t minutes is thus given by
V = 10, 000, 000t cubic centimeters. Assuming that
the resulting oil spill has a circular shape with radius r and a constant thickness h, the volume can
also be expressed as V = πr2 h. If h = 1 centimeter,
express the radius as a function of time and compute
the radius of the spill after 2 hours.
87. Survival Function Based on a mathematical model
developed by DeMoivre, the percentage of newborns
surviving to age x years can be approximated by
s(x) =
12, 500 − 100x
125
a) Because the output values of this function are
percentages, what should the range be?
b) According to this model, by what age will exactly one half of a group of newborns be alive?
c) According to this model, what is the maximum
age to which a human can live? [The official
world record for the oldest individual is 122
years and 164 days in the case of Jeanne Calment of France (Feb. 21, 1875-Aug. 4, 1997).]
12
CHAPTER 1. FUNCTIONS
d) What should we consider the domain of this
function to be?
88. Accident Claim Model In the event of a motor vehicle accident in which there is personal injury, the
party not at fault is entitled to compensation for
damage to the vehicle, lost wages, medical bills, and
pain and suffering. It’s relatively simple to put a
dollar amount on vehicle damage, lost wages, and
medical bills, but quantifying pain and suffering can
be difficult. One rule of thumb commonly used by
insurance adjusters for simple cases is to place the
value of pain and suffering at triple the total of lost
wages and medical bills. Suppose you are not at
fault in an accident that causes $20,000 of damage
to your Lamborghini Diablo and forces you to miss 8
hours of work as a corporate attorney (pay: $500 per
hour). In addition, emergency treatment and physical therapy for your sore neck totals M dollars.
a) Use the rule of thumb to express the dollar
amount of your pain and suffering in terms of
M.
b) Find a function S(M ) that gives the total
amount of the settlement as a function of M ,
the total medical bill.
c) Find the domain and range of S.
d) Estimate the total settlement if the medical bills
are $1000.
e) If the settlement amount is $100,000, then what
were the medical bills?
f) In the unlikely event that you were lying about
having a sore neck, how much profit did you
make on every dollar spent on unnecessary medical expenses?
89. The Fibonacci Sequence The Fibonacci sequence,
named after the Italian mathematician Leonardo of
Pisa, is perhaps the most famous sequence in all of
mathematics. It is an example of a recursively defined sequence, one in which the first few terms are
given and every successive term can be computed
using a formula involving one or more of the preceding terms. The Fibonacci sequence, which first
appeared in Fibonacci’s book “Liber Abaci” (“The
Book of the Abacus”) in 1202, can be described as
the sequence for which the first two terms are 1 and
each successive term is found by adding together the
previous two terms. In other words,
a1 = 1
a2 = 1
an = an−1 + an−2 for n ≥ 3
a. Find a3 . . . , a10 .
b. The Fibonacci sequence is present in the family tree of a male honeybee. As first observed
by Johann Dzierzon in 1845, the male honeybee hatches from an unfertilized egg and so has
a mother but no father. [Source: Colin Butler, “The World of the Honeybee” (New York:
Macmillian, 1955).] Female honeybees, on the
other hand, hatch from fertilized eggs and so have
a mother and a father.
i. Find the number of grandparents, greatgrandparents, and great-great-grandparents
for a male honeybee. (Hint: A family tree
may be helpful.)
ii. Show that if bn is the number of ancestors
n generations ago for a male honeybee, then
bn = bn−1 + bn−2 .
iii. Write bn in terms of an , the nth term of the
Fibonacci sequence.
90. Geometric Sequences and Population Growth Suppose that a population is undergoing a constant
growth rate of k people per thousand per year. In
other words, for each group of 1000 people in the
population, there will be an increase of people after
1 year. Let Pn represent the population in year n.
Find an expression for Pn+1 in terms of Pn .
a. Suppose that a population is undergoing a constant growth reate of k people per thousand per
year. In other words, for each group of 1000 people in the population, there will be an increase of
people after 1 year.
b. A geometric sequence is one for which successive
terms have a common ratio, that is a sequence an
for which
an = ran−1
for some real number r. Show that the sequence
Pn is geometric.
c. Find a closed formula for Pn , that is, a formula
that allows us to compute Pn directly without
first computing P1 , P2 , . . . , Pn−1 .
d. The population of Las Vegas, Nevada, grew at
an incredible rate near the turn of the last century. In 2000, its population was 478,434 and was
growing at the rate of about 58 people per thousand. Find a formula for Pn , the population of
Las Vegas in year 2000 + n, and use it to estimate
the population of Vegas in the year 2050.
13
1.2. ALGEBRA OF FUNCTIONS
1.2 Algebra of Functions
In this section, we will review some of the basic algebraic combinations and
transformations of functions that will be of particular use in calculus.
Algebraic Combinations of Functions
Just as numbers can be combined by the basic arithmetic operations of addition,
subtraction, multiplication, and division, so too can functions. For example, we can
define the sum of two functions f and g by the formula (f + g)(x) = f (x) + g(x).
The operations of subtraction, multiplication, and division of functions are defined
in a similar fashion, as follows.
Arithmetic Operations on Functions
(f + g)(x) = f (x) + g(x)
(f − g)(x) = f (x) − g(x)
(f g)(x) = f (x)g(x)
f (x)
f
g (x) = g(x) , g(x) 6= 0
In keeping with our convention of assuming that the domain (unless otherwise
specified) is the set of real numbers for which the function “makes sense,” the
domain of f + g is the set of real numbers x for which both f (x) and g(x) are
defined. Thus, the domain of f + g will consist of those real numbers that are in
the domains of both f and g. Similarly, the domains of f − g and f g consist of all
real numbers that are contained in the domains of both f and g. However, since
the expression f (x)/g(x) is not defined when g(x) = 0, the domain of f /g consists
of those real numbers x in the domains of both f and g such that g(x) 6= 0.
Example 1 Finding combinations
√ of functions
Let f (x) = x2 and g(x) = x − 1. Compute each of the following combinations of
functions and find their domains.
f
a. (f + g)(x) b. (f g)(x) c.
g (x)
Solution
a.
(f + g)(x) = f (x) + g(x)
√
= x2 + x − 1
Since the domain
of f (x) = x2 is the set of all real numbers, whereas the domain
√
of g(x) = x − 1 is [1, ∞), the domain of f + g is [1, ∞).
b.
(f g)(x) = f (x)g(x)
√
= x2 x − 1
The domain of f g is the same as that of f + g, namely [1, ∞).
14
CHAPTER 1. FUNCTIONS
c.
f
f (x)
(x) =
g
g(x)
x2
=√
x−1
The only difference between √
the domains found in parts a and b and the domain
of f /g is that, since g(x) = x − 1 is now in the denominator, we cannot allow
x = 1 in the domain. Thus, the domain of f /g is (1, ∞).
Composition of Functions
Functions can also be combined by composing or “stringing” them together—using
the output of one function as the input for another. The composition of two functions f and g, written f ◦ g, is defined by the formula (f ◦ g)(x) = f (g(x)). The
function f ◦ g takes an input value x and produces f (g(x)) as the corresponding
output value. Since an input x is first substituted into g, each input x must be in
the domain of g. We also see that because we are using g(x) as an input for f , g(x)
must be in the domain of f .
Composition of Function
The composition of the functions f and g, written f ◦ g, is defined by
(f ◦ g)(x) = f (g(x))
The domain of f ◦ g consists of those real numbers x in the domain of g such that
g(x) is in the domain of f .
Example 2 Computing the composition of two functions
Given that f (x) = 2x2 + 4x + 5 and g(x) = 2x + 1, compute (f ◦ g)(x).
Solution
(f ◦ g)(x) = f (g(x))
= f (2x + 1)
Replacing g(x) with 2x + 1
= 2(2x + 1)2 + 4(2x + 1) + 5
Substituting 2x + 1 for each x in the formula for f (x)
= 2(4x2 + 4x + 1) + 8x + 4 + 5
= 8x2 + 16x + 11
Example 3 Computing
compositions of functions
√
Let f (x) = x and g(x) = x2 + 1. Compute each of the following compositions and
find their domains.
a. (f ◦ g)(x)
b. (g ◦ f )(x)
Solution
a.
(f ◦ g)(x) = f (g(x))
= f (x2 + 1)
p
= x2 + 1
15
1.2. ALGEBRA OF FUNCTIONS
Since x2 + 1 is defined for all real numbers and is always positive, the domain of
f ◦ g is the set of all real numbers.
b.
(g ◦ f )(x) = g(f (x))
√ =g x
√ 2
=
x + 1 for x ≥ 0
= x + 1 for x ≥ 0
Note that although x + 1 is defined
for all real numbers x, (g ◦ f )(x) is not, since
√
the inside function, f (x) = x, is defined only for nonnegative x. Thus, the
domain of g ◦ f is the set of all nonnegative real numbers.
Example 4 Computing compositions of functions defined numerically
Suppose that f and g are defined by Table 1.3. Compute (f ◦ g)(x) and (g ◦ f )(x)
for x = 0, 1, 2, 3.
Table 1.3
x
0
1
2
3
f (x)
−2
0
2
4
Solution By definition, (f ◦ g)(x) = f (g(x)). Thus, using the appropriate values
from Table 1.3, we have
g(x)
1
2
5
10
(f ◦ g)(0) = f (g(0)) = f (1) = 0
(f ◦ g)(1) = f (g(1)) = f (2) = 2
But
(f ◦ g)(2) = f (g(2)) = f (5)
and f (5) is not defined in Table 1.3. Thus, (f ◦ g)(2) is undefined. Similarly,
(f ◦ g)(3) is undefined. The computations for g ◦ f are much the same as those for
f ◦ g.
Table 1.4
(g ◦ f )(0) = g(f (0)) = g(−2), which is undefined
x
(f ◦ g)(x)
(g ◦ f )(x)
(g ◦ f )(1) = g(f (1)) = g(0) = 1
0
1
2
3
0
2
undefined
undefined
undefined
1
5
undefined
(g ◦ f )(2) = g(f (2)) = g(2) = 5
(g ◦ f )(3) = g(f (3)) = g(4), which is undefined
Table 1.4 summarizes the values for f ◦ g and g ◦ f .
Transformations of Functions
y = x3
y = (x − 5)3
(2, 8)
(7, 8)
7
5
We have considered how two functions can be combined to form a new function.
Next, we investigate how the graph of a function is changed or transformed by
making slight changes in the algebraic definition of the function. An understanding
of the effect of these transformations on familiar functions can also assist us in
sketching the graphs of more complicated functions.
3
1
-3
(−1, −1) -3
(1, 1)
1
3
(6, 1)
5 7
(4, −1)
-5
(−2, −8)
-7
(3, −8)
Figure 1.14
Translations
Many graphs are easily sketched by recognizing that they are merely shifts or
translations of familiar graphs. For example, consider the graphs of y = x3 and
y = (x − 5)3 shown in Figure 1.14. It is evident from the graphs that the y-values
are the same whenever the x-values for y = (x − 5)3 are 5 greater than those for
y = x3 . The net effect is that the graph of y = (x − 5)3 has the same shape as the
graph of y = x3 , but it is translated (shifted) 5 units to the right.
16
CHAPTER 1. FUNCTIONS
More generally, if h > 0, replacing x with x − h in an equation results in a shift
of the graph to the right h units. Thus, we conclude that the graph of y = f (x − h)
can be found by shifting the graph of y = f (x) to the right h units. Similarly, if
k > 0 and we replace y by y − k in an equation, then the graph is shifted up k
units. Consequently, the graph of y − k = f (x) or, equivalently, y = f (x) + k can
be found by shifting the graph if y = f (x) up k units. We summarize translations
as follows:
y = f (x)
Transformation
y = f (x) + k
y = f (x) − k
y = f (x − h)
y = f (x + h)
k
h
y = f (x − h) + k
Effect on the graph of y = f (x), assuming h, k > 0
Translation k units up
Translation k units down
Translation h units to the right
Translation h units to the left
Vertical and horizontal translations can also be combined. For example, if h, k >
0, then the graph of y = f (x − h) + k is a translation of the graph of y = f (x) h
units to the right and k units up. See Figure 1.15.
Figure 1.15
Example 5 Graphing translated
functions
√
Use the graph of f (x) = x to sketch the graph of the given function.
a. q(x) =
√
x+3
b. r(x) =
√
x+2
c. s(x) =
√
x−1−2
Solution
√
a. Since q(x) = x + 3 = f (x) + 3, we know that the graph of q is obtained by
translating the graph of f up 3 units, as shown in Figure 1.16.
√
b. Since r(x) = x + 2 = f (x + 2), the graph of r is obtained by translating the
graph of f to the left 2 units, as shown in Figure 1.17.
c. Here s(x) = f (x − 1) − 2, so the graph of s is obtained by translating the graph
of f down 2 units and 1 unit to the right, as shown in Figure 1.18.
4
3
q(x) =
√
x+3
2
1
-4 -3 -2 -1
-1
f (x) =
1
2
3
4
√
4
3
r(x) =
√
4
x+2
2
x
1
-4 -3 -2 -1
-1
f (x) =
1
2
3
4
√
3
2
x
-4 -3 -2 -1
-1
-2
-2
-2
-3
-3
-3
-4
-4
-4
Figure 1.16
Figure 1.17
f (x) =
1
1
2
s(x) =
Figure 1.18
3
√
√
x
4
x−1+2
17
1.2. ALGEBRA OF FUNCTIONS
Reflections
As Figure 1.19 illustrates, the reflection about the x-axis of the point (x, y) is the
point (x, −y). Thus, if we substitute −y for y in an equation, the new graph will
be a reflection of the old graph about the x-axis. Similarly, as suggested by Figure
1.20, the reflection about the y-axis of the point (x, y) is the point (−x, y), and
so substitution of −x for x in an equation causes a reflection about the y-axis.
Reflection about the origin is defined by reflecting about both the x-axis and the
y-axis. Thus, the reflection about the origin of the point (x, y) is the point (−x, −y)
(see Figure 1.21), and it follows that if we substitute −x for x and −y for y in an
equation, the new graph will be the reflection of the old graph about the origin.
(x, y)
(−x, y)
(x, y)
(x, −y)
Reflection about the x-axis
(x, y)
(−x, −y)
Reflection about the y-axis
Figure 1.19
Figure 1.20
Reflection about the origin
Figure 1.21
When these principles are applied to the graph of a function y = f (x), we obtain
the following rules for reflections.
Transformation
y = −f (x)
y = f (−x)
y = −f (−x)
Effect on the graph of y = f (x)
Reflection about the x-axis
Reflection about the y-axis
Reflection about the origin
4
3
y = f (x)
2
1
-4 -3 -2 -1
-1
1
2
3
Example 6 Graphing translated and reflected functions
The graph of a function f is shown in Figure 1.22. Sketch the graphs of the following
functions.
4
-2
-3
a. h(x) = f (−x) − 3
b. h(x) = −f (x + 5)
-4
Figure 1.22
Solution
a. The graph of h(x) = f (−x) − 3 is obtained by reflecting the graph of f about
the y-axis and translating 3 units down, as shown in Figures 1.23 and 1.24.
