2 FUNCTIONS AND GRAPHS
EXERCISE 2-1
Things to remember:
1.
POINT-BY-POINT PLOTTING
To sketch the graph of an equation in two variables, plot
enough points from its solution set in a rectangular
coordinate system so that the total graph is apparent and
then connect these points with a smooth curve.
2.
A FUNCTION is a correspondence between one set of elements,
called the DOMAIN, and a second set of elements, called the
RANGE, such that to each element in the domain there
corresponds one and only one element in the range.
3.
EQUATIONS AND FUNCTIONS
Given an equation in two variables. If there corresponds
exactly one value of the dependent variable (output) to each
value of the independent variable (input), then the equation
specifies a function. If there is more than one output for at
least one input, then the equation does not specify a
function.
4.
VERTICAL LINE TEST FOR A FUNCTION
An equation specifies a function if each vertical line in the
coordinate system passes through at most one point on the
graph of the equation. If any vertical line passes through
two or more points on the graph of an equation, then the
equation does not specify a function.
5.
AGREEMENT ON DOMAINS AND RANGES
If a function is specified by an equation and the domain is
not given explicitly, then assume that the domain is the set
of all real number replacements of the independent variable
(inputs) that produce real values for the dependent variable
(outputs). The range is the set of all outputs corresponding
to input values.
In many applied problems, the domain is determined by
practical considerations within the problem.
EXERCISE 2-1
31
6.
FUNCTION NOTATION — THE SYMBOL f(x)
For any element x in the domain of the function f, the symbol
f(x) represents the element in the range of f corresponding
to x in the domain of f. If x is an input value, then f(x)
is the corresponding output value. If x is an element which
is not in the domain of f, then f is NOT DEFINED at x and
f(x) DOES NOT EXIST.
x
1. y = x + 1: y
"4
"3
"2
"1
0
1
2
3
x
3. x = y2: y
4
5
0
0
1
±1
y
4
±2
9
±3
16
±4
y
5
5
!
!
x
–5
x
5
–10
–5
x
5. y = x3: y
10
–5
"2
"8
"1
"1
0
0
1
1
2
8
7. xy = -6:
x "6 "3
y
1
2
y
"1
6
1
"6
3
"2
6
"1
y
50
6
!
!
x
–5
5
x
–6
6
–50
–6
9. The table specifies a function, since for each domain value there
corresponds one and only one range value.
11. The table does not specify a function, since more than one range
value corresponds to a given domain value. (Range values 5, 6
correspond to domain value 3; range values 6, 7 correspond to domain
value 4.)
13. This is a function.
15. The graph specifies a function; each vertical line in the plane
intersects the graph in at most one point.
32
CHAPTER 2
FUNCTIONS AND GRAPHS
17. The graph does not specify a function. There are vertical lines
which intersect the graph in more than one point. For example, the
y-axis intersects the graph in three points.
19. The graph specifies a function.
21. y = 3 – 7x
or
y = -7x + 3; linear function
23. y = -6; constant function
1
; neither linear nor constant
x
25. y = 8x – 1 +
1+ x
1" x
1
1
1
1
1
5
+
=
+
+ x - x = - x + ; linear function
2
3
2
3
2
3
6
6
!
29. y = x2 + (1 – x)(1 + x) + 1 = x2 + 1 – x2 + 1 = 2; constant function
27. y =
!
!
!
!
!
!
!
!
31. (A) This correspondence is a function; there is exactly one
representative for each congressional district.
(B) This correspondence is not a function; there are two senators
representing Pennsylvania.
y
5
33. f(x) = 1 – x: Since f is a linear
function, we only need to plot
two points.
x
"2
2
f(x)
3
"1
x
–5
5
–5
!
35. f(x) = x2 – 1:
x "3 "2 "1
f(x)
8
3
0
0
"1
1
0
2
3
37. f(x) = 4 – x3:
x "2 "1 0
f(x) 12
5 4
3
8
y
1
3
2
"4
y
50
10
!
!
x
x
–5
5
–10
–5
5
–50
EXERCISE 2-1
33
8
x
"8 "4
"1 "2
39. f(x) =
x
f(x)
8
"2
"4
1
8
"1
"8
2
4
4
2
8
1
!
–8
8
!
