Download - Calculus I 2010 Midterm Solutions
Calculus I - Spring 2010
Midterm Exam, March 3, 2010
In the following problems you are required to show all your work and provide the necessary explana-
tions everywhere to get full credit.
1. Find the domain of the function f(x) =x + 1
x2 − 3.
Solution: This function does not exist at points where the denominator is 0, that is at x = −√
3and x =
√3. Therefore the domain is
{
x | x 6= ±√
3}
or(
−∞,−√
3)
∪(
−√
3,√
3)
∪(√
3,∞)
2. Let f(x) =x − 4
|4 − x| . Find the value of each limit, if it exists. (If it does not exist, enter
NONE.)
(a) limx→4−
f(x) = −1
Solution: We have
limx→4−
x − 4
|4 − x| = limx→4−
x − 4
4 − x= lim
x→4−
x − 4
−(x − 4)= −1
(b) limx→4+
f(x) = 1
Solution: We have
limx→4+
x − 4
|4 − x| = limx→4+
x − 4
−(4 − x)= lim
x→4+
x − 4
x − 4= 1
(c) limx→4
f(x) does not exist
Solution: Since
limx→4−
x − 4
|4 − x| 6= limx→4+
x − 4
|4 − x|it follows that lim
x→4f(x) does not exist.
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3. Use the given graph of f(x) to find the value of each limit, if it exists. (If it does not exist,enter NONE.)
(a) limx→−4−
f(x) = 2
(b) limx→−4+
f(x) = 2
(c) limx→−4
f(x) = 2
(d) limx→1−
f(x) = 4
(e) limx→1+
f(x) = −2
(f) limx→1
f(x) NONE, since limx→1−
f(x) 6= limx→1+
f(x)
(g) limx→6−
f(x) = 5
(h) limx→6+
f(x) = 5
(i) limx→6
f(x) = 5
2
4. For each of the following, decide if the limit exists. If it does, find the limit. If it does not,decide also if the limit is ∞, −∞, or neither.
(a) limx→
√5
x2 − 5
x −√
5= 2
√5
Solution: We have
limx→
√5
x2 − 5
x −√
5= lim
x→√
5
(x −√
5)(x +√
5)
x −√
5= lim
x→√
5(x +
√5) =
√5 +
√5 = 2
√5
(b) limx→1
x2 − 4x + 3
(x − 1)2does not exist and neither ∞ nor −∞
Solution: We have
limx→1
x2 − 4x + 3
(x − 1)2= lim
x→1
(x − 1)(x − 3)
(x − 1)2= lim
x→1
x − 3
x − 1
Note that
limx→1−
x − 3
x − 1= ∞ and lim
x→1+
x − 3
x − 1= −∞
Since
limx→1−
x − 3
x − 16= lim
x→1+
x − 3
x − 1
it follows that limx→1
x2 − 4x + 3
(x − 1)2does not exist and neither ∞ nor −∞.