18
CHAPTER 1. FUNCTIONS
4
4
3
y = f (−x)
3
y = f (x)
2
y = f (−x)
-4 -3 -2 -1
-1
y = f (x)
2
1
1
1
2
3
4
-4 -3 -2 -1
-1
-2
1
2
3
4
-2
-3
h(x) = f (−x) − 3
-4
Figure 1.23
-3
-4
Figure 1.24
b. The graph of h(x) = −f (x + 5) is obtained by reflecting the graph of f about
the x-axis and translating 5 units to the left, as shown in Figures 1.25 and 1.26.
4
4
3
3
y = f (x)
2
-4 -3 -2 -1
-1
y = f (x)
2
1
1
1
2
-2
3
4
y = −f (x)
-3
-4
-4 -3 -2 -1
-1
h(x) = −f (x + 5)
1
2
-2
3
4
y = −f (x)
-3
-4
4
3
Figure 1.25
Figure 1.26
2
1
-4 -3 -2 -1
-1
1 2 3 4
-2
Even and Odd Functions
-3
-4
Figure 1.27: An even function
4
3
2
1
-4 -3 -2 -1
-1
1 2 3 4
-2
-3
-4
Figure 1.28: An odd function
If the graph of an equation remains the same after reflecting about the y-axis, then
the graph is said to be symmetric with respect to the y-axis. To test an
equation for symmetry with respect to the y-axis, we simply replace x with −x in
the equation and check to see whether the resulting equation is equivalent to the
original one. Thus, for the graph of a function f to be symmetric with respect to
the y-axis, f (−x) = f (x). Such functions are said to be even. Examples of even
functions include polynomials having only even degrees (thus the name) and the
cosine function. Figure 1.27 shows the graph of an even function.
Similarly, functions that are symmetric with respect to the origin—that is, functions whose graphs remain the same after reflecting about the origin—can be shown
to satisfy the equation f (−x) = −f (x) and are said to be odd. Examples of odd
functions include, not surprisingly, polynomial functions involving only terms of
odd degree as well as the sine function. The graph of an odd function is shown in
Figure 1.28.
We summarize these facts as follows.
19
1.2. ALGEBRA OF FUNCTIONS
Even and Odd Functions
Even functions
f (−x) = f (x); graphs are symmetric with respect to the y-axis.
Odd functions
f (−x) = −f (x); graphs are symmetric with respect to the origin.
Example 7 Investigating even and odd functions
Determine if the given functions are even, odd, or neither.
a. f (x) =
x2
x4 +2
b. g(x) = x3 − 3x
c. h(x) = x + 1
Solution
a. Since f (x) =
x2
x4 +2 ,
it follows that
(−x)2
(−x)4 + 2
x2
= 4
x +2
= f (x)
f (−x) =
Thus, f is an even function.
b.
g(−x) = (−x)3 − 3(−x)
= −x3 + 3x
= −(x3 − 3x)
= −g(x)
So g is an odd function.
y = f (x)
c. Since h(−x) = −x + 1 is neither equal to h(x) nor to −h(x), h is neither even
nor odd.
Example 8 Using even/odd to complete a graph
A portion of the graph of a function f is shown in Figure 1.29. Complete the graph
assuming that
a. f is even
Figure 1.29
b. f is odd
Solution
a. An even function is symmetric with respect to the y-axis, so we reflect about the
y-axis to obtain the graph shown in Figure 1.30.
b. An odd function is symmetric with respect to the origin. Reflecting about the
origin gives us Figure 1.31.
20
CHAPTER 1. FUNCTIONS
y = f (x)
y = f (x)
Figure 1.30
a. f is even
Figure 1.31
b. f is odd
1.2 Exercises
Exercises 1–6 Answer true or false.
a. (f + g)(x)
b. (f − g)(x)
c. (f g)(x)
1. (f ◦ g)(x) = f (x)g(x) for all real numbers x.
13. f (x) = x, g(x) = 5
2. If f (2) = g(2), then 2 is a zero of f − g.
14. f (x) = 3x + 2, g(x) = x − 7
3. If f (x) and g(x) are defined for all real numbers,
then the domain of f /g is the set of all real numbers.
15. f (x) = x2 , g(x) = 3x + 1
4. For any two functions f and g, f ◦ g = g ◦ f .
17. f (x) = x, g(x) = x
2
18. f (x) = , g(x) = 3x
x
5. If f has a turning point at (2, 5) and h is defined by
h(x) = f (x − 2), then h has a turning point at (2, 3).
6. If the graph of y = g(x) is the reflection about the
x-axis of the graph of y = f (x) and f (5) = 8, then
we can conclude that g(−5) = 8.
Exercises 7–12 Give an example of each.
d.
f
g
(x)
16. f (x) = 2x + 5, g(x) = x2 + 1
19. f (x) = 2x − 2, g(x) =
2
x+5
1
1
, g(x) = −
x
x
√
21. f (x) = x − 2, g(x) = x − 4
√
22. f (x) = 3x, g(x) = 2x − 6
20. f (x) =
7. The formula for a function f such that f (x + c) =
f (x) for all real numbers c
Exercises 23–40 Compute both (f ◦ g)(x) and (g ◦ f )(x).
Specify the domain if it is anything other than all real
8. The formulas
for
a
pair
of
functions
f
and
g
such
f
2
numbers.
that
(x) = x
g
9. The formulas for a pair of functions f and g such
that f ◦ g 6= g ◦ f
10. The formulas for a pair of functions f and g such
thatf ◦ g = g ◦ f
11. The formula for a function f whose graph remains
unchanged after reflecting about the y-axis
12. The formula for a function f whose graph remains
unchanged after reflecting about the origin
Exercises 13–22 Find the following combinations. Specify
the domain if it is anything other than all real numbers.
23. f (x) = 2x + 3, g(x) = 4x − 5
3−x
24. f (x) = 3 − 2x, g(x) =
2
25. f (x) = x2 , g(x) = 2x + 7
26. f (x) = 3x − 4, g(x) = x3
3
1
27. f (x) = , g(x) =
x
3x
4
28. f (x) =
, g(x) = 2x + 4
x+1
1
29. f (x) =
, g(x) = x2 − 3x
x+2
21
1.2. ALGEBRA OF FUNCTIONS
30. f (x) = x2 + 3x, g(x) = x − 5
a) h(x) = (x + 1)3
31. f (x) = 7, g(x) = x2 + 3x + 1
1
1
32. f (x) = , g(x) =
x
x
√
33. f (x) = x, g(x) = x4
b) h(x) = x3 − 4
c) h(x) = −x3
d) h(x) = −(x − 2)3
e) h(x) = −x3 + 3
34. f (x) = |x|, g(x) = 4x − 1
√
35. f (x) = 3x + 5, g(x) = x − 2
√
36. f (x) = 2x + 3, g(x) = 2x − 5
r
2
3 2
37. f (x) = 3
, g(x) =
−8
x +8
x
1
38. f (x) = 2
, g(x) = 3
x −9
39.
x
1
2
3
4
f (x)
2
4
6
8
Exercises 43–44 A function f and its graph are given. For
each translated or reflected graph, determine the corresponding function g.
43. f (x) = |x|
4
3
g(x)
1
2
4
8
2
f (x) = |x|
1
-4 -3 -2 -1
-1
1
2
-2
40.
x
0
5
10
15
f (x)
10
15
20
10
g(x)
5
10
15
0
-3
-4
a)
4
2
1
-4 -3 -2 -1
-1
4
3
f (x) =
2
√
3
1
2
3
4
-2
x
-3
-4
1
-4 -3 -2 -1
-1
1
2
3
4
b)
4
-2
3
-3
y = g(x)
2
-4
a)
b)
c)
d)
e)
y = g(x)
3
Exercises 41–42 A function f and its graph are given. Use
the graph of f to find the graph of h.
√
41. f (x) = 3 x
1
√
h(x) = 3 x + 2
√
h(x) = 3 x − 4
√
h(x) = − 3 x
√
h(x) = 3 −x
√
h(x) = − 3 x − 3
-4 -3 -2 -1
-1
1
2
3
4
-2
-3
-4
c)
42. f (x) = x3
4
4
3
3
f (x) = x3
2
-2
-3
-4
y = g(x)
1
1
-4 -3 -2 -1
-1
2
1
2
3
4
-4 -3 -2 -1
-1
-2
-3
-4
1
2
3
4
3
4
22
CHAPTER 1. FUNCTIONS
44. f (x) =
√
9 − x2
5
4
3
2
1
-5 -4 -3 -2 -1
-1
-2
-3
-4
-5
f (x) =
√
a)
b)
c)
d)
e)
f)
9 − x2
1 2 3 4 5
46.
h(x) = f (x − 2)
h(x) = f (x) + 3
h(x) = f (x + 1) − 4
h(x) = f (−x)
h(x) = −f (x) + 2
h(x) = −f (x − 1)
(−4, 4)
(4, 4)
4
3
2
a)
1
5
4
3
2
1
-4 -3 -2 -1
-1
y = g(x)
y = f (x)
1
2
3
4
-2
-3
-5 -4 -3 -2 -1
-1
-2
-3
-4
-5
1 2 3 4 5
5
4
3
2
1
y = g(x)
-5 -4 -3 -2 -1
-1
-2
-3
-4
-5
1 2 3 4 5
-4
a)
b)
c)
d)
e)
f)
b)
h(x) = f (x) − 2
h(x) = f (x + 3)
h(x) = f (x + 1) − 4
h(x) = f (−x)
h(x) = −f (x) + 2
h(x) = −f (x − 1)
47.
4
3
y = f (x)
2
1
-4 -3 -2 -1
-1
c)
1
2
3
4
-2
5
4
3
2
1
-3
-4
y = g(x)
-5 -4 -3 -2 -1
-1
-2
-3
-4
-5
a)
b)
c)
d)
1 2 3 4 5
48.
(−1, 3) 4
3
Exercises 45–48 Use the graph of the function f to sketch
the graph of the function h.
2
1
45.
-4 -3 -2 -1
-1
4
(−4, 2)
3
-2
(4, 2)
1
-2
-3
-4
1
2
3
-4
y = f (x)
1
2
3
4
a)
b)
c)
d)
4
y = f (x)
-3
2
-4 -3 -2 -1
-1
h(x) = f (x − 3) + 2
h(x) = f (−x) − 1
h(x) = −f (x + 1)
h(x) = f (−x + 2) − 3
(1, −3)
h(x) = f (x + 2) − 3
h(x) = f (−x) − 1
h(x) = −f (x + 1)
h(x) = −f (x − 2) + 3
23
1.2. ALGEBRA OF FUNCTIONS
Exercises 49–50 A function f is given. Find an expression Applications
for the function h that results from the indicated transformation.
59. Functions in the Real World A pair of functions f
and g is given. In each case, find the combination
2
49. f (x) = x
(choose from f + g, f − g, g − f , f ◦ g, g ◦ f , f g, f /g,
a) Translate up 2 units
and g/f ) that has a meaningful interpretation as a
b) Reflect about the x-axis
relationship between real-world quantities, and give
c) Translate 1 unit to the left and then reflect
that interpretation.
about the y-axis
√
a) f (x) = recommended intake (in calories) for a
50. f (x) = x
man weighing x pounds; g(x) = recommended
a) Translate 3 units to the left
intake of dietary fiber (in grams) for a person
b) Reflect about the y-axis
consuming x calories
c) Reflect about the x-axis, then translate down 2
units, and then reflect about the y-axis
b) f (x) = quantity sold for a certain model of car
when the price is set at x dollars; g(x) = rebate
Exercises 51–56 Determine if the given function is even,
(in dollars) when the price is set at x dollars
odd, or neither.
c) f (x) = total number of miles traveled by all U.S.
51. f (x) = x4 − 4x2
vehicles on the road in year x; g(x) = number
52. f (x) = 2x3 − 8x + 1
of U.S. vehicles on the road in year x
√
53. f (x) = x x2 + 1
d) f (x) = anticipated reduction in the number of
54. f (x) = 4x2x+1
fatal accidents in the U.S. if the minimum age
2
+1
for operating a motor vehicle were increased by
55. f (x) = xx+1
x years in all states; g(x) = anticipated reduc1
1
56. f (x) = 2 |x + 1| + 2 |x − 1|
tion in the number of nonfatal accidents in the
Exercises 57–58 Complete the graph of f assuming that
U.S. if the minimum age for operating a motor
vehicle were increased by x years in all states
a. f is even
b. f is odd
e) f (x) = expected before-tax income for those
with x years of higher education; g(x) = expected after-tax income for those with x years
of higher education
57.
y = f (x)
60. Temperature Conversion The functions f and g are
defined and interpreted in Table 1.5.
58.
a) Compute (f ◦ g) (x).
b) Compute (g ◦ f ) (x).
c) Which of the functions f ◦ g and g ◦ f has a
meaningful real-world interpretation? For that
function, interpret both the input and the output.
y = f (x)
Table 1.5
Function
f (x) = 59 (x − 32)
g(x) = x + 273
Interpretation of input
temperature in degrees Fahrenheit
temperature in degrees Celsius
Interpretation of output
temperature in degrees Celsius
temperature in degrees Kelvin
24
CHAPTER 1. FUNCTIONS
1.3 Library of Functions
To fully exploit the power of functions, it is not sufficient to merely master
general concepts such as notation, graphs, and combinations. Instead, one must
also gain familiarity with a few particularly useful and frequently occurring classes
of functions. In this section, we discuss a number of such classes, including linear,
polynomial, rational, trigonometric, and exponential.
Lines and Linear Functions
Lines can be represented using a variety of equation forms. The most general
form is ax + by + c = 0. Although this form does not immediately convey any
important information about a line, it is useful because of the fact that any line
can be represented with an equation of this form. For lines that are not vertical, an
equation form that includes information about the slope of the line is usually much
more useful. Recall that the slope of a nonvertical line passing through the points
(x1 , y1 ) and (x2 , y2 ) is
y2 − y1
∆y
=
m=
∆x
x2 − x1
A line with slope m that passes through the point (x1 , y1 ) has equation y − y1 =
m(x − x1 ). This is called the point-slope form. A line with slope m that passes
through the y-intercept (0, b) has equation y = mx + b. This is called the slopeintercept form. Two other facts to remember are that parallel lines have the same
slope, whereas perpendicular lines have slopes that are “negative reciprocals” of
each other.
Example 1 Finding the equation of a line
Find an equation of the line that passes through (−3, 5) and is perpendicular to the
line with equation 2x + y = 10.
Solution We begin by finding the slope of the line with equation 2x + y = 10. To
do this, we write its equation in slope-intercept form as y = −2x + 10. Thus, we
can conclude that the line y = −2x + 10 has slope −2. Since the line we are looking
for is to be perpendicular to y = −2x + 10, it must have slope 21 . Using a slope of
1
2 and the point (−3, 5), we apply the point-slope equation to arrive at
10
8
y=
(−3, 5)
+
13
2
2
-3
-1
y−5=
1
(x − (−3))
2
y = −2x + 10
4
-5
1
x
2
6
1
3
5
After some minor rearranging, we can write the line in slope-intercept form, as
follows.
1
13
y = x+
2
2
Both lines are graphed in Figure 1.32.
Figure 1.32
The slope-intercept form for a nonvertical line is of special significance because
it expresses y as a function of x. In fact, functions of the form f (x) = mx + b
are usually referred to as linear functions. Linear functions appear frequently in
real-world settings.
