Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Contents
mth.121.01.00
mth.121.02.01
mth.121.02.02
mth.121.02.03
mth.121.02.04
mth.121.02.05
mth.121.02.06
mth.121.02.07
mth.121.02.08
mth.121.03.01
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raf
t
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.10
2
13
21
32
35
41
47
54
59
61
The question presented here are representative of problems students will see
online or from the textbook. Errors should be reported to
Ron Bannon [email protected].
Fir
st
D
This document may be shared with students and instructors of MTH 121 at
Essex County College, Newark, NJ.
[email protected]
LATEX 2ε
Calculus I
MTH-121
0.1
Essex County College
Division of Mathematics
Assessments
Course Overview
mth.121.01.00
raf
t
This assignment is a review of material covered in MTH 119 and 120. This material will not
be covered in class and you may need to revisit your old mathematics textbooks if you need
to refresh your memory. However, if you are not capable of completing this assignment you
may need to consider dropping mth 121 until you able to do do this material. The material
in these assignments will not be measured, but instructors may include these questions on
assessments.
1. Write the inequality |2x + 2| < 4 in the form a < x < b for some numbers a, b.
Solution:
st
D
|2x + 2| < 4
−4 < 2x + 2 < 4
−6 < 2x < 2
−3 < x < 1
2. Describe the set as a union of finite or infinite intervals.
{ x : |x − 5| > 3 }
Solution: |x − 5| > 3 implies that x − 5 > 3 or x − 5 < −3; solving these simple linear
inequalities gives:
(−∞, 2) ∪ (8, ∞)
Fir
3. Describe the following set as an interval.
x
<0
x:
x+5
Solution: This is rational inequality and you should use a number line along with
sign-analysis to arrive at this solution:
(−5, 0)
4. Express r1 = 0.54 and r2 = 0.433 as fractions.
Page 2 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution: These are infinite geometric series and you may use a variety of methods of
arriving at the following results. I am showing two methods, one of which was certainly
covered in MTH 120.
This method may have been discussed in MTH 120, but this method may also have
been learned in middle school. I actually prefer this method.
raf
t
r1 = 0.54
100 · r1 = 54.54
99 · r1 = 54
54
r1 =
99
6
=
11
This method was done when you covered infinite geometric series in MTH 120.
st
D
r2 = 0.433
r2 = 0.4 + 0.03 + 0.003 + 0.0003 + · · ·
∞
4
3 X 1
r2 =
+
10 100 i=0 10i
3
1
4
+
·
10 100 1 − 1/10
13
=
30
=
5. Find the equation of the circle with center (2, 2):
(a) with radius r = 2;
Fir
Solution:
(x − 2)2 + (y − 2)2 = 4
(b) that passes through (1, −1).
Solution:
(x − 2)2 + (y − 2)2 = 10
6. Find the domain and range of the function.
f (x) =
1
4+x
Page 3 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution: The domain is all real numbers except −4; the range is all real numbers
except 0.
7. Let f (x) = x3 − x.
(a) Find the zeros of f (x).
(b) Is f (x) an odd function?
Solution:
raf
t
Solution: Factor f (x) = x3 − x = x (x2 − 1) = x (x − 1) (x + 1); gives these
roots:
x = 0, x = ±1.
Yes
st
D
f (x) = −f (−x)
x3 − x = − (−x)3 − (−x)
= − −x3 + x
= x3 − x
(c) Determine the intervals on which f (x) is positive.
Solution: Factor and use simple sign analysis.
(−1, 0) ∪ (1, ∞)
(d) Graph f (x).
Fir
Solution: Your graph (Figure 1, page 5). should look roughly the same, and you
should be able to do this without using a calculator.
8. Find the slope, the y-intercept, and the x-intercept of the line with the given equation.
y = 6x + 12
Solution:
(a) slope: m = 6
(b) y-intercept: (0, 12)
(c) x-intercept: (−2, 0)
Page 4 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
1
-1
0
-1
1
2
raf
t
-2
Figure 1: f (x) = x3 − x.
