Transcript ConcepTest - Piscataway High School

The amplitude and period of the graph of
the periodic function in Figure 1.19 are
ConcepTest • Section 1.5 • Question 1
ANSWER
(d). The maximum value of the function is 5,
and the minimum value is –1, so the
5  ( 1)
amplitude is 2 = 3. The function repeats
itself after 2 units, so the period is 2.
COMMENT:
Point out that the function is oscillating about the line y = 2.
Have the students find a formula for the function shown in
Figure 1.19.
ConcepTest • Section 1.5 • Answer 1
The amplitude and period of the graph of
the periodic function in Figure 1.20 are
ConcepTest • Section 1.5 • Question 2
ANSWER
(c). The maximum value of the function is 5,
5 1
and the minimum 1, so the amplitude is 2 = 2.
The function repeats itself after ½ unit, so the
period is ½.
COMMENT:
It is easiest to find the period using the extreme values of the
function.
ConcepTest • Section 1.5 • Answer 2
Which of the following could describe the
x 
graph in Figure 1.21.
(a ) y  3 sin   
2
(b)
(c )


y  3 sin  2 x  
2

y  3 cos( 2 x)
 x
(d ) y  3 cos 
2
(e) y  3 sin( 2 x)
 x
( f ) y  3 sin  
2
ConcepTest • Section 1.5 • Question 3
2
ANSWER
(a) and (d). Note that (b), (c) and (e) have
period π, with the rest having period 4π.
Answer (f) has y(0) = 0. (a) and (d) could
describe the graph.
COMMENT:
The fact that the same graph may have more than one analytic
representation could be emphasized here.
ConcepTest • Section 1.5 • Answer 3
Figure 1.22 shows the graph of which of
the following functions?
(a)
(b)
(c )
(d )
( e)
ConcepTest • Section 1.5 • Question 4
y  cos( x   / 6)
y  cos( x   / 6)
y  sin( x   / 6)
y  sin( x   / 6)
None of these
ANSWER
(b)
COMMENT:
You could ask what the graphs of the other choices look like.
ConcepTest • Section 1.5 • Answer 4
Which of the graphs are those of sin(2t) and sin(3t)?
(a)
(b)
(c)
(d)
(I) = sin(2t) and (II) = sin(3t)
(I) = sin(2t) and (III) = sin(3t)
(II) = sin(2t) and (III) = sin(3t)
(III) = sin(2t) and (IV) = sin(3t)
ConcepTest • Section 1.5 • Question 5
ANSWER
(b). The period is the time needed for the
function to execute one complete cycle. For
sin(2t), this will be π and for sin(3t), this will
be 2π/3.
COMMENT:
You could ask what equations describe the graphs of (II) and
(IV). This question is the same as Question 5 in Section 1.3.
ConcepTest • Section 1.5 • Answer 5
Consider a point on the unit circle (shown in Figure 1.23)
starting at an angle of zero and rotating counterclockwise at a
constant rate. Which of the graphs represents the horizontal
component of this point as a function of time.
ConcepTest • Section 1.5 • Question 6
ANSWER
(b)
COMMENT:
Use this question to make the connection between the unit circle
and the graphs of trigonometric functions stronger.
Follow-up Question. Which of the graphs represents that of
the vertical component of this point as a function of time?
Answer. (c).
ConcepTest • Section 1.5 • Answer 6
Which of the following is the approximate value for
the sine and cosine of angles A and B in Figure 1.24?
(a)
(b)
(c)
(d)
sin A ≈ 0.5, cos A ≈ 0.85, sin B ≈ – 0.7, cos B ≈ 0.7
sin A ≈ 0.85, cos A ≈ 0.5, sin B ≈ – 0.7, cos B ≈ 0.7
sin A ≈ 0.5, cos A ≈ 0.85, sin B ≈ 0.7, cos B ≈ 0.7
sin A ≈ 0.85, cos A ≈ 0.5, sin B ≈ – 0.7, cos B ≈ 0.7
ConcepTest • Section 1.5 • Question 7
ANSWER
(b)
COMMENT:
You could use this question to make the connection between the
unit circle and the trigonometric functions stronger.
You could also identify other points on the circle and ask for
values of sine or cosine.
ConcepTest • Section 1.5 • Answer 7
Figure 1.25 shows the graph of which of
the following functions?
(a)
(b)
(c )
(d )
( e)
ConcepTest • Section 1.5 • Question 8
y  sin( x   / 3)
y  cos( x   / 3)
y  sin( x   / 3)
y  cos( x   / 3)
None of these
ANSWER
(c)
COMMENT:
You could ask what the graphs of the other choices look like.
ConcepTest • Section 1.5 • Answer 8
Which of the following are properties of both
x
y  arcsin x and y 
?
2
1 x
(a) y(0) = 0
(b) Increasing everywhere
(c) Concave down for x < 0
(d) Domain: [–1, 1]
(e) Range: (– ∞, ∞)
ConcepTest • Section 1.5 • Question 9
ANSWER
(a), (b), and (c). The domain of y 
x
1 x2
is
(–1, 1) and the range of arcsin x is [– π/2, π/2].
COMMENT:
You may want to have the students sketch graphs of both functions.
Follow-up Question. What is the concavity for each function when
x > 0?
Answer. Both functions are concave up when x > 0.
ConcepTest • Section 1.5 • Answer 9
Which of the following have the same
domain?
(a)
(b)
(c )
(d )
arcsin x
arctan x
ln x
x
e
( e)
1 x
ConcepTest • Section 1.5 • Question 10
2
ANSWER
(a) and (e), also (b) and (d). Note that the domain
of arcsin x and 1 - x is [1,1], the domain of
x
arctan x and e is (, ).
2
COMMENT:
Students could be thinking of the range of sin x for (a) and the range of
tan x for (b).
Follow-up Question. Does changing (c) to ln |x| affect the
answer?
Answer. No, because the domain of ln |x| is all real numbers except
x = 0.
ConcepTest • Section 1.5 • Answer 10
Which of the following have the same
range?
(a)
(b)
(c )
(d )
arcsin x
arctan x
ln x
x
e
( e)
1 x
ConcepTest • Section 1.5 • Question 11
2
ANSWER
  
