Engineering Math I – Fall 2014
Quiz #9 Solutions
Problem 1 (4 pts). Find the exact value of sin−1 (2).
(a) 2
(b)
5π
2
(c)
π
8
(d) None of the above.
Solution. The correct answer is (d). This is because sin−1 (2) is the unique angle θ between
−π/2 and π/2 such that sin θ = 2. But −1 ≤ sin φ ≤ 1 for all angles φ, so this is
impossible.
Problem 2 (4 pts). What is the domain of cos−1 (2x − 1)?
(a) All real numbers x.
(b) [−1, 1]
(c) [0, 1]
(d) None of the above.
Solution. The correct answer is (c). Since 1 ≤ sin θ ≤ 1 for all θ, we must have −1 ≤
2x − 1 ≤ 1. That is, 0 ≤ 2x ≤ 2, which means that 0 ≤ x ≤ 1.
MATH 151:549-551 – Fall 2014 Quiz #9 Solutions
2
PART II: Free response. Read each problem carefully and work it out in the space
provided. Circle your final answer.
Problem 3 (6 pts). Find the derivative of each function.
√
(a) (2 pts) G(t) = cos−1 2t − 1
√
Solution. The derivative of the inverse cosine function is −1/ 1 − x2 . Thus, by
the chain rule, we have
1
1
− 12
·2
G0 (t) = − q
·
(2t
−
1)
√
2
2
1 − ( 2t − 1)
1
= −p
√
1 − (2t − 1) 2t − 1
1
√
= −√
2 − 2t 2t − 1
1
= −√
.
6t − 4t2 − 2
(b) (2 pts) y = tan−1 (sin x)
Solution. The derivative of the inverse tangent function is 1/(1 + x2 ). Thus, by
the chain rule, we have
y0 =
1
cos x
· cos x =
.
2
1 + sin x
1 + sin2 x
(c) (2 pt) y = sin−1 (x2 − 2x).
√
Solution. The derivative of the inverse sin function is 1/ 1 − x2 . Thus, by the
chain rule, we have
1
y0 = p
· 2x − 2
1 − (x2 − 2x)2
2x − 2
= p
1 − (x4 − 4x3 + 4x2 )
2(x − 1)
= p
−(x − 1)2 (x2 − 2x − 1)
2
= √
.
2
−x + 2x + 1
MATH 151:549-551 – Fall 2014 Quiz #9 Solutions
3
Problem 4 (6 pts). Find the limit.
(a) (2 pts) limx→2−
√ln x
2−x
−
Solution. The
√ limit as x → 2 of ln x is a positive constant, whereas the limit as
−
x → 2 of 2 − x is 0. Thus
ln 2
ln x
=
= ∞.
lim− √
x→2
0
2−x
(b) (2 pts) limx→0
2x−sin−1 x
2x+tan−1 x
Solution. Observe limx→0
L’Hôpital’s Rule to get
2x−sin−1 x
2x+tan−1 x
is an indeterminate of the form 00 . So we apply
1
2 − √1−x2
2−1
1
2x − sin−1 x
=
lim
=
= .
lim
1
−1
x→0 2x + tan
x x→0 2 + 1+x2
2+1
3
3
(c) (2 pt) limx→∞ x5 e−x .
3
Solution. Observe that limx→∞ x5 e−x is an indeterminate of the form ∞ · 0. We
3
3
3
rewrite x5 e−x as x5 /ex , and now we have limx→∞ x5 /ex as an indeterminate of
∞
. Hence we may apply L’Hôpital’s Rule to get
the form ∞
3
5x4
3.
x→∞ 3x2 ex
lim x5 /ex = lim
x→∞
∞
, so we apply L’Hôpital’s
The right-hand side is still an indeterminate of the form ∞
Rule a second time to get (using the product rule in the denominator)
5x4
20x3
=
lim
3
3.
x→∞ 3x2 ex
x→∞ (6x + 9x4 )ex
lim
Again, the right-hand side is an indeterminate of the form
rule three more times to get
20x3
3
x→∞ (6x2 + 9x4 )ex
lim
∞
.
∞
We apply L’Hôpital’s
60x2
3
x→∞ (6 + 54x3 + 27x6 )ex
120x
= lim
3
2
x→∞ (180x + 324x5 + 81x8 )ex
120
= lim
3 = 0.
4
x→∞ (360x + 2160x + 1620x7 + 243x10 )ex
=
lim
MATH 151:549-551 – Fall 2014 Quiz #9 Solutions
4
Bonus Problem (2 pts). State the domains of each function and its derivative from
Problem 3.
Solution.
√
[ 21 , 1]. Note that we must have 0 ≤ 2t − 1 ≤ 1, which implies
(a)
that 12 ≤ t ≤ 1.
0
Domain of G (t): ( 12 , 1). Note that 6t − 4t2 − 2 = −2(t − 1)(2t − 1), and that
6t − 4t2 − 2 is positive for 21 ≤ t ≤ 1.
Domain of y:
(−∞, ∞), since both sin x and tan−1 x are defined all on of
(b)
the real numbers.
0
Domain of y :
(−∞, ∞) since cos x and sin x are defined on all real numbers,
and √
1 + sin2√
x is never 0.
Domain of y:
[1 − 2, 1 + 2]. We must have −1 ≤ x2 − 2x ≤ 1. Note that
(c)
x2 − 2x + 1 = (x − 1)2 and that
√
√
x2 − 2x − 1 = (x − 2 − 1)(x + 2 − 1).
Domain of G(t):
Hence
1−
Domain of y 0 :
√
√
2 ≤ x ≤ 1 + 2.
√
√
2
(1 − 2, 1 + 2), since√we must have
√ −x + 2x + 1 > 0, and
2
−x + 2x + 1 = (−x + 2 + 1)(x + 2 − 1).