Solution:
st
D
9. Find the equation of the line with the given description: slope 4, y-intercept 2.
y = 4x + 2
10. Find the equation of the line shown (Figure 2, page 5).
4
3
2
Fir
1
-3
-2
-1
0
1
2
3
4
5
-1
Figure 2: Partial graph of a linear function.
Solution:
3
y =− x+3
4
Page 5 of 70
6
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
11. Find b so that the points (5, −3), (6, 3), and (b, 9) lie on a line.
Solution: You’ll need to solve (simple slope formula)
for b, and you’ll get
raf
t
6
6
=
1
b−6
b = 7.
12. Find the roots of the quadratic polynomial.
6x2 − 7x − 5
Solution: Factor 6x2 − 7x − 5 = (3x − 5) (2x + 1).
1
5
or x =
2
3
st
D
x=−
13. Complete the square and find the minimum or maximum value of the quadratic function.
y = x2 + 4x + 1
Solution:
y = x2 + 4x + 1
f (x) = (x + 2)2 − 3
The minimum occurs at x = −2 and the minimal value of f is f (−2) = −3.
Fir
14. Determine the domain of the function.
f (x) = x1/6
Solution:
x≥0
15. Determine the domain of the function.
f (x) = x3 + 9x − 10
Page 6 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution: All real numbers.
16. Determine the domain of the function.
1
t+2
raf
t
g (t) =
Solution: All real numbers except t = −2.
17. Determine the domain of the function.
f (x) =
x + x−1
(x − 3) (x + 8)
Solution: All real numbers except x = −8, 0, 3.
Solution:
st
D
18. Show that f (x) = x2 + 3x−1 and g (x) = 5x3 − 4x + x−2 are rational functions by writing
each as a quotient of polynomials.
f (x) = x2 + 3x−1
x3 + 3
=
x
g (x) = 5x3 − 4x + x−2
5x5 − 4x3 + 1
=
x2
Fir
19. Calculate the composite functions f ◦ g and g ◦ f , and determine their domains.
√
f (x) = x, g (x) = x + 2
Solution:
√
x + 2,
(f ◦ g) (x) =
√
(g ◦ f ) (x) =
x + 2,
x ≥ −2
x≥0
20. Find all values of c such that the domain of
f (x) =
x2
x+5
+ 2cx + 25
is R (all real numbers).
Page 7 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution: The discriminant need to be negative, that is
Simple sign analysis yields:
raf
t
4c2 − 100 < 0
c2 − 25 < 0
(c + 5) (c − 5) < 0
−5 < c < 5.
21. Solve sin θ = sin 2θ for π ≤ θ < 2π.
Solution:
=
=
=
=
sin 2θ
2 sin θ cos θ
0
0
st
D
sin θ
sin θ
sin θ − 2 sin θ cos θ
sin θ (1 − 2 cos θ)
Use the zero product rule to get:
sin θ = 0
1
cos θ =
2
which yields
θ = π,
5π
.
3
Fir
22. Solve the triangle if a = 7.15, b = 1.98 and C = 135.44◦ .1
Solution: Use the Law of Cosines to get c, that is c2 = a2 + b2 − 2ab cos C. Then the
Law of Sines
sin A
sin B
sin C
=
=
a
b
c
to get A and B.
1
c ≈ 8.67
A ≈ 35.34◦
B ≈ 9.22◦
Round your answer to two decimal places.
Page 8 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
23. Is f (x) = x2 + 3 one-to-one?
Solution: No. It’s a parabola and fails the horizontal line test.
24. Consider the following function.
x−2
x+2
(a) Show that f (x) is invertible by finding its inverse.
Solution:
raf
t
f (x) =
f (x) =
y =
x =
=
=
=
st
D
xy + 2x
2x + 2
2x + 2
2x + 2
1−x
2x + 2
1−x
x−2
x+2
x−2
x+2
y−2
y+2
y−2
y − xy
y (1 − x)
= y
= f −1 (x)
(b) What is the domain of f (x)?
Solution:
R, x 6= −2
Fir
(c) What is the range of f −1 (x)?
Solution: This is the same as the domain of f (x).
R, f −1 (x) 6= −2
(d) What is the domain of f −1 (x)?