25
1.3. LIBRARY OF FUNCTIONS
Example 2 Linear depreciation
When computing the value of a piece of machinery for tax purposes, it is common
practice to use linear depreciation. With linear depreciation, we assume that the
value decreases at a constant rate, and so there is a linear relationship between
value and time. In other words, value is a linear function of time. Suppose that a
construction company purchases a crane for $200,000 and estimates that the crane
will have a salvage value of only $20,000 after 10 years. Let V denote the value of
the crane (in dollars) and let t denote the number of years after purchase. We will
assume that the value of the crane depreciates linearly.
a. Express V as a function of t.
b. Find the value of the crane after 6 years.
c. Interpret the slope of the graph of V .
Solution
a. The line given by the graph of V must pass through the points (0, 200000) and
(10, 20000). Thus, it must have slope
m=
20000 − 200000
∆V
=
= −18000
∆t
10 − 0
Since the point (0, 200000) is the y-intercept, we use the slope-intercept form for
the equation a line to obtain the following linear function.
V = −18000t + 200000
b. Setting t = 6, we obtain V = −18000(6) + 200000 = 92000. So the crane will be
worth $92,000 after 6 years.
c. The slope of −18000 tells us that the value of the crane will decrease at a constant
rate of $18,000 per year.
As we can generalize from Example 2, the slope of a linear function can be
interpreted as a constant rate of increase (if the slope is positive) or decrease (if the
slope is negative). This observation will turn out to be very significant as we delve
further into calculus.
Polynomial Functions
Many of the functions we have already seen, including V (t) = −18000t + 200000
from Example 2, are examples of polynomial functions. In general, we make the
following definition.
Definition of Polynomial Function
A polynomial function is a function of the form
f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0
where n is a nonnegative integer (called the degree of the polynomial), a0 , a1 ,
. . ., an are real numbers (called coefficients), and an 6= 0.
Note that a polynomial function of degree 1 has the form f (x) = a1 x+a0 , and its
graph is nothing more than a line with slope a1 . Thus, we see that linear functions
26
CHAPTER 1. FUNCTIONS
are also polynomial functions. The simplest nonlinear polynomial functions are the
quadratic functions, which have the form f (x) = a2 x2 + a1 x + a0 and are of
degree 2. The graphs of quadradic functions are parabolas. Cubic functions have
the form f (x) = a3 x3 + a2 x2 + a1 x + a0 and are of degree 3. Two useful examples
of quadratic and cubic polynomials are y = x2 and y = x3 , the graphs of which are
shown in Figures 1.33 and 1.34.
4
4
3
3
y = x2
2
1
-4 -3 -2 -1
-1
y = x3
2
1
1
2
3
4
-4 -3 -2 -1
-1
-2
-2
-3
-3
-4
-4
Figure 1.33
1
2
3
4
Figure 1.34
Polynomial functions are defined for all real numbers, and so their domain is
(−∞, ∞). Their graphs have no breaks and are “smooth” in the sense that there are
no abrupt changes in direction. Because of their simplicity, their variety, and the
smoothness of their graphs, polynomials are often used to construct mathematical
models of real-world phenomena.
Rational Functions
When two polynomial are added, subtracted, or multiplied, the result is again a
polynomial. However, when one polynomial is divided by another, the result is
generally not a polynomial. Thus, quotients of polynomials form a new class of
functions, called the rational functions. We make the following definition.
Definition of Rational Function
A rational function is a function of the form
f (x) =
p(x)
q(x)
where p(x) and q(x) are both polynomials.
Note that by taking q(x) to be the constant function q(x) = 1, we see that a
polynomial function can also be viewed as a rational function. More generally, if a
rational function has a denominator that is never zero, the function will have some
of the same nice properties that polynomials have. In particular, the domain will
include all real numbers, and the graph will be unbroken and smooth. However,
in cases where the polynomial in the denominator has zeros, the domain will not
include all real numbers, and the graph will have breaks. We’ll investigate such
breaks in detail in Chapter 2.
Power Functions
A power function is a function of the form f (x) = xa , where a is real number.
In the case where a is a positive integer, f (x) = xa is also a polynomial function
27
1.3. LIBRARY OF FUNCTIONS
(and so also a rational function). If a is a negative integer, f (x) = xa is also a
rational function (but not a polynomial). So some power functions are polynomial
functions (and so also rational functions), and some are rational functions (but not
polynomial functions). However, some power functions
√ are neither polynomial nor
rational functions. For example, f (x) = x1/2 = x is a power function, but it
is neither a polynomial nor a rational function. Some power functions, such as
f (x) = x1/2 , have restricted domains, but others have domains that include all real
numbers.
Algebraic and Transcendental Functions
A function that is formed using arithmetic combinations or compositions of polynomials, rational functions, and power functions with rational powers is called algebraic. Examples include
f (x) =
p
x2 + 1
1
x−1
√
h(x) = (x3 + 2x2 ) x
g(x) = 3x2 + √
3
A function that is not algebraic is said to be transcendental. Examples include
the trigonometric functions and the exponential functions.
Trigonometric Functions
The six trigonometric functions—sine, cosine, tangent, secant, cosecant, and cotangent—arise frequently in calculus. In virtually all instances when these functions
appear in this text, we will follow the standard convention that input values correspond to the radian measure of an angle. So, for example, the output values of
f (x) = sin x are found by treating each x as an angle measured in radians. Since
the trigonometric functions are defined by considering points around the unit circle,
their output values eventually begin to repeat. For this reason, the trigonometric
functions are said to be periodic. For example, the sine function is periodic with
period 2π, which means that sin(x + 2π) = sin x for all x. Table 1.6 summarizes
some of the key features of the sine, cosine, and tangent functions.
The trigonometric functions satisfy many important identities. For example,
the functions secant, cosecant, and cotangent are reciprocals of the functions cosine,
sine, and tangent, respectively. The following box summarizes some of the more
common identities.
Trigonometric Identities
Ratio Identities
tan x =
1
csc x
1
csc x =
sin x
sin x =
sin x
cos x
cot x =
1
sec x
1
sec x =
cos x
cos x =
cos x
sin x
1
cot x
1
cot x =
tan x
tan x =
Pythagorean Identities
sin2 x + cos2 x = 1
1 + tan2 x = sec2 x
1 + cot2 x = csc2 x
28
CHAPTER 1. FUNCTIONS
Table 1.6: Trigonometric Functions
Function
Domain
Range
Period
Graph
1
y = sin x
All reals
[−1, 1]
2π
-2π -π
π
2π
π
2π
π
2π
-1
1
y = cos x
All reals
[−1, 1]
2π
-2π -π
-1
y = tan x
All reals except
± π2 , ± 3π
2 ,...
All reals
π
-2π -π
Due to their periodic nature, the trigonometric functions, particularly sine and
cosine, are useful for modeling phenomena that have cyclical features.
Example 3 Modeling body temperature
Body temperature, like blood pressure and hormone levels, does not remain constant
but rather fluctuates according to the body’s circadian rhythm. For a healthy
subject with a typical sleep pattern, the temperature in degrees Fahrenheit t hours
after midnight is approximated by
f (t) = 0.8 cos(0.2618t + 1.5708) + 98.4
Determine the maximum and minimum body temperatures and the times of day at
which they occur.
Solution Although we could determine the maximum and minimum values, as well
as the times of day, from facts about the cosine function, we will simply use the
graph. From Figures 1.35 and 1.36 we estimate the minimum body temperature to
be 97.6◦ at t = 6 (6 A.M.) and the maximum to be 99.2◦ at t = 18 (6 P.M.).
29
1.3. LIBRARY OF FUNCTIONS
100
100
0
30
X=6.0638298
0
30
Y=97.600112
97
X=18.191489
Y=99.198994
97
Figure 1.35
Figure 1.36
Exponential Functions
Earlier in this section we considered the value of a piece of machinery and noted
that it decreased at an approximately constant rate. Many other quantities increase
or decrease not at a constant rate, but rather at a constant percentage rate. For
example:
• During the 1990s the cost of living increased by about 3% per year.
• The amount of radioactive carbon-14 in the Dead Sea Scrolls is decreasing by
about 0.012% per year.
• The number of Internet hosts grew by about 90% per year in the 1990s.
• The world population in 2000 was 6 billion, and it is growing at about 1.3%
per year.
If a quantity Q = f (t) grows at a constant percentage rate, then f (t) is what is
known as an exponential function. For example, suppose the number of people who
have heard a rumor grows at a constant percentage rate of 100% per hour. This is
just another way of saying that the number of people who have heard the rumor
doubles each hour. Thus, if 1 person starts the rumor, then 2 have heard it after
1 hour, 4 after 2 hours, 8 after 3 hours, and so on, so that after t hours, 2t people
have heard it. The function f (t) = 2t is an exponential function with base 2.
Definition of an Exponential Function Base a
The function f (x) = ax , for a > 0 and a 6= 1, is the exponential function with
base a.
x
The graphs of y = 2x and y = 21 are shown in Figures 1.37 and 1.38. The
shapes of these graphs are characteristic of exponential functions with base a > 1
and 0 < a < 1, respectively. Several important properties of exponential functions
are suggested by these graphs.
30
CHAPTER 1. FUNCTIONS
4
4
3
3
y = 2x
2
y=
1
-4 -3 -2 -1
-1
1 x
2
2
1
1
2
3
4
-4 -3 -2 -1
-1
-2
-2
-3
-3
-4
-4
1
2
3
4
Figure 1.38
Figure 1.37
Properties of Exponential Functions
Let f (x) = ax , a > 0, a 6= 1.
1. The domain of f is the set of all real numbers, and the range of f is the set
of all positive real numbers.
2. The x-axis is a horizontal asymptote for the graph of f .
3. f has y-intercept (0, 1).
4. f is increasing if a > 1 and decreasing if 0 < a < 1.
In the next example, we see how drug levels can be modeled with exponential
functions.
Example 4 Medication in the body
A 4-milligram dose of a certain medication is eliminated from the body in such a
way that the quantity remaining at a given time is 23 of the amount that was there
1 hour earlier. Thus, after t hours, the amount remaining in the body is given by
the function
t
2
g(t) = 4
3
a. How much of the medication remains after 3 hours?
b. At what time will the amount in the body be half the original quantity?
Solution
a. After 3 hours, the amount of medication remaining is given by
3
2
8
g(3) = 4
=4
≈ 1.185 mg
3
27
b. Since 4 milligrams of medication are in the body initially, we are interested in
the time at which 2 milligrams of the substance remain. Thus, we are interested
in the value of t for which
t
2
4
=2
3
Although it is possible to solve this equation algebraically using logarithms, we
will postpone the use of logarithms until a brief review in Section 3.10. Instead
we will simply approximate the solution using a graphing utility. The solution
31
1.3. LIBRARY OF FUNCTIONS
of the equation
corresponds to the x-coordinate of the point on the graph of
x
y = 4 23 with y-coordinate 2. By tracing the graph, we see in Figure 1.39 that
y ≈ 2 when x ≈ 1.7. Thus, after 1.7 hours (roughly 1 hour and 43 minutes), half
the original dose of medication remains in the body.
4
X=1.6914894
Y=2.0146681
0
0
Figure 1.39
Among all of the exponential functions, one arises so frequently that it is called
the natural exponential function. To define the natural exponential function, we
consider the graphs of y = 2x and y = 3x shown in Figures 1.40 and 1.41, and we
3 focus our attention on the tangent lines constructed at (0, 1). We will investigate
tangent lines in great detail later in the text; for now, we simply understand a
tangent line to be a line that just “kisses” a curve at a point. The slopes of the
tangent lines in Figures 1.40 and 1.41 can be shown to be approximately 0.7 and
1.1, respectively. This suggests that there is some base a between 2 and 3 for which
the slope of the tangent line to y = ax at (0, 1) is exactly 1. Such a base does exist,
and it is none other than the irrational number e, the first few decimal places of
which are 2.71828. Since 2 < e < 3, we expect the graph of y = ex to be very
similar in shape to the graphs of y = 2x and y = 3x . This is indeed the case, as we
see in Figure 1.42.
4
4
3
3
m = 0.7
y = 0.7x + 1
2
4
m = 1.1
y = 1.1x + 1
2
1
1
-4 -3 -2 -1
-1
y = ex
y = 3x
y = 2x
1
2
3
4
-4 -3 -2 -1
-1
2
1
1
2
3
4
-4 -3 -2 -1
-1
-2
-2
-2
-3
-3
-3
-4
-4
-4
Figure 1.40
m=1
y =x+1
3
Figure 1.41
1
2
3
4
Figure 1.42
Definition of the Natural Exponential Function
The natural exponential function is f (x) = ex , where e is the irrational number 2.718281 . . ..
Mathematical Models and Regression
A mathematical model is a mathematical description of the behavior of some aspect
of the real world. Mathematical models have been formulated for such diverse
phenomena as signal transmission in the human nervous system, the aerodynamics
of a hummingbird, the spread of AIDS, the deforestation of the Amazon Rain Forest,
the flow of the jet stream, the path of Halley’s Comet, and the behavior of the entire
global economy. Many of the major theories of the physical sciences are, in essence,
mathematical models. Examples include Einstein’s general theory of relativity and
string theory, which model the very fabric of space itself.
Although some sophisticated mathematical models are constructed from underlying principles, many are formed simply by finding a function that fits a given set
of data. Often the form of the function (for example, linear, quadratic, or exponential) is either known or assumed, and the given data are then used to determine the
32
CHAPTER 1. FUNCTIONS
function precisely. This is essentially the technique that was used in Example 2.
By assuming in advance that the depreciation would be linear, we were able to use
two data points to construct a linear model for estimating the value of a piece of
machinery at any time during its usable lifetime. A more general situation is one in
which we are presented with a collection of data, and our goal is to find a function
of a particular form that fits the data “as closely as possible.”
The mathematical process by which this is accomplished is called regression,
and among the various methods for describing exactly what is meant by “as closely
as possible” is the least-squares approach.
Least-Squares Best Fit
Given a collection of data points (x1 , y1 ), (x2 , y2 ), . . . , (xn , yn ) and a function form
f , the least-squares best fit is the function for which the sum of (f (xi ) − yi )2 is as
small as possible.
In cases where the collection of data is small and the function form is simple
(e.g. linear), algebraic formulas can be used to find the least-squares best fit. We
will actually see later in the text how such formulas are obtained. However, it is
usually impractical to find a least-squares best fit by hand. Fortunately, we can
rely on technology in the form of a spreadsheet or CAS (a computer algebra system
is software capable of performing symbolic mathematical computations) to perform
this task for us. Such tools can also easily compute the correlation coefficient,
usually denoted with R. Although a full discussion of the significance of R is
beyond the scope of this text, we will make use of the fact that the closer R2 is to
1, the better the fit.
Example 5 Finding a least-squares best fit
The population of the United States is tabulated every 10 years by way of a national
census. Table 1.7 gives the values that were determined for the census years from
1960 through 2000. Find the least-squares best fit for this data using each of the
following function forms. Decide which of the function forms provides the best fit
and use that function to estimate the population in 2010.
a. Linear: f (x) = ax + b
b. Exponential: f (x) = aebt
Table 1.7: U.S. Census 1960-2000
Year
Population (millions)
1960
179
1970
203
1980
227
1990
249
2000
281
Solution For both models, we will let t denote the number of years after 1960.
