mao (tm23477) – Hw 14 – gualdani – (57180)
1
This printout should have 18 questions.
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the next column or page – fnd all choices
beFore answering.
001
10.0 points
±ind the value oF
f
(1) when
f
(
x
) = 5 sin
−
1
x
+ tan
−
1
x .
1.
f
(1) =
11
4
π
correct
2.
f
(1) =
9
4
π
3.
f
(1) =
7
4
π
4.
f
(1) =
5
4
π
5.
f
(1) =
3
4
π
Explanation:
Since
sin
−
1
(1) =
π
2
,
tan
−
1
(1) =
π
4
,
we see that
f
(1) =
p
5
2
+
1
4
P
π
=
11
4
π
.
002
10.0 points
±ind the exact value oF
sin
−
1
p
√
3
2
P
in the interval
p
0
,
π
2
P
.
1.
π
6
2.
π
5
3.
π
7
4.
π
4
5.
π
3
correct
Explanation:
We have to fnd
x
so that
sin
x
=
√
3
2
,
0
< x <
π
2
.
Known trig values thus ensure that
x
=
π
3
.
003
10.0 points
SimpliFy the expression
y
= sin
±
tan
−
1
x
√
7
²
by writing it in algebraic Form.
1.
y
=
x
x
2
+ 7
2.
y
=
√
7
√
x
2
+ 7
3.
y
=
x
√
x
2

7
4.
y
=
x
√
x
2
+ 7
correct
5.
y
=
√
x
2
+ 7
√
7
Explanation:
The given expression has the Form
y
= sin
θ
where
tan
θ
=
x
√
7
,

π
2
< θ <
π
2
.
To determine the value oF sin
θ
given the value
oF tan
θ
, we can apply Pythagoras’ theorem
to the right triangle
√
7
x
θ
r
x
2
+ 7
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2
From this it follows that
y
= sin
θ
=
x
√
x
2
+ 7
.
Alternatively, we can use the trig identity
csc
2
θ
= 1 + cot
2
θ
to determine sin
θ
.
004
10.0 points
Simplify the expression
f
(
x
) = sin
(
tan
−
1
x
)
.
1.
f
(
x
) =
r
1 +
x
2
2.
f
(
x
) =
1
√
1

x
2
3.
f
(
x
) =
r
1

x
2
4.
f
(
x
) =
x
√
1 +
x
2
correct
5.
f
(
x
) =
x
√
1

x
2
6.
f
(
x
) =
1
√
1 +
x
2
Explanation:
By de±nition
f
(
x
) = sin
θ
where
x
= tan
θ
.
But then by Pythagoras,
sin
θ
=
x
√
1 +
x
2
.
Consequently,
f
(
x
) =
x
√
1 +
x
2
.
005
10.0 points
Determine if
lim
x
→∞
sin
−
1
p
3 + 2
x
4 + 2
x
P
exists, and if it does, ±nd its value.
1.
limit does not exist
2.
limit = 0
3.
limit =
π
3
4.
limit =
π
2
correct
5.
limit =
π
6
6.
limit =
π
4
Explanation:
Since
lim
x
3 + 2
x
4 + 2
x
= 1
,
we see that
lim
x
sin
−
1
p
3 + 2
x
4 + 2
x
P
exists, and that the
limit = sin
−
1
1 =
π
2
.
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 Fall '08
 schultz
 Calculus, Trigonometry, Derivative, Right triangle

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