Solution:
R, x 6= 1
(e) What is the range of f (x)?
Page 9 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution: This is the same as the domain of f −1 (x).
R, f (x) 6= 1
25. Evaluate without using a calculator.
Solution:
raf
t
tan−1 1
π
4
26. Evaluate without using a calculator.
sin−1 1
π
2
st
D
Solution:
27. Compute the value of the given function without using a calculator.
4π
arccos cos
3
Solution:
2π
3
Fir
28. Simplify by referring to the appropriate triangle or trigonometric identity.
tan cos−1 x
Solution: This is true for all −1 ≤ x ≤ 1, except x = 0.
√
1 − x2
x
29. Solve for the unknown variable.
e9x = ex+4
Page 10 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution: You’ll need to solve 9x = x + 4 for x.
30. Solve for the unknown variable.
1
2
raf
t
x=
x+8
1
3 =
3
x
Solution:
x
3
=
st
D
3x =
x =
x =
x+8
1
3
−x−8
3
−x − 8
−4
31. Calculate directly (without using a calculator).
log2 4
Solution:
log2 4 = x
4 = 2x
x = 2
Fir
32. Calculate directly (without using a calculator).
log2 82
Solution:
log2 82
82
64
x
=
=
=
=
Page 11 of 70
x
2x
2x
6
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
33. Calculate directly (without using a calculator).
log25 30 + log25
1
6
raf
t
Solution:
1
log25 30 + log25
6
1
log25 30 ·
6
log25 5
5
= x
= x
= x
= 25x
1
x =
2
34. Solve for x.2
Solution:
st
D
ln 7x2 + 4 − 3 ln x = ln 4
Fir
ln 7x2 + 4 − 3 ln x
ln 7x2 + 4 − ln x3
7x2 + 4
ln
x3
7x2 + 4
x3
7x2 + 4
0
0
= ln 4
= ln 4
= ln 4
= 4
= 4x3
= 4x3 − 7x2 − 4
= (x − 2) 4x2 + x + 2
I used the Rational Root Theorem to factor 4x3 − 7x2 − 4. The only real solution is:
2
x = 2.
Rational Root Theorem may prove helpful.
Page 12 of 70
Calculus I
MTH-121
0.2
Essex County College
Division of Mathematics
Assessments
Course Overview
mth.121.02.01
1. A wrench released from a state of rest at time t = 0 travels a distance s (t) = 4.9t2 m in
t seconds. Estimate the instantaneous velocity v at t = 8 seconds.3
Solution:
raf
t
∆s
s (t) − s (8)
=
∆t
t−8
= 4.9 (t + 8) As long as t 6= 8.
Now just take values near t = 8 to estimate the instantaneous velocity v at t = 8
seconds.
= 4.9 (t + 8)
= 78.89
= 78.449
st
D
∆s
(t)
∆t
∆s
(8.1)
∆t
∆s
(8.01)
∆t
∆s
(8.001)
∆t
∆s
(7.9)
∆t
∆s
(7.99)
∆t
∆s
(7.999)
∆t
= 78.4049
= 77.91
= 78.351
= 78.3951
Looks like we’re hitting 78.400 meters per second t = 8 seconds.
Fir
∆y
2. Compute
for the interval x ∈ [2, 8], where y = 3x − 1. What is the instantaneous
∆x
rate of change of y with respect to x at x = 6?
∆y
Solution:
= 3, pretty simple because it’s a line with slope m = 3; The instanta∆x
neous rate of change of y with respect to x at x = 6 is 3 because the slope is constant.
3. A stone is tossed in the air from ground level with an initial velocity of 15 m/s. It’s
height at time t is h (t) = 15t − 4.9t2 m. Compute the stone’s average velocity over the
time interval [0.5, 2.5].
3
Round your answer to three decimal places.
Page 13 of 70
Calculus I
MTH-121
Solution:
Essex County College
Division of Mathematics
Assessments
Course Overview
∆h
h (2.5) − h (0.5)
=
= 0.3 mps
∆t
2.5 − 0.5
raf
t
4. With an initial deposit of 200 dollars, the balance in a bank account after t years is
f (t) = 200 (1.18)t dollars. What are the units of the ROC4 of f (t)?