(c) limx→0
sin(8x)
tan(2x)= 4
Solution: We have
limx→0
sin(8x)
tan(2x)= lim
x→0
sin(8x)sin(2x)cos(2x)
= limx→0
(
cos(2x) · sin(8x)
sin(2x)
)
= limx→0
cos(2x)︸ ︷︷ ︸
=1
· limx→0
sin(8x)
sin(2x)=
= limx→0
sin(8x)
sin(2x)= lim
x→0
sin(8x)8x
· 8xsin(2x)
2x· 2x
=
[
limu→0
sin u
u= 1
]
= limx→0
1 · 8x1 · 2x = lim
x→0
8
2= 4
(d) limx→∞
5x2 − x + 1
3x2 + 2=
5
3
Solution: We have
limx→∞
5x2 − x + 1
3x2 + 2= lim
x→∞
5x2−x+1x2
3x2+2x2
= limx→∞
5x2
x2 − xx2 + 1
x2
3x2
x2 + 2x2
= limx→∞
5 − 1x
+ 1x2
3 + 2x2
=5 − 0 + 0
3 + 0=
5
3
3
(e) limx→∞
(√
x2 + 2x −√
x2 + x) =1
2
Solution: We have
limx→∞
(√
x2 + 2x −√
x2 + x) = limx→∞
√x2 + 2x −
√x2 + x
1=
= limx→∞
(√
x2 + 2x −√
x2 + x)(√
x2 + 2x +√
x2 + x)√x2 + 2x +
√x2 + x
= limx→∞
(x2 + 2x) − (x2 + x)√x2 + 2x +
√x2 + x
=
= limx→∞
x√x2 + 2x +
√x2 + x
= limx→∞
xx√
x2+2xx
+√
x2+xx
= limx→∞
1√
x2+2x√x2
+√
x2+x√x2
=
= limx→∞
1√
x2+2xx2 +
√x2+x
x2
= limx→∞
1√
x2
x2 + 2xx2 +
√x2
x2 + xx2
= limx→∞
1√
1 + 2x
+√
1 + 1x
=
=1√
1 + 0 +√
1 + 0=
1
2
(f) limx→1
1
(1 − x)2= ∞ (does not exist)
Solution: We have
limx→1−
1
(1 − x)2= lim
x→1+
1
(1 − x)2= ∞
therefore
limx→1
1
(1 − x)2= ∞ (does not exist)
5. Use the Squeeze Theorem to find the limit limx→∞
sin x
x.
Solution: We have
−1 ≤ sin x ≤ 1 =⇒ −1
x≤ sin x
x≤ 1
x
Since
limx→∞
(
−1
x
)
= limx→∞
1
x= 0
it follows that
limx→∞
sin x
x= 0
by the Squeeze Theorem.
REMARK: Do not confuse limx→∞
sin x
x= 0 and lim
x→0
sin x
x= 1.
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6. Find the horizontal and vertical asymptotes of the function f(x) =x2 + 4x − 5
x2 + x − 2.
Solution:
(a) We have
limx→−2−
x2 + 4x − 5
x2 + x − 2= lim
x→−2−
(x − 1)(x + 5)
(x − 1)(x + 2)= lim
x→−2−
x + 5
x + 2= −∞
Therefore x = −2 is the vertical asymptote.
REMARK: Note that x = 1 is NOT a vertical asymptote, since
limx→1−
x2 + 4x − 5
x2 + x − 2= lim
x→1−
(x − 1)(x + 5)
(x − 1)(x + 2)= lim
x→1−
x + 5
x + 26= ±∞
Similarly,
limx→1+
x2 + 4x − 5
x2 + x − 2= lim
x→1+
(x − 1)(x + 5)
(x − 1)(x + 2)= lim
x→1+
x + 5
x + 26= ±∞
(b) We have
limx→±∞
x2 + 4x − 5
x2 + x − 2= lim
x→±∞
x2+4x−5x2
x2+x−2x2
= limx→±∞
x2
x2 + 4xx2 − 5
x2
x2
x2 + xx2 − 2
x2
= limx→±∞
1 + 4x− 5
x2
1 + 1x− 2
x2
=1 + 0 − 0
1 + 0 − 0= 1
Therefore y = 1 is the horizontal asymptote.
7. Let f(x) = sin x.
(a) Find an equation of the tangent line to f(x) at the point (0, 0).
Solution: We have
f ′(x) = (sin x)′ = cos x =⇒ f ′(0) = cos 0 = 1
therefore slope m = 1. It follows that an equation of the tangent line to f(x) at the point (0, 0)is
y − 0 = 1 · (x − 0) =⇒ y = x
(b) Find an equation of the normal line to f(x) at the point (0, 0).
Solution: Since
mnormal = − 1
mtangent= −1
1= −1
it follows that an equation of the normal line to f(x) at the point (0, 0) is
y − 0 = −1 · (x − 0) =⇒ y = −x
5
8. Find the derivative of the function f(x) =2√x
using the definition of derivative.