–8
y
5
41. The graph
of f is:
–5
5
x
–5
43. f(x) = 100x - 5x2; g(x) = 150 + 20x, 0 ≤ x ≤ 20
(A) f(0) = 100(0) f(5) = 100(5) f(10) = 100(10)
f(15) = 100(15)
f(20) = 100(20)
f(0) - g(0) = 0
f(10) - g(10) =
f(20) - g(20) =
5(0)2 = 0, g(0) = 150 + 20(0) = 150;
5(5)2 = 500 - 125 = 375, g(5) = 150 + 20(5) = 250;
- 5(10)2 = 1000 - 500 = 500, g(10) = 150 + 20(10) = 350;
- 5(15)2 = 1500 - 1125 = 375, g(15) = 150 + 20(15) = 450;
- 5(20)2 = 2000 - 2000 = 0, g(20) = 150 + 20(20) = 550;
- 150 = -150; f(5) - g(5) = 375 - 250 = 125;
500 - 350 = 150; f(15) - g(15) = 375 - 450 = -75;
0 - 550 = -550
y
(B)
y = g(x)
500
y = f(x)
5
10
15
x
y = f(x) - g(x)
–500
45. y = f(-5) = 0
47. y = f(5) = 4
49. f(x) = 0 at x = -5, 0, 4
51. f(x) = -4 at x = -6
53. f(2) = 2·2 - 3 = 4 - 3 = 1
55. f(-1) = 2(-1) - 3 = -2 - 3 = -5
34
CHAPTER 2
FUNCTIONS AND GRAPHS
57. g(-3) = (-3)2 + 2(-3) = 9 - 6 = 3
59. f(1) + g(2) = [2·1 - 3] + [22 + 2·2] = -1 + 8 = 7
61. g(3)·f(0) = [32 + 2·3]·(2·0 - 3) = 15(-3) = -45
(!2)2 + 2(!2)
g(!2)
4 ! 4
0
63.
=
=
=
= 0
2(!2) ! 3
f(!2)
!7
!7
65. domain: all real numbers or (-∞, ∞)
67. domain: all real numbers except -4
69. domain: x ≤ 7
71. f is not defined at the values of x where x2 - 9 = 0, that is, at 3
0
and -3; f is defined at x = 2, f(2) =
= 0.
!5
73. g(x) = 2x3 - 5
75. G(x) = 2 x - x2
77. Function f multiplies the domain element by 2 and subtracts 3 from
the result.
79. Function F multiplies the cube of the! domain element by 3 and
subtracts twice the square root of the domain element from the
result.
81. Given 4x - 5y = 20. Solving for y, we have:
-5y = -4x + 20
4
y = x - 4
5
Since each input value x determines a unique output value y, the
equation specifies a function. The domain is R, the set of real
numbers.
83. Given x2 - y = 1. Solving for y, we have:
-y = -x2 + 1 or y = x2 - 1
This equation specifies a function. The domain is R, the set of real
numbers.
85. Given x + y2 = 10. Solving for y, we have:
y2 = 10 - x
y = ± 10 " x
This equation does not specify a function since each value of x,
x < 10, determines two values of y. For example, corresponding to
x = 1, we have y = 3 and y = -3; corresponding to x = 6, we have y =
!2 and y = -2.
87. Given xy - 4y = 1.
Solving for y, we have:
1
(x - 4)y = 1 or y =
x ! 4
This equation specifies a function. The domain is all real numbers
except x = 4.
EXERCISE 2-1
35
89. Given x2 + y2 = 25.
Solving for y, we have:
y2 = 25 - x2 or y = ± 25 " x 2
Thus, the equation does not specify a function since, for x = 0, we
have y = ±5, when x = 4, y = ±3, and so on.
91. Given F(t) = 4t + 7.! Then:
F(3 + h) ! F(3)
4(3 + h) + 7 ! (4 " 3 + 7)