1. Using a spreadsheet, we plot the data and find the least-squares best fit line,
shown in Figure 1.43. The equation of the line is y = 2.5x + 177.8, and we
see that R2 = 0.9957, which indicates a very good fit.
2. The least-squares exponential fit, shown in Figure 1.44, yields the equation
y = 180.37e0.0111x and R2 = 0.9977, which is even closer to 1.
We conclude that, although both models are quite good, the exponential model
gives a slightly better fit. To estimate the population in 2010, we set t = 50 to
obtain 180.37e0.0111(50) ≈ 314 million.
33
1.3. LIBRARY OF FUNCTIONS
Figure 1.43
Figure 1.44
1.3 Exercises
Exercises 1–4 Answer true or false.
13. f (x) = sin(2x + 1)
1. Every linear function is also a polynomial function.
14. f (x) =
2. Every polynomial function is also a rational function.
15. f (x) =
3. Every polynomial function is also an algebraic function.
4. It is possible for a function to be both algebraic and
transcendental.
Exercises 5–8 Give an example of each.
16. f (x) =
x−4
5
√
3x2 + πx + e
1
1+2x
Exercises 17–22 Find an equation of the line with the given
properties. Express your final answer in slope-intercept
form, if possible.
17. Passes through (1, 4) and (−2, 3)
18. Has x-intercept 4 and y-intercept 2
5. The formula for a linear function whose output value
increases 2 units whenever the input value increases
3 units
6. The formula for a rational function for which x = 2
is not in the domain
7. The formula for an algebraic function that is not a
rational function
19. Is parallel to y = 4x + 2 and has y-intercept 5
20. Passes through (−2, 5) and is parallel to the line containing (0, 3) and (4, 0)
21. Passes through (2, −4) and is perpendicular to x −
3y = 8
22. Passes through the origin and the point of intersection of 2x + y = 10 and x − 2y = 5
8. The formula for a power function that is not algebraic
Exercises 23–26 For the given data,
Exercises 9–16 Identify all categories to which the given a. perform both linear and exponential regressions and
function belongs. Choose from linear, polynomial, ratiob. decide which seems to provide the better fit.
nal, algebraic, and transcendental.
9. f (x) =
10. f (x) =
x
x2 +1
√
x2 + 1
11. f (x) = x2 + 2x + ex
√
12. f (x) = x2 + 2x + x
23.
x
y
2
5
14
17
2
7
25
36
34
24.
25.
26.
CHAPTER 1. FUNCTIONS
x
y
−2
2
5
8
0.466
2.925
8.108
26.463
x
y
2
4
6
8
10
12
5.595
7.248
10.21
14.105
17.003
22.184
x
y
0
5
10
15
20
25
3
12
38
91
163
248
Applications
27. Computer Depreciation Suppose that a computer
system purchased new in 2010 for $1200 will be
worth $500 in 2013. Assuming linear depreciation—
that is, assuming the value of the computer system
and its age are related linearly—estimate the value
of the computer in 2012.
28. Temperature Conversion The relationship between
degrees Fahrenheit (F ) and Celsius (C) is linear. At
atmospheric pressure, water freezes at 32◦ F and 0◦ C,
whereas water boils at 212◦ F and 100◦ C.
a) Express C as a linear function of F .
b) Room temperature is usually taken to be 72◦ F.
What is room temperature in degrees Celsius?
c) For decades, the average human body temperature was assumed to be 98.6◦ F. (This figure has
since been adjusted to 98.2◦ F.) What is 98.6◦ F
in degrees Celsius?
d) Interpret the slope of the line C = f (F ).
29. Growth Rate Suppose that a child were to grow at
a constant rate from conception to maturity. Thus,
if h represents the height (or length) of the child in
inches and t represents the number of years since
conception, then h and t are related linearly. Assume that at t = 0, h = 0. Assume further that the
child is fully grown at 16 years old, at which time
she is 5 feet 6 inches tall.
a) Express h as a function of t.
b) Use the result of part a to estimate the length
of the baby at birth.
c) Estimate the height of the child at age 12.
d) Interpret the slope of the line h = f (t).
30. Sales Commission Each month a salesman earns a
$2000 salary plus a 10% commission on all sales over
$10,000. Let P be the salesman’s pay and S be his
monthly sales.
a) Express P as a linear function of S (assuming
S > 10000).
b) Use the result of part a to find the salesman’s
total pay in a month when he sold $25,000.
31. Slow Growing Tree According to The Guinness Book
of World Records, the slowest growing tree is a white
cedar located on a cliff side in the Canadian Great
Lakes area. Its mass (in grams) can be modeled by
the linear equation m = 17+0.11t where t represents
the year (with t = 0 corresponding to 2000).
a) Estimate the mass of the tree in the year 3000.
b) Find the slope and y-intercept of the line with
equation m = 17 + 0.11t.
c) Interpret the values you found in part b.
32. Under Pressure Water pressure increases linearly
with depth. At a depth of 500 feet, the water pressure is 232 pounds per square inch. At 1000 feet,
the pressure is 448 pounds per square inch.
a) Express pressure P as a linear function of depth
d.
b) Interpret both the slope and the y-intercept of
the equation you found in part a.
c) Find the pressure at 36,000 feet, the depth of
the Mariana trench in the Pacific Ocean.
33. Bungee Jumping A bungee jumper leaps from a platform 89 feet above the ground. Her height above
the ground after t seconds, for 0 ≤ t ≤ 9, is given
448
16 3
2
t + 272
by h(t) = − 27
27 t − 9 t + 89. Use a graphing utility to determine how close she gets to the
ground before the first rebound and then her maximum height after the first rebound.
34. Digital TV Explosion Digital television exploded
onto the consumer electronics scene at the beginning of the twenty-first century. Based on data
from 1999–2003, we can model the average price
per unit (in dollars) t years after 1999 with the
function f (t) = 2350 − 235t, and the number of
unit sales (in thousands) can be modeled by g(t) =
146t2 + 370t + 125.
a) Neither of the models given in this problem is
valid when applied to years before 1999. In addition, one of the models is clearly invalid from
a certain point on. In which year does one of
35
1.3. LIBRARY OF FUNCTIONS
b)
c)
d)
e)
the models give impossible results, and which
model is it?
Use the functions f and g to find a function h
that models total revenue (in thousands of dollars) for digital televisions t years after 1999.
What is the degree of this polynomial?
Some industry insiders predict that revenue generated from digital television sales will eventually level off, gradually approaching a fixed
value. Explain why this is inconsistent with the
model from part b.
For what time period does the model for total
sales revenues from part b give plausible results?
Use the model from part b (and a graphing
utility) to estimate the year in which the total
revenue from the sales of digital televisions will
peak.
35. Average Temperature The average monthly temperature (in degrees Fahrenheit) for Orlando, Florida,
can be approximated using the function
πt
πt
− 13 sin
f (t) = 73 − 5.5 cos
6
6
where t = 1 corresponds to January 2009, t = 2
corresponds to February 2009, and so on.
a) Compute f (3) and f (15) and interpret your result.
b) Use a graphing utility to estimate the month
when the average temperature is highest and the
month when it is lowest.
36. Canine Blood Pressure The aortic blood pressure
(measured in mm/Hg) of a dog can be approximated
using the function
f (t) = −6.3 cos (42t) + 5 sin (42t)
− 7.8 cos (21t) + 16.2 sin (21t)
+ 115
where t is measured in seconds [Data source: Bjorn
Folkow and Eric Neil, Circulation (New York: Oxford, 1971), p. 428].
π
a) Compute f (0) and f ( 21
) and interpret your result.
b) Use a graphing utility to estimate the highest
and lowest blood pressures.
37. Spring Motion When an object hung from a certain
stretched spring (see Figure 1.45) is released, its displacement in centimeters from the equilibrium (resting) position t seconds after release is approximated
by
π
y(t) = 5 sin πt −
2
a) Find the displacement when t = 0, t = 1, and
t = 2.
b) Use a graphing utility to find the maximum displacement of the object.
Equi
libriu
m
y(t)
Figure 1.45
38. Sunset Times Sunset times in Indianapolis, Indiana,
can be roughly approximated using the function
2π
3π
f (t) = 89 sin
t+
+ 409
365
2
where t is the day of the year and f (t) is the corresponding sunset time, in minutes, after 12:00 P.M.
(noon).
a) Find the sunset time when t = 31, t = 150, and
t = 310.
b) Find the earliest and latest sunset times and estimate the days on which they occur.
39. Ferris Wheel Motion The height in feet of a certain
passenger on a Ferris wheel is given by
πt π
y(t) = 55 + 50 sin
−
15
2
where t is the time in seconds, and t = 0 coincides
with the time at which the wheel was set in motion.
a) Find the initial height of the passenger.
b) Find the maximum and minimum heights of the
passenger and the first times at which those
heights are attained.
c) How long does it take for the Ferris wheel to
make one complete revolution?
40. Weight Fluctuations A certain dieter’s weight has
been found to “yo-yo” according to the formula
π w(t) = 10 cos
t + 200
6
where t represents the number of months that have
elapsed since January 1.
a) What is the greatest weight that the dieter attains in the course of the year? When does the
dieter reach this weight?
36
b) What is the lightest weight that the dieter attains in the course of the year? When does the
dieter reach this weight?
41. Energy Consumption Energy consumption in the
United States during the years 1935–2000 can be
approximated using the exponential function Q(t) =
23.17 · 1.023t , where t denotes the year (t = 0 corresponds to 1935) and Q(t) is the energy consumed
in that year (in quadrillions of Btus). Estimate the
energy consumption in the years 1940 and 2000 and
use these values to determine the percent increase
from 1940 to 2000. What does this model predict
for energy consumption in 2010?
42. Computer Transistors Moore’s law, formulated in
1965 by Gordon Moore, cofounder of Intel, states
that the number of transistors that can be put on a
computer processor chip will approximately double
every 18–24 months. In 1971, Intel introduced the
4004 processor with 2300 transistors. According to
Moore’s law, the number of transistors t years after
1971 can be approximated by N = 2300 · 2t/2 . Use
this function to estimate the number of transistors
on a processor in 2000. The actual number of transistors on the Pentium 4 processor, introduced by
Intel in 2000, was 42 million. How well does your
estimate compare to this value? What does Moore’s
law predict for the number of transistors in the year
2005?
43. Medication Concentration A certain medication is
ingested and begins spreading throughout the bloodstream.
The concentration (in nanograms per
CHAPTER 1. FUNCTIONS
milliliter) of the medication in the bloodstream after
t hours is given by Q(t) = 4te−0.8t . Use a graphing utility to plot this function. Estimate the maximum concentration of the medication and the time
at which it occurs. What happens to the concentration as time increases beyond this point?
44. Internet Growth The following table shows the
approximate number of internet hosts during the
early years of the internet. Find the least-squares
best fit for this data using each of the given function forms. Decide which of the function forms
provides the best fit and use that function to
estimate the number of internet hosts in 2002.
Year
Internet hosts
(thousands)
1991
376
1993
1313
1995
5846
1997
21,819
1999
43,230
a) Linear: f (x) = ax + b
b) Exponential: f (x) = aebx
45. Alaska Population Population data for Alaska for
the census years from 1960 through 2000 are given
in the table. Find the least-squares best fit for this
data using each of the given function forms. Decide
which of the function forms provides the best fit and
use that function to estimate the population in 2010.
Year
Population
(thousands)
1960
226
1970
303
a) Linear: f (t) = at + b
b) Exponential: f (t) = aebt
1980
402
1990
550
2000
627
37
1.4. IMPLICIT FUNCTIONS AND CONIC SECTIONS
1.4 Implicit Functions and Conic Sections
4
3
2
x2
=
y5
1
2
3
+
5y 3
1
-4 -3 -2 -1
-1
-2
-3
-4
Figure 1.46
4
+3
Up to this point, our emphasis has been on functions for which the dependent
variable, often denoted y, has been defined explicitly in terms of the independent
variable, usually x. When given symbolically, explicitly defined functions can be
written as an equation of the form y = f (x), where it is understood that for each
x there is exactly one y. Most of our later work in calculus will also deal with
explicitly defined functions. There are times, however, when the variables x and y
are related by an equation that can’t easily (or at all) be put in the form y = f (x).
An example is the equation x2 = y 5 + 5y 3 + 3. It is not possible to solve this
equation for y in terms of x. However, the graph shown in Figure 1.46 reveals that
for any x there is exactly one y; that is, the graph passes the vertical line test.
There are also many equations involving x and y whose graphs don’t pass the
vertical line test. The circle x2 + y 2 = 25 shown in Figure 1.47 is one such example.
However, even in cases like this, if we consider almost any point on the graph and
zoom in near that point, the graph appears to look like that of a function. See
Figure 1.48 for a zoomed-in view near the point (3, 4). We say that the equation
x2 + y 2 = 25 defines y locally as a function of x.
5
4
3
2
1
-5 -4 -3 -2 -1
-1
-2
-3
-4
-5
x2 + y 2 = 25
4
1 2 3 4 5
3.5
3
Figure 1.47
2
2.5
3
3.5
Figure 1.48
Motivated by examples such as these, we say that an equation involving x and
y implicitly defines y as a function of x if the equation is not written in the form
y = f (x) but y can be viewed locally as a function of x.
4
y 2 = x3 − 8x2 + 16x
Example 1 Investigating an implicit function
Figure 1.49 shows the graph of y 2 = x3 − 8x2 + 16x.
3
2
1
-1
a. Does the graph define y as a function of x?
1
2
3
4
5
b. Find the coordinates of any points at which y is not locally a function of x.
-2
Solution
-3
-4
a. Since this graph fails the vertical line test, it does not define y as a function of
x.
Figure 1.49
b. Locally, y is a function of x except at two points: (0, 0) and (4, 0). No matter
how many times we zoom in on (0, 0), we obtain a curve that violates the vertical
38
CHAPTER 1. FUNCTIONS
line test. At (4, 0), zooming aways results in an x-shaped curve, which violates
the vertical line test.
For the remainder of this section, we will investigate a special category of implicitly defined functions, the conic sections, which include circles, parabolas, ellipses,
and hyperbolas.
Conic Sections
First studied and developed for purely theoretical reasons, the conic sections are
yet another example of what has been referred to as the “unreasonable effectiveness
of mathematics.” Who could have guessed that this collection of abstract curves,
the intellectual offspring of ancient Greeks more philosophers than mathematicians,
would provide the mathematical machinery for communicating across continents,
beaming moving images from Earth to space and back again, guiding spacecraft
to distant planets, and painlessly obliterating kidney stones with little more than
a well-placed whisper? From astronomy to economics, mass communication to
geolocation, modern applications of this ancient subject abound. A few specific
examples are cited in Table 1.8.
Table 1.8
Applied Context
Path of a projectile
Graph of a revenue function
Shape of a satellite dish
Shape of a shock wave lithotripsy reflector
Shape of a whispering room
LORAN navigation
Orbit of a comet
Shape of a telescope mirror
Type of Conic
Parabola
Parabola
Parabola
Ellipse
Ellipse
Hyperbola
Parabola, ellipse, or hyperbola
Parabola, ellipse, or hyperbola
There are several ways to study conic sections. A purely geometric approach
was developed by the ancient Greeks around 350 B.C. They described the conic
sections as curves that are formed by the intersection of a plane with a double cone,
as shown in Figure 1.50.
