Solutions to Chapter Review Questions, Chapter 1park/Fall2013/precalculus/ch1_sol.pdf · Instructor’s Solutions Manual, Chapter 1 Review Question 1 Solutions to Chapter Review Questions,

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 32 is in the

domain of f .

solution If 32 is in the domain of f , then f(3

2) is defined and makessense.


2. Suppose f is a function. Explain what it means to say that 32 is in the

range of f .

solution If 32 is in the range of f , then there exists a number x in the

domain of f such that f(x) = 32 .


3. Give an example of a function whose domain consists of five numbersand 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, f (4) = 3, f (5) = 3.

Then the range of f is the set {1,2,3}.


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 linex = c intersects the graph of f .


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 liney = c intersects the graph of f .


6. Explain how to use the vertical line test to determine whether or not aset in the plane is the graph of some function.

solution A set in the plane is the graph of some function if and onlyif every vertical line in the plane intersects the set in at most one point.


7. Sketch a curve in the coordinate plane that is not the graph of anyfunction.

solution One solution among many possibilities in the circle ofradius 1 centered at the origin, shown below:

�1 1x

�1

1

y

The curve above fails the vertical line test because the vertical linex = 0, which is the vertical axis, intersects the curve in two points.Thus this curve is not the graph of any function.


For Questions 8–15, assume that f is the function defined on the interval[1, 3] by the formula

f(x) = 1

x2 − 3x + 3.

The domain of f is the interval [1, 3], the range of f is the interval[1

3 ,43], and the graph of f is shown below.

13

23

x

1

3

1

4

3

y

The graph of f .

For each function g described below:

(a) Sketch the graph of g.

(b) Find the domain of g (the endpoints of this interval should beshown 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 beshown on the vertical axis of your sketch of the graph of g).


8. The graph of g is obtained by shifting the graph of f up 2 units.

solution

(a)

13

23

x

7

3

3

10

3

y

(b) The domain of g is the same as the domain of f . Thus the domain of gis the interval [1,3].

(c)

g(x) = f(x)+ 2 = 1x2 − 3x + 3

+ 2

(d) The range of g is obtained by adding 2 to each number in the range off . Thus the domain of g is the interval [7

3 ,103 ].


9. The graph of g is obtained by shifting the graph of f down 2 units.

solution

(a)

13

23

x

�5

3

�1

�2

3

y


(c)

g(x) = f(x)− 2 = 1x2 − 3x + 3

− 2

(d) The range of g is obtained by subtracting 2 from each number in therange of f . Thus the range of g is the interval [−5

3 ,−23].


10. The graph of g is obtained by shifting the graph of f left 2 units.

solution

(a)

�1 �1

21

x

1

3

1

4

3

y

(b) The domain of g is obtained by subtracting 2 from each number in thedomain of f . Thus the domain of g is the interval [−1,1].

(c)

g(x) = f(x + 2) = 1(x + 2)2 − 3(x + 2)+ 3

= 1x2 + x + 1

(d) The range of g is the same as the range of f . Thus the range of g is theinterval [1

3 ,43].


11. The graph of g is obtained by shifting the graph of f right 2 units.

solution

(a)

37

25

x

1

3

1

4

3

y

(b) The domain of g is obtained by adding 2 to each number in the domainof f . Thus the domain of g is the interval [3,5].

(c)

g(x) = f(x − 2) = 1(x − 2)2 − 3(x − 2)+ 3

= 1x2 − 7x + 13


3 ,43].


12. The graph of g is obtained by vertically stretching the graph of f by afactor of 3.

solution

(a)

13

23

x

1

3

4

y


(c)

g(x) = 3f(x) = 3x2 − 3x + 3

(d) The range of g is obtained by multiplying each number in the range off by 3. Thus the range of g is the interval [1,4].


13. The graph of g is obtained by horizontally stretching the graph of f bya factor of 2.

solution

(a)

2 3 6x

1

3

1

4

3

y

(b) The domain of g is obtained by multiplying each number in the domainof f by 2. Thus the domain of g is the interval [2,6].

(c)

g(x) = f(x2 ) =1

(x2 )2 − 3 · x2 + 3= 4x2 − 6x + 12


3 ,43].


14. The graph of g is obtained by reflecting the graph of f through thehorizontal axis.

solution

(a)

13

23

x�

1

3

�1

�4

3

y

(b) The domain of g is same as the domain of f . Thus the domain of g isthe interval [1,3].