=
h
h
4h
12 + 4h + 7 ! 19
=
=
= 4
h
h
93. Given Q(x) = x2 - 5x + 1.
Then:
h)2
(2 +
! 5(2 + h) + 1 ! (22 ! 5 " 2 + 1)
Q(2 + h) ! Q(2)
=
h
h
2
4 + 4h + h ! 10 ! 5h + 1 ! (!5)
h2 ! h ! 5 + 5
=
=
h
h
h(h ! 1)
=
= h - 1
h
95. f(5) = 52
1 = 24
97. f(2 + 5) = f(7) = 72 – 1 = 48
99. f(2) + f(5) = 22 – 1 + 52 – 1 = 3 + 24 = 27
101. f(f(1)) = (f(1))2 – 1 = (12 – 1) - 1 = 0 – 1 = -1
103. f(2x) = (2x)2 – 1 = 4x2 – 1
105. f(x + 1) = (x + 1)2 – 1 = x2 + 2x + 1 – 1 = x2 + 2x
107. f(x) = 4x
(A) f(x +
(B) f(x +
f(x +
(C)
- 3
h) = 4(x + h) - 3 = 4x + 4h - 3
h) - f(x) = 4x + 4h - 3 - (4x - 3) = 4h
h) ! f(x)
4h
=
= 4
h
h
109. f(x) = 4x2 - 7x + 6
(A) f(x + h) = 4(x + h)2 - 7(x + h) + 6
= 4(x2 + 2xh + h2) - 7x - 7h + 6
= 4x2 + 8xh + 4h2 - 7x - 7h + 6
(B) f(x + h) - f(x) = 4x2 + 8xh + 4h2 - 7x - 7h + 6 - (4x2 - 7x + 6)
= 8xh + 4h2 - 7h
8xh + 4h2 ! 7h
f(x + h) ! f(x)
h(8x + 4h ! 7)
(C)
=
=
= 8x + 4h - 7
h
h
h
36
CHAPTER 2
FUNCTIONS AND GRAPHS
111. f(x) = x(20 - x) = 20x - x2
(A) f(x + h) = 20(x + h) - (x + h)2 = 20x + 20h - x2 - 2xh - h2
(B) f(x + h) - f(x) = 20x + 20h - x2 - 2xh - h2 - (20x - x2)
= 20h - 2xh - h2
20h ! 2xh ! h 2
f(x + h) ! f(x)
h(20 ! 2x ! h)
(C)
=
=
= 20 - 2x - h
h
h
h
113. Given A = lw = 25.
25
Thus, l =
. Now P = 2l + 2w
w
50
! 25
= 2 " #$ + 2w =
+ 2w.
w
w
The domain is w > 0.
115. Given P = 2l + 2w = 100 or l + w = 50 and w = 50 - l .
Now A = lw = l (50 - l ) and A = 50l - l 2 .
The domain is 0 ≤ l ≤ 50. [Note: l ≤ 50 since l > 50 implies w < 0.]
p
$ 100
p(x) = 75 - 3x
p(7) = 75 - 3(7) = 75 - 21 =
p(11) = 75 - 3(11) = 75 - 33
Estimated price per chip for
of 7 million chips: $54; for
of 11 million chips: $42
117.
50
5
10 15 20
x
54;
= 42
a demand
a demand
119. (A) R(x) = x·p(x) = x(75 - 3x) = 75x - 3x2 , 1 ≤ x ≤ 20
(C)
(B)
x
1
4
8
12
16
20
R(x)
72
252
408
468
432
300
(C)
R
$500
400
300
200
100
5
10
15
20
x
121. (A) Profit: P(x) = R(x) - C(x) = 75x - 3x2 - [125 + 16x]
!
= 59x - 3x2 - 125, 1 ≤ x ≤ 20
(B)
x
1
4
8
12
16
20
P(x)
"69
63
155
151
51
"145
(C)
(C) $200
100
5
10
15
20
–100
–200
!
EXERCISE 2-1
37
12
x
123.
8
x
(A) V = (length)(width)(height)
V(x) = (12 - 2x)(8 - 2x)x
= x(8 - 2x)(12 - 2x)
(B) Domain: 0 ≤ x ≤ 4
x
(C) V(1) =
=
V(2) =
=
V(3) =
=
(12 - 2)(8 - 2)(1)
(10)(6)(1) = 60
(12 - 4)(8 - 4)(2)
(8)(4)(2) = 64
(12 - 6)(8 - 6)(3)
(6)(2)(3) = 36
Thus,
Volume
x V(x)
1
60
2
64
3
36
(D)
125. (A) The graph indicates that there is a value of x near
2, and slightly less than 2, such that V(x) = 65.
The table is shown at the right.
Thus, x = 1.9 to one decimal place.
(B)
x
1.90
1.91
1.92
1.93
1.94
1.95
1.96
y1
65.436
65.307
65.176
65.040
64.902
64.760
64.614
Thus, x = 1.93 to two
decimal places.
127. Given (w + a)(v + b) = c. Let
Then: (w + 15)(v + 1) = 90
Solving for v, we have
90
90
v + 1 =
and v =
w + 15
w + 15
75 ! w
so that v =
.
w + 15
75 ! 16
If w = 16, then v =
=
16 + 15
38
CHAPTER 2
FUNCTIONS AND GRAPHS
a = 15, b = 1, and c = 90.
- 1 =
90 ! (w + 15)
,
w + 15
59
≈ 1.9032 cm/sec.
31
x
1.7
1.8
1.9
2.0
y1
67.252
66.528
65.436
64.000