Circle
Parabola
Ellipse
Hyperbola
Figure 1.50
An algebraic approach to the study of conic sections was made possible after the
development of analytic geometry in the seventeenth century. Using this approach,
the conic sections are described as relations of the form
Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0
39
1.4. IMPLICIT FUNCTIONS AND CONIC SECTIONS
A third approach describes each conic as a locus or set of points satisfying certain
geometric properties.
(x, y)
Circles
r
A circle is defined to be the set of points that are an equal distance (the length of
the radius) from a fixed point (the center). If we denote the center by (h, k) and
the distance from the center to an arbitrary point on the circle (x, y) by r, as in
Figure 1.51, we can apply the distance formula to obtain
(h, k)
r = The distance from (h, k) to (x, y)
p
= (x − h)2 + (y − k)2
Squaring both sides, we arrive at the standard equation of a circle.
Figure 1.51
Standard Equation of a Circle
The standard form for the equation of a circle with radius r and center (h, k) is
(x + 4)2 + (y − 3)2 = 16 8
(x − h)2 + (y − k)2 = r2
7
6
5
4
(−4, 3)
Example 2 Graphing a circle
Graph the circle with equation (x + 4)2 + (y − 3)2 = 16.
4
3
2
Solution We begin by rewriting our equation in standard form.
1
-8 -7 -6 -5 -4 -3 -2 -1
-1
(x − (−4))2 + (y − 3)2 = 42
1
We then simply “read off” h = −4, k = 3, and r = 4. Thus, our circle is centered
at (−4, 3) and has radius 4. The graph is shown in Figure 1.52.
Figure 1.52
Consider the equation (x + 4)2 + (y − 3)2 = 16 from Example 2. If we expand
the terms on the left-hand side, we obtain
(x + 4)2 + (y − 3)2 = 16
x2 + 8x + 16 + y 2 − 6y + 9 = 16
x2 + y 2 + 8x − 6y = −9
Now, in this form, the center and radius of the circle are difficult to identify. Consequently, if we were given the task of graphing the equation x2 + y 2 + 8x − 6y = −9,
our first step would be to write it in standard form by reversing the squaring-out
process just shown. This is done by completing the square, which we illustrate
in the following example.
Example 3 Completing the square to graph a circle
Graph the circle with equation x2 + y 2 − 2x + 4y − 20 = 0.
4
3
2
1
-4 -3 -2 -1
-2
-3
-4
-5
-6
-7
Solution To find the center and radius, we must write the equation in standard
form. To do this, we complete the square in both x and y as follows:
x2 − 2x + y 2 + 4y = 20
1 2 3 4 5 6 7
(1, −2)
5
(x − 1)2 + (y + 2)2 = 25
Figure 1.53
2
(x − 2x +
2
) + (y + 4y +
) = 20
(x2 − 2x + 1 ) + (y 2 + 4y + 4 ) = 20 + 1 + 4
(x − 1)2 + (y + 2)2 = 25
Grouping the x and y terms
Preparing to complete the square in x and
y
Adding 1 and 4 to both sides to complete
the square in x and y
Expressing in standard form
Thus, the circle is centered at (1, −2), with radius 5. Its graph is shown in Figure
1.53.
40
CHAPTER 1. FUNCTIONS
Parabolas
Parabola
As suggested by Figure 1.54, a parabola is a curve that is formed when a cone is
cut with a plane parallel to the side of the cone. If we restrict our attention to
parabolas in the xy-coordinate system, parabolas can be described algebraically as
equations in x and y. We will consider two types of parabolas: those that open up
or down, such as y = x2 , and those that open right or left, such as x = y 2 , as shown
in Figures 1.55 and 1.56.
Figure 1.54
4
4
3
3
y = x2
2
x = y2
2
1
1
-4 -3 -2 -1
-1
1
2
3
4
-4 -3 -2 -1
-1
-2
-2
-3
-3
-4
-4
Figure 1.55
1
2
3
4
Figure 1.56
Axis of symmetry
The lowest point on a parabola that opens upward, or the highest point on a
parabola that opens downward, is called the vertex. A parabola that opens upward
or downward has a vertical axis of symmetry through its vertex, as shown in Figure
1.57. If a parabola opens left or right, its vertex is the point furthest to the right
or left, respectively, and it has a horizontal axis of symmetry through its vertex,
as we see in Figure 1.58. The standard equations of parabolas with vertical and
horizontal axes follow.
Axis of symmetry
Vertex
Vertex
Figure 1.57
Figure 1.58
41
1.4. IMPLICIT FUNCTIONS AND CONIC SECTIONS
Standard Equations Of A Parabola
The standard forms for the equation of a parabola with vertex (h, k) are given
below.
y − k = a(x − h)2
(vertical axis of symmetry)
2
(horizontal axis of symmetry)
x − h = a(y − k)
A parabola with a vertical axis opens upward if a > 0 and downward if a < 0.
A parabola with a horizontal axis opens to the right if a > 0 and to the left if
a < 0.
4
3
(−1, 2)
2
1
-4 -3 -2 -1
-1
1
2
3
4
-2
x = −1 -3
Example 4 Graphing a parabola
Graph the parabola with equation y − 2 = (x + 1)2 , identify the vertex, and find
the equation of the axis of symmetry.
-4
Solution The vertex is the point (−1, 2), and the axis of symmetry is the vertical
line x = −1. The graph is shown in Figure 1.59.
Figure 1.59
Example 5 Graphing a parabola
Graph the parabola with equation x = −2y 2 − 12y − 16, identify the vertex, and
find the equation of the axis of symmetry.
Solution Even though the equation is not in standard form, we can tell that the
axis is horizontal because of the y 2 term. To locate the vertex, we must put the
equation in standard form. This is done by completing the square.
4
3
x = −2y 2 − 12y − 16
2
x + 16 = −2y 2 − 12y
1
-4 -3 -2 -1
-1
y = −3
1
2
-2
-3
3
4
(2, −3)
x + 16 = −2(y 2 + 6y +
Adding 16 to both sides
)
x + 16 − 18 = −2(y 2 + 6y + 9 )
x − 2 = −2(y + 3)
Preparing to complete the square on the
y terms by first factoring out −2
Adding 9(−2) = −18 to both sides
2
-4
x − 2 = −2(y + 3)2
From the standard form, we see that the vertex is the point (2, −3). The axis of
symmetry is the horizontal line y = −3. The graph is shown in Figure 1.60.
Figure 1.60
Equations of parabolas are often given in the general form y = ax2 + bx + c or
x = ay 2 +by+c, as was the parabola in Example 5. In such instances, it is convenient
to have a simple technique for finding the vertex of the parabola without completing
the square. As we will see later, a tool from calculus called the derivative will be
useful for the task of finding a vertex. For now, in cases where an approximation
for the vertex is sufficient, we will simply use a graphing utility.
Example 6 Finding the maximum height of a human cannonball
The muzzle velocity of the renowned human cannonball Emanual Zacchini was
estimated to be as much as 36 feet per second (Source: “Guinness Book of World
Records”). If we assume that this muzzle velocity was possible with the cannon
pointed straight up and if we assume that Zacchini’s initial height was 5 feet, then
his height in feet after t seconds would have been given by s = −16t2 + 36t + 5.
Find Zacchini’s maximum height given these assumptions.
Solution The maximum height will occur at the vertex of the parabola s = −16t2 +
36t + 5. Plotting the graph of this equation with a graphing utility produces
42
CHAPTER 1. FUNCTIONS
30
the graph shown in Figure 1.61. The approximate coordinates of the vertex are
(1.12, 25.25). Consequently, the maximum height is 25.25 feet.
The equation of a parabola can be determined from its graph if three pieces of
information are known: the orientation of the axis of symmetry, the location of the
vertex, and the location of one additional point on the graph.
X=1.1170213 Y=25.248981
0
0
Figure 1.61
3
Example 7 Finding the equation of a parabola
Find the standard equation of the parabola that has a vertical axis of symmetry,
vertex (1, −2), and passes through the point (6, 3).
Solution Since the parabola has a vertical axis of symmetry and vertex (1, −2), its
equation must have the form
y + 2 = a(x − 1)2
for some value of a. To find the correct value for a, we use the fact that the parabola
passes through (6, 3). Thus, we set x = 6 and y = 3 and solve for a.
3 + 2 = a(6 − 1)2
5 = 25a
1
=a
5
Thus, the equation of the parabola is
y+2=
Ellipse
1
(x − 1)2
5
Ellipses and Hyperbolas
An ellipse is a curve that is formed when a cone is cut with a plane that is not
parallel to the side of the cone, as shown in Figure 1.62. Just as with parabolas,
if we restrict our attention to the xy-coordinate system, ellipses can be described
algebraically as equations in x and y. The standard form for the equation of an
ellipse centered at the origin and with orientation parallel to one of the coordinate
axes is given below.
Figure 1.62
Standard Equation Of An Ellipse Centered At The Origin
x2
y2
+ 2 =1
2
a
b
The graph of an ellipse centered at the origin can be obtained by first locating
the x- and y-intercepts and then connecting the intercepts with a smooth curve.
To find the x-intercepts, we set y = 0 and solve for x. This yields x = ±a. The
y-intercepts are found by setting x = 0 and solving for y. We obtain y = ±b. If
a > b, the graph will have a horizontal orientation, such as the one shown in Figure
1.63. If b > a, the graph will have a vertical orientation, such as the one shown in
Figure 1.64.
43
1.4. IMPLICIT FUNCTIONS AND CONIC SECTIONS
x2
52
+
(−5, 0)
y2
32
=1
5
4
(0, 3)
3
2
1
x2
22
(5, 0)
-5 -4 -3 -2 -1
1 2 3 4 5
-1
-2
-3
(0, −3)
-4
-5
Figure 1.63
+
y2
42
5
(0, 4)
4
3
2
1
(2, 0)
=1
(−2, 0)
-5 -4 -3 -2 -1
1 2 3 4 5
-1
-2
-3
-4
(0, −4)
-5
Figure 1.64
The major axis is the longer of the two line segments connecting the intercepts,
the vertices are the endpoints of the major axis, and the minor axis is the line
segment connecting the intercepts that are not vertices. Thus, the ellipse in Figure
1.63 has a horizontal major axis with vertices (5, 0) and (−5, 0) and a minor axis
connecting (0, −3) and (0, 3). The ellipse in Figure 1.64 has a vertical major axis
with vertices (0, 4) and (0, −4) and a minor axis connecting (−2, 0) and (2, 0).
Example 8 Graphing an ellipse centered at the origin
Write the equation of the ellipse 9x2 + 4y 2 = 36 in standard form, sketch its graph,
and identify the vertices.
Solution To write the equation in standard form, we divide through by 36 as follows.
2
x
4
+
y2
9
4
=13
(0, 3)
9x2 + 4y 2 = 36
4y 2
9x2
+
=1
36
36
x2
y2
+
=1
4
9
x2
y2
+
=1
22
32
2
1
-4 -3 -2 -1
-1
1
2
3
4
-2
-3
-4
(0, −3)
Thus, a = 2 and b = 3. The graph is shown in Figure 1.65. Since the major axis is
vertical, the vertices have coordinates (0, −3) and (0, 3).
Figure 1.65
Hyperbola
A hyperbola is a curve that is formed when a double cone is cut by a plane
parallel to the axis of the cone, as shown in Figure 1.66. We will restrict our
attention to hyperbolas in the xy-coordinate system and, in particular, to those
that have horizontal or vertical axes, such as the ones shown in Figures 1.67 and
1.68.
The standard forms for the equations of hyperbolas centered at the origin are
as follows.
Standard Equations Of Hyperbolas Centered At The Origin
Figure 1.66
x2
y2
−
=1
a2
b2
2
2
y
x
− 2 =1
2
b
a
(horizontal axis)
(vertical axis)
44
CHAPTER 1. FUNCTIONS
The vertices of a hyperbola centered at the origin are its intercepts. For example,
given a hyperbola with a horizontal axis and equation
y2
x2
− 2 =1
2
a
b
we locate the x-intercepts by setting y = 0 to obtain x = ±a. Thus, the vertices are
(a, 0) and (−a, 0), as shown in Figure 1.67. There are no y-intercepts since setting
x = 0 leads to nonreal solutions. A hyperbola with a vertical axis has vertices
(0, −b) and (0, b), as seen in Figure 1.68. The line segment connecting the vertices
is called the transverse axis.
x2
a2
−
y2
b2
y2
b2
=1
−
x2
a2
=1
(0, b)
(−a, 0)
(a, 0)
Transverse axis
(0, −b)
Transverse
axis
Figure 1.68
Figure 1.67
Notice in Figures 1.69 and 1.70 that the graphs of both hyperbolas approach
the lines y = (b/a) x and y = − (b/a) x as x gets large. These lines are called
asymptotes, and they serve as useful aids in sketching graphs of hyperbolas. The
asymptotes for a hyperbola centered at the origin can be located by constructing
a rectangle with sides passing through the points (−a, 0), (a, 0), (0, b), and (0, −b).
The extended diagonals of this rectangle are the asymptotes; their equations are
y = (b/a) x and y = − (b/a) x.
y = − ab x
y=
y2
b2
b
x
a
−
x2
a2
=1
(0, b)
(0, b)
(−a, 0)
(−a, 0)
(a, 0)
(0, −b)
x2
a2
−
(a, 0)
2
y y = −bx
b2 = 1 a
Figure 1.69
y=
b
x
a
(0, −b)
Figure 1.70
Asymptotes Of A Hyperbola Centered At The Origin
The asymptotes for a hyperbola centered at the origin are
y=
b
b
x and y = − x
a
a
45
1.4. IMPLICIT FUNCTIONS AND CONIC SECTIONS
Example 9 Graphing a hyperbola centered at the origin
Sketch the graph of the hyperbola
y = − 43 x 6
y=
3
x
4
4
2
(−4, 0)
-6
(4, 0)
-4
-2
2
4
6
y2
9
=1
-2
-4
-6
x2
16
−
x2
y2
−
=1
16
9
Identify the asymptotes, vertices, and transverse axis.
Solution From the
we know that the hyperbola has a horizontal
√
√ form of the equation,
axis with a = 16 = 4 and b = 9 = 3. Thus, the vertices are the x-intercepts
(−4, 0) and (4, 0), and the transverse axis is the line segment connecting these two
points. To find the asymptotes, we construct the rectangle whose sides pass through
the vertices (−4, 0) and (4, 0) on the x-axis and the points (0, −3) and (0, 3) on the
y-axis. The asymptotes are then formed by extending the diagonals of the rectangle.
The equations of the asymptotes are y = 34 x and y = − 34 x. The graph is shown in
Figure 1.71.
Figure 1.71
The standard equations of ellipses and hyperbolas centered at a point (h, k)
can be derived by applying a translation to the equations centered at the origin. In
order to move the center from (0, 0) to (h, k), we must translate h units horizontally
and k units vertically. This is accomplished by replacing x with x − h and y with
y − k. The resulting equations are as follows.