(c)

g(x) = −f(x) = − 1x2 − 3x + 3

(d) The range of g is obtained by multiplying each number in the range off by −1. Thus the range of g is the interval [−4

3 ,−13].


15. The graph of g is obtained by reflecting the graph of f through thevertical axis.

solution

(a)

�1�3

2�3

x

1

3

1

4

3

y

(b) The domain of g is obtained by multiplying each number in the domainof f by −1. Thus the domain of g is the interval [−3,−1].

(c)

g(x) = f(−x) = 1(−x)2 − 3(−x)+ 3

= 1x2 + 3x + 3

(d) The range of g is same as the range of f . Thus the range of g is theinterval [1

3 ,43].


16. Suppose f is a function with domain [1,3] and range [2,5]. Definefunctions 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 gis the interval [1,3].

(b) The range of g is obtained by multiplying each number in the range off 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 off by 4. Thus the domain of h is the interval [1

4 ,34].

(d) The range of h is the same as the range of f . Thus the range of h is theinterval [2,5].


17. Show that the sum of two odd functions (with the same domain) is anodd 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.


18. Define the composition of two functions.

solution Suppose f and g are functions. Then the composition f ◦ gis defined by the formula

(f ◦ g)(x) = f (g(x)).


19. Suppose f(x) = x2+35x2−9 . 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) = x + 35x − 9

and h(x) = x2.

Then

(g ◦ h)(x) = g(h(x)) = g(x2) = x2 + 35x2 − 9

.

Thus f = g ◦ h.


For Questions 20–23, suppose

h(x) = |2x + 3| + x2 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


21. Evaluate (f ◦ h)(−4).

solution

(f◦h)(−4) = f (h(−4)) = f(|2(−4)+3|+(−4)2) = f(21) = 3·21−5 = 58


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| + 9x2 − 30x + 25


23. Find a formula for f ◦ h.

solution

(f ◦ h)(x) = f (h(x))= f(|2x + 3| + x2)

= 3(|2x + 3| + x2)− 5

= 3|2x + 3| + 3x2 − 5

= |6x + 9| + 3x2 − 5


24. Explain how to use the horizontal line test to determine whether or nota function is one-to-one.

solution A function is one-to-one if and only if every horizontal lineintersects the graph of the function in at most one point.


25. Suppose f(x) = 2x+13x−4 . Evaluate f−1(−5).

solution To evaluate f−1(−5), we need to find a number x such thatf(x) = −5. In other words, we need to solve the equation

2x + 13x − 4

= −5

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 = 1917 , and hence f−1(−5) = 19

17 .


26. Suppose g(x) = 3+ x2x−3 . Find a formula for g−1.

solution To find a formula for g−1(y), we need to solve the equation

3+ x2x − 3

= y

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 = 9−3y7−2y , and hence f−1(y) = 9−3y

7−2y .


27. Suppose f is a one-to-one function. Explain the relationship betweenthe 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 obtainedfrom the graph of f by a reflection through the line x = y in thexy-plane.


28. Suppose f is a one-to-one function. Explain the relationship betweenthe 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 onlyif f−1(y) = x. This implies that the domain of f equals the range off−1, and the range of f equals the domain of f−1.


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 rangeof f , then f−1(x) denotes the value of the inverse function f−1 at x. Inother 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, thenf(x)−1 denotes the multiplicative inverse of f(x). In other words,

f(x)−1 = 1f(x)

.


30. The function f defined by

f(x) = x5 + 2x3 + 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.


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 hasthe desired properties:

1 2 5

1

8


32. Make up a table that defines a one-to-one function whose domainconsists of five numbers. Then sketch the graph of this function and itsinverse.

solution Define a function f whose domain is the set {1,2,3,4,5} bythe following table:

x f(x)1 32 43 24 55 1

The table for f−1 is obtained by interchanging the two columns in thetable above:

x f−1(x)3 14 22 35 41 5

Here is the graph of f :


1 2 3 4 5x

1

2

3

4

5

y

Here is the graph of f−1, which is obtained by reflecting the graphabove through the line x = y :

1 2 3 4 5x

1

2

3

4

5

y

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Solutions to Chapter Review Questions, Chapter 1park/Fall2013/precalculus/ch1_sol.pdf · Instructor’s Solutions Manual, Chapter 1 Review Question 1 Solutions to Chapter Review Questions,