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Instructor’s Solutions Manual, Chapter 1
Review Question 1
Solutions to Chapter Review Questions, Chapter 1
1. Suppose f is a function. Explain what it means to say that
domain of f .
solution If
sense.
3
2
3
2
is in the
is in the domain of f , then f ( 32 ) is defined and makes
Instructor’s Solutions Manual, Chapter 1
Review Question 2
2. Suppose f is a function. Explain what it means to say that
range of f .
3
2
is in the
solution If 32 is in the range of f , then there exists a number x in the
3
domain of f such that f (x) = 2 .
Instructor’s Solutions Manual, Chapter 1
Review Question 3
3. Give an example of a function whose domain consists of five numbers
and whose range consists of three numbers.
solution Let f be the function whose domain is the set {1, 2, 3, 4, 5},
with f defined as follows:
f (1) = 1,
f (2) = 2,
f (3) = 3,
Then the range of f is the set {1, 2, 3}.
f (4) = 3,
f (5) = 3.
Instructor’s Solutions Manual, Chapter 1
Review Question 4
4. Explain how to find the domain of a function from its graph.
solution Suppose we have the graph of a function f in the xy-plane.
The domain of f is the set of numbers c such that the vertical line
x = c intersects the graph of f .
Instructor’s Solutions Manual, Chapter 1
Review Question 5
5. Explain how to find the range of a function from its graph.
solution Suppose we have the graph of a function f in the xy-plane.
The domain of f is the set of numbers c such that the horizontal line
y = c intersects the graph of f .
Instructor’s Solutions Manual, Chapter 1
Review Question 6
6. Explain how to use the vertical line test to determine whether or not a
set in the plane is the graph of some function.
solution A set in the plane is the graph of some function if and only
if every vertical line in the plane intersects the set in at most one point.
Instructor’s Solutions Manual, Chapter 1
Review Question 7
7. Sketch a curve in the coordinate plane that is not the graph of any
function.
solution One solution among many possibilities in the circle of
radius 1 centered at the origin, shown below:
y
1
1
1
x
1
The curve above fails the vertical line test because the vertical line
x = 0, which is the vertical axis, intersects the curve in two points.
Thus this curve is not the graph of any function.
Instructor’s Solutions Manual, Chapter 1
Review Question 7
For Questions 8–15, assume that f is the function defined on the interval
[1, 3] by the formula
1
.
f (x) = 2
x − 3x + 3
The domain of f is the interval [1, 3], the range of f is the interval
1 4
[ 3 , 3 ], and the graph of f is shown below.
y
4
3
1
The graph of f .
1
3
1
3
2
x
3
For each function g described below:
(a) Sketch the graph of g.
(b) Find the domain of g (the endpoints of this interval should be
shown on the horizontal axis of your sketch of the graph of g).
(c) Give a formula for g.
(d) Find the range of g (the endpoints of this interval should be
shown on the vertical axis of your sketch of the graph of g).
Instructor’s Solutions Manual, Chapter 1
Review Question 8
8. The graph of g is obtained by shifting the graph of f up 2 units.
solution
(a)
y
10
3
3
7
3
1
3
2
x
3
(b) The domain of g is the same as the domain of f . Thus the domain of g
is the interval [1, 3].
(c)
g(x) = f (x) + 2 =
1
+2
x 2 − 3x + 3
(d) The range of g is obtained by adding 2 to each number in the range of
7 10
f . Thus the domain of g is the interval [ 3 , 3 ].
Instructor’s Solutions Manual, Chapter 1
Review Question 9
9. The graph of g is obtained by shifting the graph of f down 2 units.
solution
(a)
y
1
2
3
2
x
3
3
1
5
3
(b) The domain of g is the same as the domain of f . Thus the domain of g
is the interval [1, 3].
(c)
g(x) = f (x) − 2 =
1
−2
x 2 − 3x + 3
(d) The range of g is obtained by subtracting 2 from each number in the
5
2
range of f . Thus the range of g is the interval [− 3 , − 3 ].
Instructor’s Solutions Manual, Chapter 1
Review Question 10
10. The graph of g is obtained by shifting the graph of f left 2 units.
solution
(a)
y
4
3
1
1
3
1
1
2
x
1
(b) The domain of g is obtained by subtracting 2 from each number in the
domain of f . Thus the domain of g is the interval [−1, 1].
(c)
g(x) = f (x + 2) =
1
1
= 2
(x + 2)2 − 3(x + 2) + 3
x +x+1
(d) The range of g is the same as the range of f . Thus the range of g is the
1 4
interval [ 3 , 3 ].
Instructor’s Solutions Manual, Chapter 1
Review Question 11
11. The graph of g is obtained by shifting the graph of f right 2 units.
solution
(a)
y
4
3
1
1
3
3
7
2
x
5
(b) The domain of g is obtained by adding 2 to each number in the domain
of f . Thus the domain of g is the interval [3, 5].