Standard Equation Of An Ellipse Centered At (h, k)
(x − h)2
(y − k)2
+
=1
a2
b2
Standard Equations Of A Hyperbola Centered At (h, k)
(y − k)2
(x − h)2
−
=1
2
a
b2
(y − k)2
(x − h)2
−
=1
b2
a2
(horizontal axis)
(vertical axis)
The graph of an ellipse or hyperbola centered at (h, k) can be obtained by
viewing (h, k) as the origin of an imaginary coordinate system and then applying
the process used for an ellipse or hyperbola centered at the origin.
Example 10 Graphing an ellipse centered away from the origin
Graph the ellipse with equation
(x − 2)2
(y + 3)2
+
=1
25
4
1
-3 -2 -1
-1
-2
-3
-4
-5
(−3, −3)
1 2 3 4 5 6 7
(2, −3)
(7, −3)
(x−2)2
25
+
Figure 1.72
(y+3)2
4
=1
Identify the center, the major axis, and the vertices.
√
Solution This ellipse has center (2, −3). Since a = 25 = 5, two points on the
√ellipse
are located 5 units to the left and right of the center. Similarly, since b = 4 = 2,
two points are located 2 units above and below the center. Connecting these points
with a smooth curve yields the graph in Figure 1.72. The major axis is horizontal
and connects the vertices (−3, −3) and (7, −3).
46
CHAPTER 1. FUNCTIONS
Example 11 Graphing a hyperbola centered away from the origin
Write the equation of the hyperbola 16x2 − y 2 + 64x − 2y + 67 = 0 in standard form
and sketch its graph.
Solution In order to put the equation in standard form, we must complete the square
in x and y.
16x2 − y 2 + 64x − 2y + 67 = 0
(16x2 + 64x +
2
16(x + 4x +
) + (−y 2 − 2y +
) = −67
2
) = −67
) − (y + 2y +
2
2
16(x + 4x + 4 ) − (y + 2y + 1 ) = −67 + 64 − 1
2
4
3
(y+1)2
4
−
(x+2)2
1/4
16(x + 2) − (y + 1)
(y + 1)2
−4(x + 2)2 +
4
(y + 1)2
− 4(x + 2)2
4
(y + 1)2
(x + 2)2
−
4
1/4
2
(x + 2)2
(y + 1)
−
2
2
2
(1/2)
=1
2
1
-4 -3
-1
-1
1
2
-2
-3
-4
3
4
2
Adding 16(4) and −(1) to both sides
= −4
=1
Dividing both sides by −4
=1
=1
=1
The center of the hyperbola is at (−2, −1). The axis is vertical with a = 21 and
b = 2. The vertices are 2 units above and below the center, at (−2, −3) and (−2, 1).
To sketch the hyperbola, we construct a rectangle with center (−2, −1), horizontal
sides 2 units above and below the center, and vertical sides 12 units to the left and
to the right of the center. The asymptotes are formed by extending the diagonals,
as shown in Figure 1.73.
Figure 1.73
1.4 Exercises
Exercises 1–6 Answer true or false.
Exercises 7–10 For the given graph,
1. The graph of x = ay 2 + by + c is a parabola with a
a. determine if y can be defined as a function of x and,
horizontal axis of symmetry.
if not,
2. The vertex of a parabola lies on its axis of symmetry.
3. Any ellipse centered at the origin is symmetric with b. find the coordinates of any points at which y is not
locally a function of x.
respect to the origin.
4. The points (±a, 0) and (0, ±b) lie on the graph of
the hyperbola with equation
7.
x2
y2
−
=1
a2
b2
4
3
2
5. If an equation defines y implicitly as a function of
x, then the graph of the equation passes the vertical
line test.
6. If an equation defines y implicitly as a function of
x and defines y locally as a function of x at a point
P , then if we zoom in close enough on P , the graph
passes the vertical line test near P .
1
-4 -3 -2 -1
-1
-2
-3
-4
1 2 3 4
47
1.4. IMPLICIT FUNCTIONS AND CONIC SECTIONS
22. Center (−3, −6) and passes through the point
(−3, 0)
23. A diameter with endpoints (−4, 2) and (2, 8)
24. A diameter with endpoints (−3, 6) and (5, −6)
8.
4
3
2
1
-4 -3 -2 -1
-1
1 2 3 4
-2
-3
-4
9.
4
3
2
1
-4 -3 -2 -1
-1
1 2 3 4
-2
Exercises 25–32 Find the vertex of the parabola and sketch
its graph.
25.
26.
27.
28.
29.
30.
31.
32.
y = 2x2
x = −4y 2
y − 3 = 3(x + 5)2
x + 4 = −(y − 2)2
x2 + y − 6x + 10 = 0
y 2 + x − 6y + 10 = 0
x = 2y 2 + 12y + 2
y = −x2 − x + 1
Exercises 33–38 Find the equation of the parabola with the
given properties.
-3
-4
10.
4
3
2
1
-4 -3 -2 -1
-1
1 2 3 4
-2
-3
-4
Exercises 11–18 Find the center and radius of the circle
and sketch its graph.
Exercises 39–48 Sketch the graph of the ellipse and identify its center, vertices, and major and minor axes.
11. x2 + y 2 = 25
12. x2 + y 2 = 10
2
33. Vertical axis of symmetry, vertex (0, 0), and passes
through (−1, 2)
34. Horizontal axis of symmetry, vertex (0, 0), and
passes through (4, 3)
35. Horizontal axis of symmetry, vertex (−3, 4), and
passes through (−5, 0)
36. Vertical axis of symmetry, vertex (−2, −4), and
passes through (0, −6)
37. Vertical axis of symmetry and passes through (0, 0),
(−1, 3), and (4, 8)
38. Horizontal axis of symmetry and passes through
(0, 0), (−2, −10), and (2, 2)
39.
2
13. (x + 4) + (y − 1) = 9
x2
9
+
2
y2
25
y2
9
=1
15. x2 − 4x + y 2 + 6y − 3 = 0
40. x16 +
=1
2
41. 4x + 9y 2 = 36
42. 16x2 + 9y 2 = 144
16. x2 + y 2 + 6x + 6y + 8 = 0
43.
14. (x − 3)2 + (y + 1)2 = 16
17. 4x2 + 12x + 4y 2 + 8y = 3 (Hint: First divide through
by the coefficient of x2 .)
18. 9x2 + 9y 2 + 36x − 12y + 31 = 0 (Hint: First divide
through by the coefficient of x2 .)
Exercises 19–24 Find the equation of the circle with the
given properties.
19. Center (−1, 3) and radius 4
20. Center (4, 0) and radius 2
21. Center (4, −3) and passes through the origin
44.
45.
46.
47.
48.
(x+3)2
4
(x−2)2
25
2
(y−1)2
16
(y+2)2
+ 9
2
+
=1
=1
9x + 4y − 18x − 24y + 9 = 0
25x2 + 16y 2 − 200x − 32y + 16 = 0
x2 + 36y 2 + 4x − 72y + 4 = 0
16x2 + y 2 + 4y − 12 = 0
Exercises 49–58 Sketch the graph of the hyperbola and
identify its center, vertices, transverse axis, and asymptotes.
49.
x2
9
−
y2
16
=1
48
50.
CHAPTER 1. FUNCTIONS
y2
25
x2
16
−
69. Cliff Diving The highest regularly performed dives
are off La Quebrada in Acapulco, Mexico (Source:
“Guinness Book of World Records”). The take-off
point is 87.5 feet above the water, and because of
the presence of base rocks, the divers must land 27
feet out from the take-off point (see Figure 1.74).
=1
2
2
51. 4y − 9x = 36
52. 25x2 − 16y 2 = 400
53.
54.
(x−1)2
4
2
(y+3)
16
2
−
−
(y−4)2
9
(x+2)
25
2
2
=1
=1
55. 4x − 9y − 16x + 54y − 101 = 0
56. 25x2 − 16y 2 − 50x + 160y + 25 = 0
57. 4x2 − 25y 2 − 32x + 164 = 0
58. x2 − 4y 2 + 4x − 24y − 36 = 0
Exercises 59–66 Identify the conic section as either a circle, parabola, ellipse, or hyperbola.
a) Assuming the path of a diver is a parabola with
the vertex located at the take-off point, choose
a coordinate system with the origin at water
level directly below the take-off point and find
an equation for the path.
b) Suppose that 12 feet below the take-off point,
a large rock juts out 9 feet from the vertical
line passing through the take-off point. By what
horizontal distance will the diver clear the rock?
59. 2x2 − x − 3y + 4 = 0
9 ft
60. 3x2 + 2y 2 = 7x − 7y − 5
61. 4x2 + 4y 2 = 2x − 6y + 10
12 ft
62. 4y 2 − 9x + 23y − 9 = 0
63. −5x2 − 11x = y 2
115 ft
64. x2 − 2y 2 + 5 = 0
65. −2x2 + 4y 2 − 15x + 8 = 0
66. −3x2 + 2 = 3y 2 − 5y
Applications
30 ft
67. Path of a Ball A ball is thrown at an angle of 45◦
and with an initial velocity of 80 feet per second. Its
path is parabolic and satisfies the equation
y=−
x2
+x
200
where y is the height in feet when the ball is a horizontal distance of x feet from where it was thrown.
Find the maximum height of the ball and the total horizontal distance it travels before hitting the
ground.
68. Stunt Car The stunt coordinator for a movie is planning a stunt with a car speeding off the top floor of
a parking garage. It can be shown that the path of
the falling car will be parabolic with an equation of
the form
16
y = h − 2 x2
v
where h is the height of the garage in feet, v is the
velocity of the car in feet per second, and y is the
height of the car in feet when its horizontal distance
away from the edge of the garage is x feet. If the
height of the garage is 100 feet and the car must land
200 feet from the base of the garage, determine the
required velocity of the car.
Figure 1.74
70. Suspension Bridge When a cable is suspended between two towers of a suspension bridge in such a
way that it bears a uniform load, the shape of the
cable will be parabolic.
a) Find the equation of the shape of the cable if
the towers are 200 feet apart and 40 feet high
and the cable touches the road midway between
the towers (see Figure 1.75).
b) Use your equation to determine the total length
of cable needed for the vertical sections of cable
spaced at 20-foot intervals between the towers.
y
40 ft
x
100 ft
Figure 1.75
49
1.4. IMPLICIT FUNCTIONS AND CONIC SECTIONS
71. London Eye Rising 450 feet above the Thames River,
the London Eye, also called the Millennium Wheel,
was the world’s largest Ferris wheel in 2004. The
outer rim is about 440 feet in diameter (see Figure
1.76).
a) Set up a coordinate system with the origin directly below the center of the wheel and find an
equation for the circular edge of the wheel.
b) What is the horizontal distance from the center
of the wheel at the instant a passenger is 300
feet high?
440
230
The London Eye
Figure 1.76
72. Earth’s Orbit The Earth’s orbit is elliptical with
major axis length 186 million miles and minor axis
length 185.8 million miles. Write an equation for the
path of the orbit, assuming the center is located at
the origin. It may surprise you to find out the Sun is
not at the center of the orbit. The Sun is, in fact, on
the major axis, a distance of 4.3 million miles from
the center. Sketch a graph of the orbit of the Earth
and locate the Sun on the major axis.
73. Mars’ Orbit The orbit of Mars is elliptical with major axis length 283.5 million miles and minor axis
length 278.5 million miles. Write an equation for
the path of the orbit, assuming the center is located
at the origin. The Sun is on the major axis, a distance of 26.5 million miles from the center. Sketch
a graph of the orbit of Mars and locate the Sun on
the major axis.
50
CHAPTER 1. FUNCTIONS
1.5 Polar Functions
Thus far, we have specified points in the plane using rectangular coordinates.
With this system, the location of a point is determined by its x- and y-coordinates.
The x-coordinate tells us how far over, and the y-coordinate tells us how far up
or down we must move from the origin in order to reach the point. In the polar
coordinate system, by contrast, the position of a point is specified by r- and θcoordinates; r tells us how far out from the origin we must travel, and θ tells us in
what direction.
Polar Coordinates
P (r, θ)
r
θ
O
Polar axis
Pole
The polar coordinate system is defined in terms of a fixed point O, called the origin
or pole, and a half-line or ray with endpoint O, called the polar axis. To each point
P in the plane, we assign polar coordinates (r, θ), where r is the directed distance
from O to P and θ is the directed angle between the polar axis and the segment OP
(Figure 1.77). Thus, for example, the point P with rectangular coordinates (0, 3)
can be expressed as 3, π2 in polar coordinates since P lies 3 units from the origin
and π2 is the angle made between the polar axis and OP , as shown in Figure 1.78.
Figure 1.77
Polar Coordinates Notation
The point P lying a directed distance of r units from O such that θ is the directed
angle from the polar axis to OP is said to have polar coordinates (r, θ).
3 P
Rectangular: (0, 3)
Polar: 3, π2
2
1
O
π
2
1
2
Figure 1.78
3
There is an important distinction between rectangular and polar coordinates.
Although there is only one way of specifying a point in rectangular coordinates,
there are infinitely many ways of specifying a point in polar coordinates. This
ambiguity arises from two different sources. First of all, the angle θ associated with
a point P is defined as a directed angle from the polar axis to the segment OP ,
which says that θ is an angle that, in standard position, has terminal side OP . But
as we have already seen, infinitely many angles are coterminal with a given angle.
In the case of the point P (0, 3) in Figure 1.78, not only can θ be taken to be π2 , it
3π
can also be taken to be 5π
2 (Figure 1.79), − 2 (Figure 1.80), and so on. Nowhere
is this ambiguity more evident than at the pole itself, which has polar coordinates
(0, θ) for any value of θ. In other words, at the pole, θ can be anything!
A second source of ambiguity follows from the fact that r is a directed distance
(as opposed to a “distance”) and so may be negative. For negative values of r, the
distance is measured along the line segment formed by extending the terminal side
of θ to the opposite side of O. (This is like describing a point 2 miles to the north
as being −2 miles to the south.) Figure 1.81 shows the point P (0, 3) described in
this way.
51
1.5. POLAR FUNCTIONS
P 3,
3
5π
2
3
2
1
5π
2
O
1
2
3
2
3
2
1
1
1
2
3
O
P −3, − π2
1
2
3
− π2
Figure 1.79
1
2
O
− 3π
2
P 3, − 3π
2
Figure 1.80
Figure 1.81
3
When working with the polar coordinate system, it is sometimes convenient to
use a grid of concentric circles with the pole at the common center, the polar axis
directed to the right, and line segments radiating from the pole at regular angular
intervals, as shown in Figure 1.82.
Figure 1.82
Example 1 Plotting points in polar coordinates
Plot the indicated point.
a. 2, π6
b. 3, − 2π
c. − 52 , 3π
3
4
Solution The points are shown in Figures 1.83–1.85.
1
2
(2,
π
)
6
3
4
1
2
3
4
1
2
Figure 1.84
4
(− 25 , 3π
)
4
)
(3, − 2π
3
Figure 1.83
3
Figure 1.85
Polar Curves
A polar equation is an equation relating r and θ. The graph of a polar equation
consists of the collection of all points (r, θ) such that r and θ satisfy the equation.
A polar function is defined by a polar equation of the form r = f (θ). Just as the
simplest and most useful categories of rectangular functions have been given special
names and studied in great detail (polynomial, rational, exponential, logarithmic,
and so forth), so too have many polar functions been classified and explored in
great detail. There are polar functions bearing names such as lemniscate, cardioid,
limaçon, spiral, and rose. We will avoid an in-depth exploration of these exotic
curves; instead, our focus will be on general techniques that apply to all polar
functions.
