Zachary Scherr
1
Math 103
Exam 1 Prep
Practice Problems
1. Simplify each of the following expressions:
√ (a) cos−1 −2 2
Solution: We must find an angle θ in the interval [0, π] with cos(θ) =
so
√
− 2
3π
π
= cos
= cos π −
2
4
4
√ so cos−1 −2 2 = 3π
4 .
(b)
1
8
3π
6
cos2
+
1
8
sin2
3π
6
√
− 2
2 .
Now, cos
π
4
√
=
Solution: We have
1
cos2
8
3π
6
1
+ sin2
8
3π
6
1
=
8
cos
2
3π
6
+ sin
2
3π
6
=
1
.
8
(c) e−2 ln( 7 )
5
Solution: We have
e−2 ln( 7 )
5
=
=
1
5
e2 ln( 7 )
1
2
ln 5
e (7)
1
=
e
=
=
(d)
ln( 25
49 )
1
25
49
49
25
1
ln( 16
)
ln(2)+ln(2)
Solution:
1
ln 16
ln(2) + ln(2)
=
− ln (16)
ln(2) + ln(2)
ln(16)
−
ln(4)
− log4 (16)
=
−2
=
=
since 16 = 42 .
2. For each of the following parts, compute the corresponding limit or explain why it doesn’t exist.
2
2
Zachary Scherr
Math 103
Exam 1 Prep
(a)
lim p
x→∞
x+4
4x2
+ sin(x) + 1
Solution:
lim p
x→∞
x+4
lim p
·
4x2 + sin(x) + 1
=
4x2 + sin(x) + 1
x→∞
1 + x4
lim q
x→∞
4 + sin(x)
x2 +
=
=
x→∞
1
x2
1
√
4
1
.
2
=
It’s worth remarking that lim
1
x
1
x
x+4
sin(x)
= 0 since
x2
−1
x2
≤
sin(x)
x2
≤
1
x2 .
(b)
lim
1
h+2
−
1
2
h
h→0
Solution:
lim
h→0
1
h+2
−
1
2
h
=
=
lim
2
2h+4
h+2
2h+4
h
h→0
lim
−
−h
2h+4
h
−1
= lim
h→0 2h + 4
1
= − .
4
h→0
(c)
x3 − 6x2 + 11x − 6
x→1
x2 − 3x + 2
lim
Solution: Plugging x = 1 in to the numerator and denominator gives 00 . So we should try to
factor. We have
x3 − 6x2 + 11x − 6
x→1
x2 − 3x + 2
lim
(x − 1)(x2 − 5x + 6)
x→1
(x − 1)(x − 2)
2
(x − 5x + 6)
= lim
x→1
x−2
2
1 −5·1+6
=
1−2
= −2.
=
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lim
Zachary Scherr
Math 103
Exam 1 Prep
(d)
x−2
x→2 |x − 2|
lim
Solution: The limit does not exist. We have
lim+
x→2
yet
lim
x→2−
x−2
x−2
= lim+
=1
|x − 2| x→2 x − 2
x−2
x−2
= lim
= −1.
|x − 2| x→2− 2 − x
(e)
4
lim x cos
x→0
3x2 + 2
x+1
2 +2
Solution: The function f (x) = x4 cos 3xx+1
is continuous at 0, so this limit is just f (0)
which is 0. If instead this question had been more complicated, like
1
lim x4 cos
,
x→0
x3
we could have used the squeeze theorem. Since −1 ≤ cos(θ) ≤ 1 for any angle θ, we get
1
−x4 ≤ x4 cos
≤ x4
x3
and
lim −x4 = lim x4 = 0
x→0
x→0
showing that
4
lim x cos
x→0
1
x3
= 0.
3. Suppose f and g are two functions given by the following tables
x
f (x)
x
g(x)
−1
3
0
2
−1
−2
2
0
0
0
3
−1
2
3
2
3
2
5
−3
5
4
6
2
6
5
(a) Does f (x) have an inverse? Why or why not.
Solution: f (x) does not have an inverse because f (0) = f (6).
(b) Compute g ◦ f (−1)
Solution:
g ◦ f (−1) = g(f (−1)) = g(3) = 2.
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Zachary Scherr
Math 103
Exam 1 Prep
(c) Compute g −1 (4)
Solution:
g −1 (4) = 5.
(d) What is the domain of f ◦ g?
Solution: The domain of f (x) is {−1, 0, 2, 3, 5, 6}, so we just need to see which x values have
g(x) in that set. These are:
{0, 3, 6}.
4. Consider the function
f (x) =
x2 − 4x + 3
.
3x2 − 15x + 12
(a) Find the horizontal asymptote of f (x).
Solution: Since
x2 − 4x + 3
x2 − 4x + 3
1
=
lim
= ,
x→−∞ 3x2 − 15x + 12
x→∞ 3x2 − 15x + 12
3
lim
we have that y =
1
3
is the only horizontal asymptote of this rational function.
(b) Find all the vertical asymptotes of f (x).
Solution: We must factor. Since
x2 − 4x + 3 = (x − 1)(x − 3)
and
3x2 − 15x + 12 = 3(x2 − 5x + 4) = 3(x − 1)(x − 4)
we see there is a vertical asymptote only at x = 4.
(c) For which values of x is f (x) continuous?
Solution: Since this is a rational function, it is continuous everywhere on its domain. Thus is
is continuous everywhere except x = 1, x = 4. In interval notation:
(−∞, 1) ∪ (1, 4) ∪ (4, ∞).
5. Suppose I invest 1000 dollars in a stock whose value increases 2% per year.
(a) Write down the formula for a function P (t), giving the amount of value of the stock after t years.
Solution:
P (t) = 1000(1.02)t .
(b) After how many years will the original value of the stock triple? You do not have to simplify your
answer.
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Zachary Scherr
Math 103
Exam 1 Prep
Solution: We must solve 3000 = 1000(1.02)t for t. Dividing through by 1000 gives 3 = (1.02)t
and taking natural log of both sides gives
ln(3) = t ln(1.02)
showing that
t=
ln(3)
.
ln(1.02)
6. Consider the function p(x) = 2x3 − 5x2 − 10x + 5. Explain why p(x) necessarily has a root on the interval
[−1, 2].
Solution: p(x) is continuous everywhere since it is a polynomial. Since p(−1) = −2 − 5 + 10 + 5 =
8 > 0 and p(2) = 16 − 20 − 20 + 5 = −19 < 0, the intermediate value theorem guarantees that
p(x) = 0 somewhere in [−1, 2].
7. Consider the following piece-wise defined function:

2

x<2
1 + x ,
f (x) = x + a + b, 2 ≤ x < 4

√
x + 5a, x ≥ 4
For what values of a and b is f (x) continuous everywhere on its domain?
Solution: The individual pieces here are continuous on their domains, so we just have to make sure
that the pieces touch at the endpoints. This means that we need
1 + (2)2 = 2 + a + b
and
4+a+b=
√
4 + 5a.
The first equation tells us that a + b = 3, so we can plug this into the second equation to get
√
4 + 5a = 4 + a + b = 4 + 3 = 7
or
4 + 5a = 49.
Solving gives a = 9, and since a + b = 3 we get b = −6.
8. Draw a graph of a function y = f (x) satisfying the following properties
• lim f (x) = 4
x→2−
• lim+ f (x) = 7
x→2
•
lim f (x) = −2
x→−∞
• lim f (x) = 1
x→∞
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Zachary Scherr
Math 103
Exam 1 Prep
• f (x) has a vertical asymptote at x = −3
• f (x) is continuous on [2, ∞)
Solution: There are many correct answers. Here is one of them:
9. The distance that a bicycle travels, t seconds after starting, is given by
t (seconds)
d (meters)
0
0
1
1.4
2
5.1
3
10.7
4
17.7
5
25.8
(a) Find the average velocity of the bicycle in [2, 4]
Solution: The average velocity is
17.7−5.1
4−2
= 6.3 m/s.
(b) Estimate the instantaneous velocity of the bicycle at t = 3 (there is no one right answer to this)
Solution: The average velocity over [2, 3] is 5.6 m/s while the average velocity over [3, 4] is
7 m/s. The instantaneous velocity is probably somewhere in between, so a reasonable guess
might be the average of those two numbers which is 6.3 m/s.
(c) At what time does the bicycle appear to be traveling fastest?
Solution: It appears from the numbers that the velocity keeps increasing since the average
velocities are increasing. A reasonable guess is that it is traveling fastest at t = 5.
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