Solution: Dollars per year.
5. Let P (x) = 8x2 − 4. Estimate5 the instantaneous rate of change of this function at the
point x = 2.
Solution:
P (x) − P (2)
x−2
= 8 (x + 2) As long as x 6= 2.
=
st
D
∆P
∆x
Now just take values near x = 2 to estimate the instantaneous rate of change at x = 2.
Fir
∆P
(x)
∆x
∆P
(2.1)
∆x
∆P
(2.01)
∆x
∆P
(2.001)
∆x
∆P
(1.9)
∆x
∆P
(1.99)
∆x
∆P
(1.999)
∆x
= 8 (x + 2)
= 32.8
= 32.08
= 32.008
= 31.2
= 31.92
= 31.992
Looks like we’re hitting 32 as the instantaneous rate of change of this function at the
point x = 2.
6. Estimate6 the instantaneous rate of change of the function y (t) =
t = 1.
4
ROC = rate of change
Round your answer to the nearest whole number.
6
Round your answer to three decimal places.
5
Page 14 of 70
√
6t + 7 at the point
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution:
As long as t 6= 1.
raf
t
∆y
y (t) − y (1)
=
∆x
t−1 √
√
6t + 7 − 13
=
t−1
st
D
Now just take values near t = 1 to estimate the instantaneous rate of change at t = 1.
√
√
∆y
6t + 7 − 13
(t) =
∆t
t−1
∆y
(1.1) ≈ 0.823
∆t
∆y
(1.01) ≈ 0.831
∆t
∆y
(1.001) ≈ 0.832
∆t
∆y
(0.9) ≈ 0.842
∆t
∆y
(0.99) ≈ 0.833
∆t
∆y
(0.999) ≈ 0.832
∆t
Looks like we’re hitting 0.832 as the instantaneous rate of change of this function at
the point t = 1.
7. Estimate7 the instantaneous rate of change of the function f (x) = 9ex at the point
x = e.
Solution:
Fir
∆f
f (x) − f (e)
=
∆x
x−e
9ex − 9ee
=
As long as x 6= e.
x−e
7
Round your answer to three decimal places.
Page 15 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Now just take values near x = e to estimate the instantaneous rate of change at x = e.
∆f
9ex − 9ee
(x) =
∆x
x−e
≈ 143.441
≈ 137.073
raf
t
∆f
(e + 0.1)
∆x
∆f
(e + 0.01)
∆x
∆f
(e + 0.001)
∆x
∆f
(e + 0.0001)
∆x
∆f
(e − 0.1)
∆x
∆f
(e − 0.01)
∆x
∆f
(e − 0.001)
∆x
∆f
(e − 0.0001)
∆x
≈ 136.457
≈ 136.395
≈ 129.791
≈ 135.709
st
D
≈ 136.320
≈ 136.382
Looks like we’re hitting 136.388 as the instantaneous rate of change of this function at
the point x = e.
8. Estimate8 the instantaneous rate of change at the point x = 7, for f (x) = ln x.
Solution:
Fir
∆f
f (x) − f (7)
=
∆x
x−7
ln x − ln 7
As long as x 6= 7.
=
x−7
8
Round your answer to three decimal places.
Page 16 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Now just take values near x = 7 to estimate the instantaneous rate of change at x = 7.
=
ln x − ln 7
x−7
≈ 0.142
≈ 0.143
raf
t
∆f
(x)
∆x
∆f
(7.1)
∆x
∆f
(7.01)
∆x
∆f
(7.001)
∆x
∆f
(6.9)
∆x
∆f
(6.99)
∆x
∆f
(6.999)
∆x
≈ 0.143
≈ 0.144
≈ 0.143
≈ 0.143
st
D
Looks like we’re hitting 0.143 as the instantaneous rate of change of this function at
the point x = 7.
9. Estimate9 the instantaneous rate of change of function f (x) = 3 arctan x at the point
x = π/4.
Solution: You need to set your calculator to radian mode!
f (x) − f (π/4)
∆f
=
∆x
x − π/4
3 arctan x − 3 arctan π/4
=
x − π/4
As long as x 6= π/4.