(c)
g(x) = f (x − 2) =
1
1
= 2
(x − 2)2 − 3(x − 2) + 3
x − 7x + 13
(d) The range of g is the same as the range of f . Thus the range of g is the
1 4
interval [ 3 , 3 ].
Instructor’s Solutions Manual, Chapter 1
Review Question 12
12. The graph of g is obtained by vertically stretching the graph of f by a
factor of 3.
solution
(a)
y
4
3
1
1
x
3
3
2
(b) The domain of g is the same as the domain of f . Thus the domain of g
is the interval [1, 3].
(c)
g(x) = 3f (x) =
x2
3
− 3x + 3
(d) The range of g is obtained by multiplying each number in the range of
f by 3. Thus the range of g is the interval [1, 4].
Instructor’s Solutions Manual, Chapter 1
Review Question 13
13. The graph of g is obtained by horizontally stretching the graph of f by
a factor of 2.
solution
(a)
y
4
3
1
1
3
2
3
6
x
(b) The domain of g is obtained by multiplying each number in the domain
of f by 2. Thus the domain of g is the interval [2, 6].
(c)
x
g(x) = f ( 2 ) =
1
x
( 2 )2 − 3 ·
x
2
+3
=
4
x 2 − 6x + 12
(d) The range of g is the same as the range of f . Thus the range of g is the
1 4
interval [ 3 , 3 ].
Instructor’s Solutions Manual, Chapter 1
Review Question 14
14. The graph of g is obtained by reflecting the graph of f through the
horizontal axis.
solution
(a)
y
1
3
1
3
2
x
3
1
4
3
(b) The domain of g is same as the domain of f . Thus the domain of g is
the interval [1, 3].
(c)
g(x) = −f (x) = −
1
x 2 − 3x + 3
(d) The range of g is obtained by multiplying each number in the range of
4
1
f by −1. Thus the range of g is the interval [− 3 , − 3 ].
Instructor’s Solutions Manual, Chapter 1
Review Question 15
15. The graph of g is obtained by reflecting the graph of f through the
vertical axis.
solution
(a)
y
4
3
1
1
3
3
3
2
x
1
(b) The domain of g is obtained by multiplying each number in the domain
of f by −1. Thus the domain of g is the interval [−3, −1].
(c)
g(x) = f (−x) =
(−x)2
1
1
= 2
− 3(−x) + 3
x + 3x + 3
(d) The range of g is same as the range of f . Thus the range of g is the
1 4
interval [ 3 , 3 ].
Instructor’s Solutions Manual, Chapter 1
Review Question 16
16. Suppose f is a function with domain [1, 3] and range [2, 5]. Define
functions g and h by
g(x) = 3f (x)
and h(x) = f (4x).
(a) What is the domain of g?
(b) What is the range of g?
(c) What is the domain of h?
(d) What is the range of h?
solution
(a) The domain of g is the same as the domain of f . Thus the domain of g
is the interval [1, 3].
(b) The range of g is obtained by multiplying each number in the range of
f by 3. Thus the range of g is the interval [6, 15].
(c) The domain of h is obtained by dividing each number in the domain of
f by 4. Thus the domain of h is the interval [ 14 , 34 ].
(d) The range of h is the same as the range of f . Thus the range of h is the
interval [2, 5].
Instructor’s Solutions Manual, Chapter 1
Review Question 17
17. Show that the sum of two odd functions (with the same domain) is an
odd function.
solution Suppose f and g are odd functions with the same domain.
Let x be a number in the domain of f and g. Then
(f + g)(−x) = f (−x) + g(−x) = −f (x) − g(x) = −(f + g)(x).
Thus f + g is an odd function.
Instructor’s Solutions Manual, Chapter 1
Review Question 18
18. Define the composition of two functions.
solution Suppose f and g are functions. Then the composition f ◦ g
is defined by the formula
(f ◦ g)(x) = f g(x) .
Instructor’s Solutions Manual, Chapter 1
Review Question 19
x 2 +3
19. Suppose f (x) = 5x 2 −9 . Find two functions g and h, each simpler than
f , such that f = g ◦ h.
solution Define functions g and h by the formulas
g(x) =
Then
x+3
5x − 9
and h(x) = x 2 .
x2 + 3
.
(g ◦ h)(x) = g h(x) = g(x 2 ) =
5x 2 − 9
Thus f = g ◦ h.
Instructor’s Solutions Manual, Chapter 1
Review Question 20
For Questions 20–23, suppose
h(x) = |2x + 3| + x 2
and
f (x) = 3x − 5.