52
CHAPTER 1. FUNCTIONS
Example 2 Sketching a polar curve
Sketch the graph of r = 4.
r=4
1
2
3
Solution Here θ is unrestricted, and so we are interested in the collection of points
for which r, the distance to the origin, is 4. This is just a circle centered at the
origin with radius 4, as shown in Figure 1.86.
4
Example 3 Sketching a polar curve
Sketch the graph of θ = π3 .
Figure 1.86
Solution In this case, r is unrestricted, and θ is fixed at π3 . Thus, we are looking for
the set of points for which θ is π3 . By considering both positive and negative values
of r, we obtain the line shown in Figure 1.87.
θ=
1
2
3
π
3
In each of the previous two examples, we were able to sketch the graph by
using a straightforward analysis of the polar equation. However, in cases where the
equation is complex and unfamiliar, this may not be possible. In such cases, we
may be able to shed some light on the graph by plotting a few points, as we do in
the following examples.
4
Example 4 Sketching a polar curve
Sketch the graph of r = θ for θ ≥ 0.
Figure 1.87
Solution We begin by making a table of values.
θ
0
r
0
π
3
π
3
≈ 1.05
π
2
π
2
π
≈ 1.57
π ≈ 3.14
3π
2
3π
2
≈ 4.71
2π
3π
4π
2π ≈ 6.28
3π ≈ 9.42
4π ≈ 12.57
Plotting these points and connecting them with a smooth curve, we obtain the
spiral shown in Figure 1.88.
r=θ
3
6
9 12
Example 5 Sketching a polar curve
Sketch the graph of r = 2 sin θ.
Solution We begin by making a table of values for 0 ≤ θ ≤ π and plotting the
corresponding points (Figure 1.89). By connecting the points with a smooth curve,
we obtain the graph shown in Figure 1.90.
Figure 1.88
θ
0
π
6
r = 2 sin θ
0
1
π
π
4
√
3
√
2
π
2
3
2
2π
3π
√3
√4
3
2
5π
6
π
1
0
Note that the graph appears to be a circle (see Exercise 55 for the first step in
the confirmation of this). Moreover, if θ > π or θ < 0, the corresponding value for
r would simply yield a point on the existing graph. For example, if θ = 7π
6 , then
π
r = −1, and the point (−1, 7π
)
is
equivalent
to
(1,
).
6
6
53
1.5. POLAR FUNCTIONS
r = 2 sin θ
1
2
1
Figure 1.89
2
Figure 1.90
Some graphs can be fairly difficult to plot, in part because of the complications
arising from negative r values. The next example illustrates the process by which
a complicated polar graph can be plotted.
Example 6 Sketching a polar curve
Sketch the graph of r = 3 cos 2θ.
Solution Because of the complexity of producing a plot of this graph, we form it in
several stages. We begin by plotting points for convenient values of θ between 0
and π4 ; connecting the points with a smooth curve gives the graph shown in Figure
1.91.
θ
0
r = 3 cos 2θ
3
π
12
√
3 3
2
≈ 2.60
π
8√
3 2
2
≈ 2.12
π
6
3
2
Next, we consider points for which θ lies between
θ
π
4
π
3
r = 3 cos 2θ
0
− 32 = −1.5
3π
8 √
−322
≈ −2.12
π
4
= 1.5
π
4
and
0
π
2.
5π
12 √
−323
π
2
≈ −2.60
−3
Notice that each of these points has a negative r value. Thus, even though the
angles θ all have terminal sides in the first quadrant, the corresponding points lie in
the third quadrant, as shown in Figure 1.92. If this process is continued to include
all values of θ between 0 and 2π, we obtain the graph shown in Figure 1.93, called
a four-petaled rose.
54
CHAPTER 1. FUNCTIONS
1
2
3
1
Figure 1.91
2
3
1
Figure 1.92
2
3
Figure 1.93
For complicated polar graphs, we will rely on a graphing utility.
Example 7 Plotting a polar curve with a graphing utility
Use a graphing utility to plot the graphs of r = 2 cos θ and r = 1 + 2 sin θ. Estimate
the points of intersection.
Solution Using [0, 2π] as an interval for θ, we select an appropriate viewing window
and zoom in to obtain the graph shown in Figure 1.94. The graph of r = 2 cos θ is
a circle of radius 1 centered at (1, 0). The graph of r = 1 + 2 sin θ is known as a
limaçon with an inner loop.
One of the intersection points is clearly at the pole. The other two can be
estimated by tracing the graph, as shown in Figures 1.95 and 1.96.
4
-3
4
3
4
-3
3
R=1.8477591
T=.39269908
-3
3
R=.76536686
T=1.1780972
-2
-2
-2
Figure 1.94
Figure 1.95
Figure 1.96
Conversion
At times, it is desirable to convert from one coordinate system to another. The following relationships between rectangular and polar coordinates are easily obtained
by considering Figure 1.97 for the case that θ is acute and r is positive. With little
effort, it can be shown that these relationships hold for all θ.
55
1.5. POLAR FUNCTIONS
Coordinate Relationships
(x, y)
The rectangular coordinates (x, y) and polar coordinates (r, θ) of a point P are
related as follows:
x = r cos θ
r
θ
x
Figure 1.97
y
2
2
r =x +y
2
y = r sin θ
y
tan θ = , x 6= 0
x
(1.1)
(1.2)
Equations 1.1 enable us to find the rectangular coordinates of a point if the
polar coordinates are known.
Example 8 Converting from polar to rectangular coordinates
Convert the given point from polar to rectangular coordinates.
a. (2, π6 )
b. 3, − 2π
3
Solution
a. We have r = 2 and θ =
π
6.
Thus,
π
x = 2 cos
√6
3
=2·
√ 2
= 3
y = 2 sin
=2·
π
6
1
2
=1
and so the rectangular coordinates are
√
3, 1 .
b. With r = 3 and θ = − 2π
3 , we have
2π
y = 3 sin −
3
√ !
3
=3· −
2
√
3 3
3
=−
=−
2
2
√ Thus, the rectangular coordinates are − 32 , − 3 2 3 .
2π
x = 3 cos −
3
1
=3· −
2
It is slightly more difficult to convert from rectangular to polar coordinates because of the nonuniqueness of polar coordinates. The value for r can be determined
using the equation r2 = x2 + y 2 . An angle θ can be found by using the equation
tan θ = xy and knowledge of the quadrant in which θ lies.
Example 9 Converting from rectangular to polar coordinates
Convert the given point from rectangular to polar coordinates.
a.
1 1
3, 3
Solution
√ b. −2, 2 3
c. (−1, −3)
56
CHAPTER 1. FUNCTIONS
a. We begin by finding r. For convenience, we take r to be positive.
p
r = x2 + y 2
s 2
2
1
1
=
+
3
3
r
√
2
2
=
=
9
3
From the fact that tan θ = xy , we have
tan θ =
1/3
=1
1/3
Thus, θ is an angle in the first quadrant with
equal to 1, and so we take
√ tangent
θ = π4 . The point in polar coordinates is 32 , π4 .
b. We find r and tan θ as in part a.
q
r=
=
√
√
(−2)2 + (2 3)2
4 + 12
=4
√
2 3
tan θ =
−2
√
=− 3
√
Since tan π3 = 3, we conclude that θ has a reference angle of π3 . Because the
2π
point is in the second quadrant, we have θ = 2π
in
3 . Thus, the point is 4, 3
polar coordinates.
c. Once again, r and tan θ are easily found.
q
2
2
r = (−1) + (−3)
√
= 1+9
√
= 10
−3
−1
=3
tan θ =
Thus, θ is an angle in the third quadrant with tan θ = 3. Using the tan−1 key
on a calculator, we find that the reference angle is tan−1 3 ≈ 1.2490. Since θ is
in the third quadrant, it follows that
θ ≈ π + 1.2490 ≈ 4.3906
√
Thus, the polar coordinates are approximately
10, 4.3906 .
It is often the case that an equation expressed in terms of one coordinate system
is much more complex than when expressed in terms of the other system. In the
following examples, we see how it is possible to convert equations from one system
to the other.
57
1.5. POLAR FUNCTIONS
Example 10 Converting an equation from rectangular to polar form
Find a polar equation for the line y = mx + b, for b 6= 0.
Solution Replacing x with r cos θ and y with r sin θ, we have
r sin θ = m (r cos θ) + b
Solving for r gives us
r sin θ − mr cos θ = b
r(sin θ − m cos θ) = b
r=
b
sin θ − m cos θ
Example 11 Converting an equation from polar to rectangular form
Convert the polar equation r = 2 cos θ to an equivalent equation in x and y.
Solution In order to exploit the relation x = r cos θ, we begin by multiplying both
sides of r = 2 cos θ by r.
r · r = r · 2 cos θ
r2 = 2r cos θ
Substituting x2 + y 2 for r2 and x for r cos θ gives
x2 + y 2 = 2x
(x − 1)2 + y 2 = 1
1
This, by the way, is the equation of a circle of radius 1 centered at (1, 0). To see
this, we complete the square as follows:
x2 + y 2 = 2x
1
2
x2 − 2x + y 2 = 0
(x2 − 2x + 1) + y 2 = 1
-1
(x − 1)2 + y 2 = 1
The graph is shown in Figure 1.98.
Figure 1.98
1.5 Exercises
Exercises 1–4 Answer true or false.
1. Each point in the plane can be represented in rectangular coordinates in only one way.
2. Each point in the plane can be represented in polar
coordinates in only one way.
3. For any point with rectangular coordinates (x, y),
a corresponding value for θ is always given by θ =
tan−1 xy .
4. The point with polar coordinates (0, 2.76) is called
the pole.
Exercises 5–8 Give an example of each.
5. A polar equation whose graph is a circle centered at
the origin
6. A polar equation whose graph is a line
58
CHAPTER 1. FUNCTIONS
7. A polar equation whose graph is a spiral
8. A polar equation whose graph is a circle centered
away from the origin
Exercises 49–58 Convert the given polar equation to rectangular form.
49.
51.
53.
55.
57.
r=7
θ = 2π
3
r cos θ = 1
r = 2 sin θ
r = cos 2θ
50.
52.
54.
56.
58.
θ = π4
r = −3
r sin θ = 4r cos θ + 2
r = 2 cos θ + 3 sin θ
r = sin 2θ
Exercises 9–18 Plot the point given in polar coordinates
and find the corresponding rectangular coordinates.
10. 3, 3π
9. 2, − π2
2 11. 1, 3π
12. −4, π4 4 Exercises 59–64 Using a graphing utility as necessary,
π
13. −2, 6
14. 3, − π3
√
match each polar equation with the appropriate descrip15. 23 , − 5π
16. 2 2, 7π
6
4 tion. Choose from:
17. (−3, −π)
18. − 25 , − 7π
2
i. Cardioid (a heart shaped curve)
Exercises 19–28 Plot the point given in rectangular coorii. 4-petaled rose
dinates and find two sets of polar coordinates for which
iii. 5-petaled rose
0 ≤ θ < 2π, one with r > 0 and the other with r < 0.
iv. Limaçon with an inner loop (see Example 7)
19. (4, 0)
20. (0, −3)
v. Spiral
21. (−3, 3)
22. (2, 2)√ √ 23. 1, − 3
24. −2 3, 2
vi. Lemniscate (a curve in the shape of a figure 8)
25. (−4, −3)
26. (12, −5)
59. r = 3 cos 2θ
27. (3, 1)
28. (−2, 4)
60. r2 = 4 cos 2θ (Hint: Solve for r and
Exercises 29–42 Find and plot at least four points satisfyplot two branches.)
ing the given polar equation and then sketch the graph of
61. r = 2 + 4 cos θ
the equation.
62. r = 4 (1 − sin θ)
29. r = 3
30. r = −2
63. r = 3 sin 5θ
31. θ = 2π
32. θ = 5π
3
6
33. r = 2θ, θ ≥ 0
34. r = 1 + θ, θ ≥ 0
64. r = πθ
35. r = 3 sin θ
36. r = −2 cos θ
Exercises 65–70 Use a graphing utility to plot the graphs of
37. r = 2 − 4 cos θ
38. r = 2 sin θ + 2
the given polar equations. Estimate the polar coordinates
39. r = sin 2θ
40. r = cos 4θ
of the point(s) of intersection.
41. r = 2 cos 3θ
42. r = 3 sin θ
2
Exercises 43–48 Convert the given rectangular equation to
polar form.
65.
67.
43. y = 3
45. y + x = 0
47. x2 + y 2 = 9
44. x = √
−1
46. y = 3 x
48. x2 − y 2 = 1
69.
r
r
r
r
r
r
=2
= 4 sin θ
= 2 cos θ
= 1 − sin θ
= sin 2θ
= sin θ
66.
68.
70.
r
r
r
r
r
r
= 2 sin θ
= 4 cos θ
=3
= 2 + 3 cos θ
= cos 3θ
= sin θ
59
1.6. PARAMETRIC FUNCTIONS
1.6 Parametric Functions
A wide variety of curves can be represented as algebraic equations in x and y.
For example, a circle with radius 2 centered at the origin has equation x2 + y 2 = 4,
a line with slope 1 and y-intercept 3 has equation y = x + 3, and so forth. But there
are many curves—particularly, paths arising from complex motion—for which it is
more natural to view x and y as functions of a third variable t, called a parameter.
The curve is then defined by the parametric equations x = f (t) and y = g(t). The
graph of the parametric equations x = f (t) and y = g(t) consists of all points of
the form (f (t), g(t)). The orientation of the graph is determined by the direction of
motion as t increases, and when desired, it is indicated on the graph with arrows.
Example 1 Graphing parametric equations
Graph the relation defined by the following parametric equations.
x=t+1
y = 2t − 1
Solution This set of parametric equations is relatively simple, so we just plot a few
points by selecting convenient values of t and computing the corresponding x- and
y-values.
x=t+1
y = 2t − 1
4
t=2
3
2
t=1
1
-4 -3 -2 -1
-1
1
2
3
t=0
-2
-3 t = −1
-4
4
t
−1
0
1
2
x=t+1
0
1
2
3
y = 2t − 1
−3
−1
1
3
Connecting the points (0, −3), (1, −1), and so forth gives us the graph shown in
Figure 1.99. Note that the arrow indicates the orientation—that is, the direction
of motion as t increases. Note also that the graph appears to be a line with slope
2 and y-intercept −3, and, if so, would have equation y = 2x − 3. We verify this in
Example 3.
Figure 1.99
It is quite possible for different sets of parametric equations to have identical
graphs. In fact, just as there are infinitely many possible journeys along a single road, so too are there infinitely many parameterizations for a given curve, as
suggested by the following examples.
Example 2 Graphing parametric equations
Graph the given parametric equations.
a. x = 3t, y = 6t
b. x = −t + 3, y = −2t + 6
c. x = t2 , y = 2t2
Solution
a. We begin by selecting some convenient t-values and computing the corresponding
x- and y-values.
60
CHAPTER 1. FUNCTIONS
t
−2
−1
0
1
2
x = 3t
−6
−3
0
3
6
y = 6t
−12
−6
0
6
12
Plotting the points, we obtain the graph shown in Figure 1.100. We note that
since the y-coordinate (6t) is twice the x-coordinate (3t), the graph is simply the
line y = 2x.
b. This time we note from the outset that the y-value (−2t + 6) is twice the x-value
(−t + 3). Thus, just as in part a, the parametric equations define the line y = 2x.