(a) and (b). The range of (a) and (b) is  , .
 2 2
COMMENT:
Follow-up Question. Does changing (c) to |ln x| affect the
answer?
Answer. No, the range of |ln x| is [0, ∞) while the range of ex
is (0, ∞).
ConcepTest • Section 1.5 • Answer 11
If y = arcsin x, then cos y =
1 x
1
(a)
(c )
( e)
2
x 1
2
(b)
(d )
1
x
ConcepTest • Section 1.5 • Question 12
x 1
1
2
1 x
2
ANSWER
(a). If y  arcsin x, then 

2
2
2
cos y  1  sin y  1  x .
 y

2
giving
COMMENT:
You could also solve this problem by drawing a right triangle
with hypotenuse of 1 and labeling the side opposite angle y as x.
ConcepTest • Section 1.5 • Answer 12
If y = arcsin x, then tan y =
1 x2
x
x
(a)
(c )
( e)
x2 1
(b)
(d )
1
x
ConcepTest • Section 1.5 • Question 13
x2 1
x
x
1 x2
ANSWER
(d). If y  arcsin x, then 

2
 y

2
giving
sin y
sin y
x
tan y 


.
2
2
cos y
1  sin y
1 x
COMMENT:
You could also solve this problem by drawing a right triangle
with hypotenuse of 1 and labeling the side opposite angle y as x.
ConcepTest • Section 1.5 • Answer 13
If y = arctan x, then cos y =
1 x
x
1
(a)
(c )
( e)
2
x2 1
(b)
(d )
1
x
ConcepTest • Section 1.5 • Question 14
x
x2 1
1 x2
ANSWER
(c). If y  arctan x, then 
1
tan y  1 
, cosy 
2
cos y
2

2
 y

2
. Because
1

2
tan y  1
1
.
2
x 1
COMMENT:
You could also solve this problem by drawing a right triangle
with the side opposite angle y labeled as x and the side adjacent
to angle y labeled as 1.
ConcepTest • Section 1.5 • Answer 14
If y = arctan x, then sin y =
1 x
x
1
(a)
(c )
( e)
1
x
2
x2 1
(b)
(d )
ConcepTest • Section 1.5 • Question 15
x
x 1
1 x2
2
ANSWER
(b). If y  arctan x, then 

2
 y

2
and
1
sin y  tan y cos y. Because tan y  1 
,
2
cos y
2
cosy 
1
and sin y 
2
tan y  1
x
x 1
2
.
COMMENT:
You could also solve this problem by drawing a right triangle
with the side opposite angle y labeled as x and the side adjacent
to angle y labeled as 1.
ConcepTest • Section 1.5 • Answer 15