20. Evaluate (h ◦ f )(3).
solution
(h ◦ f )(3) = h f (3) = h(3 · 3 − 5) = h(4) = |2 · 4 + 3| + 42 = 27
Instructor’s Solutions Manual, Chapter 1
Review Question 21
21. Evaluate (f ◦ h)(−4).
solution
(f ◦h)(−4) = f h(−4) = f (|2(−4)+3|+(−4)2 ) = f (21) = 3·21−5 = 58
Instructor’s Solutions Manual, Chapter 1
Review Question 22
22. Find a formula for h ◦ f .
solution
(h ◦ f )(x) = h f (x)
= h(3x − 5)
= |2(3x − 5) + 3| + (3x − 5)2
= |6x − 7| + 9x 2 − 30x + 25
Instructor’s Solutions Manual, Chapter 1
23. Find a formula for f ◦ h.
solution
(f ◦ h)(x) = f h(x)
= f (|2x + 3| + x 2 )
= 3(|2x + 3| + x 2 ) − 5
= 3|2x + 3| + 3x 2 − 5
= |6x + 9| + 3x 2 − 5
Review Question 23
Instructor’s Solutions Manual, Chapter 1
Review Question 24
24. Explain how to use the horizontal line test to determine whether or not
a function is one-to-one.
solution A function is one-to-one if and only if every horizontal line
intersects the graph of the function in at most one point.
Instructor’s Solutions Manual, Chapter 1
25. Suppose f (x) =
2x+1
3x−4 .
Review Question 25
Evaluate f −1 (−5).
solution To evaluate f −1 (−5), we need to find a number x such that
f (x) = −5. In other words, we need to solve the equation
2x + 1
= −5
3x − 4
for x. To do this, multiply both sides by 3x − 4, getting
2x + 1 = −15x + 20,
which is equivalent to the equation
17x = 19.
Thus x =
19
17 ,
and hence f −1 (−5) =
19
17 .
Instructor’s Solutions Manual, Chapter 1
26. Suppose g(x) = 3 +
x
2x−3 .
Review Question 26
Find a formula for g −1 .
solution To find a formula for g −1 (y), we need to solve the equation
3+
x
=y
2x − 3
for x. To do this, multiply both sides by 2x − 3, getting
6x − 9 + x = 2xy − 3y,
which is equivalent to the equation
7x − 2xy = 9 − 3y,
which can be rewritten as
x(7 − 2y) = 9 − 3y.
Thus x =
9−3y
7−2y ,
and hence f −1 (y) =
9−3y
7−2y .
Instructor’s Solutions Manual, Chapter 1
Review Question 27
27. Suppose f is a one-to-one function. Explain the relationship between
the graph of f and the graph of f −1 .
solution A point (x, y) is on the graph of f if and only if the point
(y, x) is on the graph of f −1 . Thus the graph of f −1 can be obtained
from the graph of f by a reflection through the line x = y in the
xy-plane.
Instructor’s Solutions Manual, Chapter 1
Review Question 28
28. Suppose f is a one-to-one function. Explain the relationship between
the domain and range of f and the domain and range of f −1 .
solution Suppose x and y are numbers. Then f (x) = y if and only
if f −1 (y) = x. This implies that the domain of f equals the range of
f −1 , and the range of f equals the domain of f −1 .
Instructor’s Solutions Manual, Chapter 1
Review Question 29
29. Explain the different meanings of the notations f −1 (x) and f (x)−1 .
solution If f is a one-to-one function and x is a number in the range
of f , then f −1 (x) denotes the value of the inverse function f −1 at x. In
other words, f −1 (x) is the number t such that f (t) = x.
In contrast, if x is a number in the domain of x and f (x) = 0, then
f (x)−1 denotes the multiplicative inverse of f (x). In other words,
f (x)−1 =
1
.
f (x)
Instructor’s Solutions Manual, Chapter 1
Review Question 30
30. The function f defined by
f (x) = x 5 + 2x 3 + 2
is one-to-one (here the domain of f is the set of real numbers).
Compute f −1 (y) for four different values of y of your choice.
solution Note that
f (−1) = −1,
f (0) = 2,
f (1) = 5,
f (2) = 50.
Thus
f −1 (−1) = −1,
f −1 (2) = 0,
f −1 (5) = 1,
f −1 (50) = 2.
Instructor’s Solutions Manual, Chapter 1
Review Question 31
31. Draw the graph of a function that is decreasing on the interval [1, 2]
and increasing on the interval [2, 5].
solution The graph of the function defined by f (x) = (x − 2)2 has
the desired properties:
8
1
1
2
5
Instructor’s Solutions Manual, Chapter 1
Review Question 32
32. Make up a table that defines a one-to-one function whose domain
consists of five numbers. Then sketch the graph of this function and its
inverse.
solution Define a function f whose domain is the set {1, 2, 3, 4, 5} by
the following table:
x f (x)
1
3
4
2
2
3
5
4
1
5
The table for f −1 is obtained by interchanging the two columns in the
table above:
x f −1 (x)
3
1
2
4
3
2
4
5
5
1
Here is the graph of f :
Instructor’s Solutions Manual, Chapter 1
Review Question 32
y
5
4
3
2
1
1
2
3
4
5
x
Here is the graph of f −1 , which is obtained by reflecting the graph
above through the line x = y:
y
5
4
3
2
1
1
2
3
4
5
x