However, since the x- and y-values both decrease as t increases, the orientation
is in the direction opposite that in part a. The resulting graph is shown in Figure
1.101.
c. Again, the y-coordinate is twice the x-coordinate. But in this case, both x = t2
and y = 2t2 are nonnegative for all values of t. Thus, the graph consists of just
the portion of the line y = 2x for which x ≥ 0 and y ≥ 0, as shown in Figure
1.102. An orientation arrow is omitted since the direction is toward the origin
as t increases through negative values but away from the origin as t increases
through positive values.
12
12
9
9
x = 3t
y = 6t
6
3
-12 -9 -6 -3
-3
3
6
9 12
12
9
x=t+3
y = 2t + 6
6
3
-12 -9 -6 -3
-3
3
6
9 12
3
-12 -9 -6 -3
-3
-6
-6
-9
-9
-9
-12
-12
-12
Figure 1.100
Figure 1.101
x = t2
y = 2t2
6
3
6
9 12
-6
Figure 1.102
Eliminating the Parameter
As we have seen in the previous examples, it is sometimes possible to find an
equation involving only x and y that is equivalent to a given set of parametric
equations. We refer to the process of producing an equation in x and y from a
set of parametric equations as eliminating the parameter. This process can
be simple, tricky, or downright impossible (sometimes the parameter cannot be
eliminated at all). The most straightforward technique is substitution—solving one
equation for t and substituting into the second equation, as demonstrated in the
following example.
61
1.6. PARAMETRIC FUNCTIONS
Example 3 Eliminating the parameter by substitution
Eliminate the parameter.
x=t+1
y = 2t − 1
Solution We solve the first equation for t and substitute into the second, as follows:
x=t+1
The parametric equation for x
t=x−1
Solving for t
y = 2t − 1
The parametric equation for y
y = 2(x − 1) − 1
Substituting t = x − 1
y = 2x − 3
Simplifying
Another standard trick for eliminating the parameter is to multiply the equations by appropriate constants and then add in such a way that the parameter is
eliminated. Not surprisingly, this technique is called elimination.
Example 4 Eliminating the parameter
Eliminate the parameter.
x = −t3 + 1
y = 3t3
Solution To eliminate t, we multiply the first equation by 3 and add to the second.
3x = −3t3 + 3
Multiplying the first equation by 3
3
+ y = 3t
3x + y = 0t3 + 3
Adding to eliminate t
3x + y = 3
y = −3x + 3
Simplifying
Parametric equations for conic sections often involve trigonometric functions. In
particular, ellipses and circles are generally parameterized using the sine and cosine
functions. In such cases, trigonometric identities, especially cos2 t + sin2 t = 1, are
often helpful in eliminating the parameter.
Example 5 Eliminating the parameter
Eliminate the parameter t and graph the relation defined by the following parametric equations:
x = 3 cos t − 2
y = 5 sin t + 1
Solution We first solve the equations for cos t and sin t.
x = 3 cos t − 2
x+2
cos t =
3
y = 5 sin t + 1
y−1
sin t =
5
62
CHAPTER 1. FUNCTIONS
Next, we apply the identity cos2 t + sin2 t = 1.
6
4
x = 3 cos t − 2
y = 5 sin t + 1
2
-6
-4
-2
2
4
6
-2
-4
-6
cos2 t + sin2 t = 1
2 2
x+2
y−1
+
=1
3
5
(y − 1)2
(x + 2)2
+
=1
32
52
Substituting for cos t and sin t
Simplifying
This is the equation of an ellipse centered at (−2, 1) with vertical major axis of
length 2 · 5 = 10 and minor axis of length 2 · 3 = 6. The graph is shown in Figure
1.103. Note that when t = 0, we obtain the point (1, 1), whereas when t = π2 , we
obtain (−2, 6). Thus, the graph is oriented counterclockwise.
Figure 1.103
Applications
Thus far, we have attached no special meaning to the parameter t; the parameter
has served simply as a device for defining a relation between x and y. In practice,
however, parametric equations generally arise in contexts in which the parameter
has some real-world interpretation—time, angle of rotation, distance traveled, and
so forth.
Consider the case of a child riding on a merry-go-round with a 10-foot radius.
If we define a coordinate system with the center of the carousel as the origin, then
the child’s path is given by x2 + y 2 = 102 . From this equation we could determine,
for example, that the child will pass through the point (6, 8), but we wouldn’t know
how fast she was traveling, what direction she was going, or how many times she
passed through this point. Whether she rotates clockwise or counterclockwise, goes
around once or a thousand times, travels at 1 foot per second, or breaks the sound
barrier, the equation of her path will still be x2 + y 2 = 102 .
To describe the girl’s position at any time, we could give her x- and y-coordinates
as functions of t, the length of time that she has been on the ride. Her position
would then be given by the parametric equations
x = f (t)
y = g(t)
In the next example, we illustrate how parametric equations can be used to describe
two very different 8-second carousel rides.
Example 6 Carousel motion
Describe the motion of a girl traveling on a merry-go-round according to the given
set of parametric equations. Assume that t represents the elapsed time for the ride
in seconds and x and y are in feet.
a.
x = 10 cos 21 πt
y = 10 sin 12 πt
b.
x = 10 cos 50πt
y = −10 sin 50πt
Solution
a. We begin by finding the girl’s location at conveniently chosen time intervals.
63
1.6. PARAMETRIC FUNCTIONS
t
0
1
2
3
4
..
.
x = 10 cos 12 πt
10
0
−10
0
10
..
.
y = 10 sin 12 πt
0
10
0
−10
0
..
.
8
10
0
Plotting these points as shown in Figure 1.104, we see that the girl begins at the
point (10, 0) and travels counterclockwise at the steady rate of one revolution
every 4 seconds. In the indicated 8-second time interval, she makes a total of
two revolutions.
b. Since functions of the form f (t) = a cos bt or g(t) = a sin bt have periods of 2π
|b| ,
2π
the x- and y-values will repeat themselves every |50π| = 0.04 second. Thus, we
suspect that the girl is moving at the rate of one revolution every 0.04 second. To
determine the direction of her motion, we’ll sample her position at 0.01-second
time intervals.
t
0
0.01
0.02
0.03
0.04
..
.
x = 10 cos 50πt
10
0
−10
0
10
..
.
y = −10 sin 50πt
0
−10
0
10
0
..
.
Plotting these points as shown in Figure 1.105, we see that the girl begins at
(10, 0) and travels clockwise at the rate of one revolution for every 0.04 second
(about Mach 1.6) for 8 seconds, making a total of 200 revolutions in all.
t = 2, 6
10
8
6
4
2
-10 -8 -6 -4 -2
-2
-4
-6
-8
-10
t = 1, 5
t = 0, 4, 8
2 4 6 8 10
t = 3, 7
Figure 1.104
t = 0.02, 0.06
10
8
6
4
2
-10 -8 -6 -4 -2
-2
-4
-6
-8
-10
t = 0.03, 0.07
t = 0, 0.04, 0.08
2 4 6 8 10
t = 0.01, 0.05
Figure 1.105
In situations such as the one described in Example 6 where the position of an
object can be described using parametric equations, we will eventually be able to
apply techniques of calculus to find the velocity of the object.
64
CHAPTER 1. FUNCTIONS
1.6 Exercises
Exercises 1–4 Answer true or false.
1. If t seconds after the cue ball has been struck the
location of the 8-ball on a pool table is given by
(f1 (t), g1 (t)), and the location of the 7-ball is given
by (f2 (t), g2 (t)), and the graphs of these two sets
of parametric equations intersect, then the 8- and
7-balls must collide.
2. If t seconds after the cue ball has been struck the
location of the 8-ball on a pool table is given by
(f1 (t), g1 (t)), and the location of the 7-ball is given
by (f2 (t), g2 (t)), and for some value of t, f1 (t) =
f2 (t) and g1 (t) = g2 (t), then the 8- and 7-balls must
collide.
3. No portion of the curve with parametric equations
x = t2 + 1 and y = g(t) lies in the second quadrant.
4. If f (2) = g(2) = 0, then we can conclude that the
curve given by x = f (t) and y = g(t) passes through
the origin.
Exercises 5–8 Give an example of each.
Exercises 13–20 Eliminate the parameter and then graph.
13.
x=3+t
y = −2t + 2
14.
x = 3t3 − 4
y = 6t3 + 1
15.
x=t+1
y = t2 + 2t + 1
16.
x = t2
y =t−1
17.
x = 4 cos t
y = 4 sin t
18.
x = 2 cos t
y = 3 sin t
19.
x = 1 − 2 sin t
y = 2 + 3 cos t
20.
x = 2 + 3 sin t
y = −3 − 3 cos t
Exercises 21–26 For the given parametric equations:
5. Parametric equations for the portion of the graph of a. Eliminate the parameter and graph the resulting equation in x and y.
y = x2 lying in the first quadrant
6. Two different sets of parametric equations for the b. Use a graphing utility to plot the original parametric
equations.
line y = 2x
c.
Explain any differences between the graphs in parts a
7. Parametric equations for a curve for which y cannot
and
b.
be defined as a function of x
8. A parameterization for the curve y = f (x)
Exercises 9–10 Complete the table and use it to sketch the
graph of the parametric equations.
9.
10.
x=t−3
y = 2t + 1
x = 3 − 2t
y = 2t − 1
t
−1
0
1
2
x
t
−1
0
1
2
x
y
12. x = −t2 + t, y = 2t
x = t2 − 2
y = t2
22.
x = −t2
y = t4
23.
x = et
y = e2t
24.
x = |2t|
y = |t|
25.
x = cos t
y = 4 cos t
26.
x = sin t
y = 2 sin2 t
y
Exercises 11–12 Find and plot at least four points on the
graph of the parametric equations and then sketch the
graph.
11. x = t2 , y = −3t + 2
21.
Exercises 27–32 Use a graphing utility as necessary to
match the parametric equations with the appropriate description. Choose from:
i. Path of a point on a basketball, shot from the freethrow line with a slight backspin
ii. Path of a reflector strapped to a cyclist’s ankle
iii. The letter A
65
1.6. PARAMETRIC FUNCTIONS
iv. Path of a point on a quarter rolling around the outside of a second quarter
v. A spiral
vi. Path of a ray of light inside a mirrored room
Applications
37. Elliptical Cross Trainer The motion of the heel of
the foot when using an elliptical cross trainer is approximated by the parametric equations
x = 6.1 cos t − 0.95 sin t − 0.19
27.
x=
10
t+1
10
t+1
y = 2.1 cos t + 2.7 sin t + 7.9
cos t
where y gives the height of the heel above the floor
in inches. Estimate the maximum and minimum
heights.
38. Earth Orbit Earth travels around the Sun along
an elliptical path approximated by the parametric
equations
x = 1.496 cos 0.172t2 − 6.459t − 3.071 − 0.025
y = −1.49579 sin 0.172t2 − 6.459t − 3.071
y=
sin t
0 ≤ t ≤ 50
28.
29.
x = 0.75 sin
8π
3 t + 26.4t
8π
3 t + 1.25
y = 0.75 cos
0 ≤ t ≤ 10
x = 8 − 8 t − 2 12 t − 1
y = 8 − 8 7 t − 2 7 t − 1
11
22
0 ≤ t ≤ 11
(Hint: The floor function is defined by bxc = the
greatest integer less than or equal to x. So b3.2c = 3
and b−5.4c = −6.)
30.
x=
4
9
|t − 9| − |t − 6| − |t − 5| + 4 |t − 4|
−3 |t − 3| + 13
9 |t|
y = 2 |t − 6| − 2 |t − 5| + 2 |t − 3| − 4 |t − 2| + 2 |t|
0≤t≤9
31.
x = 2 sin t + sin (2t)
y = 2 cos t + cos (2t)
0 ≤ t ≤ 2π
32.
x=
5
12
5
12
cos (4πt) +
y = sin (4πt) −
0 ≤ t ≤ 1.5
√
30 34
17 t √
16t2 + 761734 t
+6
Exercises 33–36 Use a graphing utility to plot the graph
and estimate the indicated point(s).
33.
where t is measured in years (t = 0 corresponds to
the beginning of 2005) and x and y are measured
in 1011 meters. Find the dates on which Earth is
closest to (perihelion) and furthest from (aphelion)
the Sun.
39. Double Ferris Wheel The position of a rider on a
double Ferris wheel (see the photo) is approximated
by the parametric equations
πt
2πt
x = 20 sin
+ 10 sin
10
5
2πt
πt
− 10 cos
y = 35 − 20 cos
10
5
where x represents the horizontal distance in feet
from the base, y is the height in feet, and t is elapsed
time in seconds from the start of the ride.
a) Find the initial position of the rider.
b) How long does it take for the rider to return to
the starting position?
c) Find the greatest height and the first two times
at which this height is attained.
x = t3 − 3t2 + 2t + 2
y = −t2 + 2t + 3
Crossing point (where the curve intersects itself)
34.
x = cos t + cos (2t) + 3
y = sin t + sin (2t) − 4
Crossing point (where the curve intersects itself)
35.
x = 2 cos3 t
y = 4 sin3 t
Intercepts
36.
x = t2 − t − 6
y = t2 − 4t − 5
Intercepts
A double Ferris wheel
66
CHAPTER 1. FUNCTIONS
40. Bouncing Golf Ball The position of a golf ball t seconds after it has been struck can be approximated
by
t − 2 − btc
+2
x = 200
2btc
y = 25 (t − btc) (btc − t + 1) 24−btc
where x represents the horizontal distance traveled
and y is the height, with both measured in feet.
(Hint: The floor function is defined by bxc = the
greatest integer less than or equal to x. So b3.2c = 3
and b−5.4c = −6.)
a) What horizontal distance does the ball travel?
b) What is the ball’s maximum height?
c) What is the maximum height attained by the
ball after its third bounce?
41. Daily Commute In parts a through e, parametric
equations for an employee’s daily commute from
home to work for a 5-day workweek are given, but
in no particular order. In addition, the employee’s
account of each day’s commute is included. Use the
trace feature of a graphing utility to find the day of
the week corresponding to the commutes described
in parts a through e.
a) x = −4 + t, y = 5; 0 ≤ t ≤ 8
b) x = −4 + 0.1t, y = 5; 0 ≤ t ≤ 80
c) x = 21 (|t| − |t − 4| + |t − 8| − |t − 12|), y = 5;
0 ≤ t ≤ 12
d) x = 4 sin π8 t − π2 , y = 5; 0 ≤ t ≤ 24
e) x = −4 + t + sin (πt), y = 5; 0 ≤ t ≤ 8
Monday: I stopped off to get gas about halfway to
work, but other than that, I went pretty fast.
Tuesday: I sailed to work—there was almost no traffic.
Wednesday: My transmission acted up on me. The
car kept lurching into reverse, and I would find myself hurtling backwards for several hundred yards before I could get her back in drive again.
Thursday: It was such a nice day that I jogged to
work.
Friday: I got to work in good time but then realized
I had forgotten some important papers, and so I had
to go back home to get them.
In general, what type of information about a journey can be conveyed by parametric equations that
cannot be conveyed by an equation in x and y?