Solution: We have
f ′(x) = limh→0
f(x + h) − f(x)
h= lim
h→0
2√x+h
− 2√x
h= lim
h→0
2√
x−2√
x+h√x√
x+h
h= lim
h→0
2√
x − 2√
x + h
h√
x√
x + h=
= limh→0
(2√
x − 2√
x + h)(2√
x + 2√
x + h)
h√
x√
x + h(2√
x + 2√
x + h)= lim
h→0
4x − 4(x + h)
h√
x√
x + h(2√
x + 2√
x + h)=
= limh→0
−4h
h√
x√
x + h(2√
x + 2√
x + h)= lim
h→0
−4√x√
x + h(2√
x + 2√
x + h)=
=−4√
x√
x + 0(2√
x + 2√
x + 0)=
−4√x√
x(2√
x + 2√
x)=
−4
x · 4√x= − 1
x√
x= −x−3/2
9. Let f(x) = sin(sin x). Find f ′(x) and f ′′(x).
Solution: We have
f ′(x) = [sin(sin x)]′ = cos(sin x) · (sin x)′ = cos(sin x) · cos x
andf ′′(x) = [cos(sin x) · cos x]′ = [cos(sin x)]′ · cos x + cos(sin x) · (cos x)′ =
= − sin(sin x) · (sin x)′ · cos x + cos(sin x) · (− sin x) =
= − sin(sin x) · cos x · cos x − cos(sin x) · sin x =
= − sin(sin x) · cos2 x − cos(sin x) · sin x
10. Use limits to explain why the function
f(x) =
x2 − x + 2 if x ≤ 2
1
x + 2if x > 2
is discontinuous at x = 2.
Solution: We have
limx→2−
f(x) = limx→2−
(x2 − x + 2) = 22 − 2 + 2 = 4 and limx→2+
f(x) = limx→2+
1
x + 2=
1
2 + 2=
1
4
Sincelim
x→2−f(x) 6= lim
x→2+f(x)
it follows that f(x) is discontinuous at x = 2.
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11. Differentiate the following functions:
(a) f(x) = 7√
2x
Solution: We have
f ′(x) = (7√
2x)′ = 7(√
2x)′ = 7((2x)1/2)′ = 7 · 1
2(2x)1/2−1 · (2x)′ =
= 7 · 1
2(2x)−1/2 · 2 = 7 · (2x)−1/2 =
7√2x
(b) f(x) = x(1 − x)8
Solution: We have
f ′(x) = [x(1 − x)8]′ = x′(1 − x)8 + x[(1 − x)8]′ = (1 − x)8 + x · 8(1 − x)7 · (1 − x)′ =
= (1 − x)8 + 8x(1 − x)7 · (−1) = (1 − x)8 − 8x(1 − x)7 = (1 − x)7(1 − 9x)
(c) f(x) =1 − 3
√x
5x3
Solution: We have
f ′(x) =
(1 − 3
√x
5x3
)′
=
(1 − 3x1/2
5x3
)′
=
(1
5x3− 3x1/2
5x3
)′
=
(1
5x−3 − 3
5x1/2−3
)′
=
=
(1
5x−3 − 3
5x−5/2
)′
=1
5
(x−3
)′ − 3
5
(x−5/2
)′=
1
5· (−3)x−3−1 − 3
5·(
−5
2
)
x−5/2−1 =
= −3
5x−4 +
3
2x−7/2
(d) f(x) =2 cos x − 4 sin x
x + 1
Solution: We have
f ′(x) =
(2 cos x − 4 sin x
x + 1
)′
=(2 cos x − 4 sin x)′ · (x + 1) − (2 cos x − 4 sin x) · (x + 1)′
(x + 1)2=
=(−2 sin x − 4 cosx) · (x + 1) − 2 cos x + 4 sinx
(x + 1)2
(e) f(x) = sec2(1 +√
5 − x)
Solution: We have
f ′(x) = [sec2(1 +√
5 − x)]′ = 2 sec(1 +√
5 − x) · (sec(1 +√
5 − x))′ =
= 2 sec(1 +√
5 − x) · sec(1 +√
5 − x) tan(1 +√
5 − x) · (1 +√
5 − x)′ =
= 2 sec(1 +√
5 − x) · sec(1 +√
5 − x) tan(1 +√
5 − x) · 1
2(5 − x)−1/2 · (−1) =
= − sec2(1 +√
5 − x) · tan(1 +√
5 − x) · (5 − x)−1/2
7