Now just take values near x = π/4 to estimate the instantaneous rate of change at
Round your answer to three decimal places.
Fir
9
Page 17 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
x = π/4.
∆f
3 arctan x − 3 arctan π/4
(x) =
∆x
x − π/4
raf
t
≈ 1.767
≈ 1.846
≈ 1.855
≈ 1.856
≈ 1.947
≈ 1.864
≈ 1.856
st
D
∆f
(π/4 + 0.1)
∆x
∆f
(π/4 + 0.01)
∆x
∆f
(π/4 + 0.001)
∆x
∆f
(π/4 + 0.0001)
∆x
∆f
(π/4 − 0.1)
∆x
∆f
(pi/4 − 0.01)
∆x
∆f
(π/4 − 0.001)
∆x
∆f
(π/4 − 0.0001)
∆x
≈ 1.856
Looks like we’re hitting 1.856 as the instantaneous rate of change of this function at
the point x = π/4.
t1
t2
Distance
Fir
Distance
10. The graphs in the figure (Figure 3, page 18) represent the positions of moving particles
as a function of time.
t 3 Time
(A)
Time
(B)
FIGURE 11
Figure 3: Position as a function of time.
n
(a) Do the instantaneous velocities at times t1 , t2 , t3 in (A) form an increasing or
decreasing, or a constant sequence?
e value of the independent
variable increases, we note that the slope of the tangent lines decrea
(A) displays position as a function of time, the slope of each tangent line is equal to the velocity of th
ntly, the velocities at t1 , t2 , t3 form a decreasing sequence.
Page 18 of 70
d on the solution to part (a), the velocity of the particle is decreasing; hence, the particle is slowing d
were to draw several lines tangent to the graph in Figure 11(B), we would find that the slopes would be i
Calculus I
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Essex County College
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Course Overview
Solution: Decreasing.
(b) Is the particle speeding up or slowing down or moving at constant speed in (A).
Solution: Slowing down.
raf
t
(c) Is the particle speeding up or slowing down or moving at constant speed in (B).
Solution: Speeding up.
11. The graphs (Figure 4, page 19) in figure
represent
positions
of moving
particles
SEC
T I O N 2.1 the Limits,
Rates ofsChange,
and Tangent
Lines as
85
functions of time t. Match each graph with a description:
25. The
graphs
in Figureup
13 represent the positions s of moving particles as functions of time t. Match each graph with
(a)
Speeding
a description:
(b) Speeding
up and then slowing down
(a) Speeding
up
(b) Speeding
up and then
slowing down
(c) Slowing
down
(c) Slowing down
(d) Slowing down and then speeding up
(d) Slowing down and then speeding up
s
s
s
st
D
s
t
(A)
t
(B)
(C)
t
t
(D)
FIGURE 13
Figure 4: Each graph represent the positions s of moving particles as functions of time t.
solution When a particle is speeding up over a time interval, its graph is bent upward over that interval. When a
particle is slowing down, its graph is bent downward over that interval. Accordingly,
• In graph (A), the particle is (c) slowing down.
Solution: When a particle is speeding up over a time interval, its graph is bent upward
over that interval. When a particle is slowing down, its graph is bent downward over
In graph (C), the particle is (d) slowing down and then speeding up.
that interval. Accordingly,
• In graph (B), the particle is (b) speeding up and then slowing down.
•
• In graph (D), the particle is (a) speeding up.
Fir
(a) In graph (A), the particle is (c) slowing down.
26. An epidemiologist finds that the percentage N(t) of susceptible children who were infected on day t during the first
three weeks
measles (B),
outbreak
given, to aisreasonable
approximation,
the formula
(Figure
14)
(b) of
Ina graph
theisparticle
(b) speeding
up andbythen
slowing
down.
2
100t
(c) In graph(C), the particleN(t)
is (d)
= slowing down and then speeding up.
t 3 + 5t 2 − 100t + 380
(d) In graph (D), the particle is (a) speeding up.
% Infected
20
12. Consider the function of the following
graph (Figure 5, page 20). (Round your answers
15
to two decimal places.)
