Answers and Explanations 179
Answers
101–200
153.
f x
x
( )
cos
= −
−
(
)
2
1
2
4
π
For the function f x
A
B x C
B
D
( )
cos
=
−
( )
+ , the amplitude is A , the period is 2π
B
, the
phase shift is C
B
, and the midline is y = D. The function has a period of 4π, so one possible value of B is B = 1
2
. Think of the graph as a cosine graph that’s flipped about the
x-axis and shifted to the right; that means there’s a phase shift of π
2
to the right, so
C
B
= π
2
. The line y = 0 is the midline of the function, so D = 0. Last, the amplitude is 2, so
you can use A = –2 because the function increases for values of x that are slightly
larger than π
2
. Therefore, the following function describes the graph:
f x
x
x
( )
cos
cos
= −
−
( )
= −
−
(
)
2
1
2
2
2
1
2
4
π
π
Note that you can also use other functions to describe this graph.
154.
f x
x
( )
cos
= −
−
(
) −
2
1
2
4
1
π
For the function f x
A
B x C
B
D
( )
cos
=
−
( )
+ , the amplitude is A , the period is 2π
B
, the
phase shift is C
B
, and the midline is y = D. The function has a period of 4π, so one
possible value of B is B = 1
2
. Think of the graph as a cosine graph that’s flipped about
the x-axis and shifted to the right and down; there’s a phase shift of π
2
to the right so
that C
B
= π
2
. The line y = –1 is the midline of the function, so D = –1. Last, the amplitude
is 2, so you can use A = –2 because the function increases for values of x that are
slightly larger than π
2
. Therefore, the following function describes the given graph.
f x
x
x
( )
cos
cos
= −
−
( )
−
= −
−
(
) −
2
1
2
2
1
2
1
2
4
1
π
π
Note that you can also use other functions to describe this graph.
155.
π
3
To evaluate sin
−
=
1
3
2
x, find the solution of 3
2
= sin x, where − ≤ ≤
π
π
2
2
x
. Because
sin π
3
3
2
=
, you have sin
−
=
1
3
2
3
π .
Answers
101–200
153.
f x
x
( )
cos
= −
−
(
)
2
1
2
4
π
For the function f x
A
B x C
B
D
( )
cos
=
−
( )
+ , the amplitude is A , the period is 2π
B
, the
phase shift is C
B
, and the midline is y = D. The function has a period of 4π, so one possible value of B is B = 1
2
. Think of the graph as a cosine graph that’s flipped about the
x-axis and shifted to the right; that means there’s a phase shift of π
2
to the right, so
C
B
= π
2
. The line y = 0 is the midline of the function, so D = 0. Last, the amplitude is 2, so
you can use A = –2 because the function increases for values of x that are slightly
larger than π
2
. Therefore, the following function describes the graph:
f x
x
x
( )
cos
cos
= −
−
( )
= −
−
(
)
2
1
2
2
2
1
2
4
π
π
Note that you can also use other functions to describe this graph.
154.
f x
x
( )
cos
= −
−
(
) −
2
1
2
4
1
π
For the function f x
A
B x C
B
D
( )
cos
=
−
( )
+ , the amplitude is A , the period is 2π
B
, the
phase shift is C
B
, and the midline is y = D. The function has a period of 4π, so one
possible value of B is B = 1
2
. Think of the graph as a cosine graph that’s flipped about
the x-axis and shifted to the right and down; there’s a phase shift of π
2
to the right so
that C
B
= π
2
. The line y = –1 is the midline of the function, so D = –1. Last, the amplitude
is 2, so you can use A = –2 because the function increases for values of x that are
slightly larger than π
2
. Therefore, the following function describes the given graph.
f x
x
x
( )
cos
cos
= −
−
( )
−
= −
−
(
) −
2
1
2
2
1
2
1
2
4
1
π
π
Note that you can also use other functions to describe this graph.
155.
π
3
To evaluate sin
−
=
1
3
2
x, find the solution of 3
2
= sin x, where − ≤ ≤
π
π
2
2
x
. Because
sin π
3
3
2
=
, you have sin
−
=
1
3
2
3
π .