10
5
(a) Find the average rate of change
over [0, 1]
2 4 6 8 10 12 14 16 18 20
Time (days)
FIGURE 14 Graph of N(t).
Page
of 70 in infected children over the intervals [4, 6] and
(a) Draw the secant line whose slope is the average
rate19
of change
[12, 14]. Then compute these average rates (in units of percent per day).
(b) Is the rate of decline greater at t = 8 or t = 16?
(c) Estimate the rate of change of N(t) on day 12.
hich the rate of change is positive
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
y
0.25
0.15
0.1
0.05
0.2
0.4
raf
t
0.2
0.6
0.8
1
x
− x). average rate of change i
re is no change between xFigure
= 5:0Graph
andofxf (x)
==1.x (1The
st
D
1 ; the instantaneous rate of chan
he graph of f (x)
is
horizontal
at
x
=
Solution: 0
2
the instantaneous
of change
at xis
= 0.5.
positive at(b)allFind
points
whererate
the
graph
rising, because the slope of th
Solution: 0
for all x between
x = 0 and x = 0.5.
(c) Choose the intervals of x at which the rate of change is positive.
A. (0, 0.5)
Solution:
Correct choice.
This is the only
interval
wherethe
the rate
of
h in Figure 16 has the
following
property:
For
all
x,
average
rate
change is positive.
eous rate of change
at 1)x. Explain.
B. (0.5,
Fir
C. (0, 1)
D. never
y
y
x
(A)
x
(B)
FIGURE 16
Page 20 of 70
Calculus I
MTH-121
0.3
Essex County College
Division of Mathematics
Assessments
Course Overview
mth.121.02.02
1. Fill in the table, where10
f (x) =
x3 − 27
.
x2 − 9
(a) f (3.0020) ≈
raf
t
Solution: 4.502
(b) f (3.0010) ≈
Solution: 4.501
(c) f (3.0005) ≈
Solution: 4.500
(d) f (3.0001) ≈
Solution: 4.500
st
D
(e) f (2.9980) ≈
Solution: 4.499
(f) f (2.9990) ≈
Solution: 4.499
(g) f (2.9995) ≈
Solution: 4.500
(h) f (2.9999) ≈
Solution: 4.500
Guess the value of
lim f (x) =
Fir
x→3
Solution: 4.5 = 9/2
2. Fill in the table, where11
f (x) = x · ln 9x.
(a) f (1.0000) ≈
10
11
Round your answers to three decimal places.
Round your answers to three decimal places.
Page 21 of 70
Calculus I
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Essex County College
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Solution: 2.197
(b) f (0.5000) ≈
Solution: 0.752
(c) f (0.1000) ≈
raf
t
Solution: −0.011
(d) f (0.0500) ≈
Solution: −0.040
(e) f (0.0100) ≈
Solution: −0.024
(f) f (0.0050) ≈
Solution: −0.016
st
D
(g) f (0.0010) ≈
Solution: −0.005
(h) f (0.0005) ≈
Solution: −0.003
Guess the value of
lim f (x) =
x→0+
Solution:
lim f (x) = 0
Fir
x→0+
3. Evaluate the limit.
lim
x→4.7
√
7
Solution:
lim
x→4.7
√
7=
√
7
Page 22 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
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Course Overview
4. Evaluate the limit.
lim 3x
x→2
Solution:
5. Evaluate the limit.
raf
t
lim 3x = 6
x→2
lim (3x + 8)
x→6
Solution:
lim (3x + 8) = 26
st
D
x→6
6. Evaluate the limit.
lim (5x − 8)
x→4
Solution:
lim (5x − 8) = 12
x→4
7. Evaluate the limit.
lim 3x2 − 5
x→3
Fir
Solution:
lim 3x2 − 5 = 22
x→3
8. Estimate the limit numerically or state that the limit doesn’t exist. (Round your answer
to four decimal places. If an answer does not exist, state DNE.)
√
x−3
lim
x→9 x − 9
Page 23 of 70
Calculus I
MTH-121
Essex County College
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Solution:
f (x) =
≈
≈
≈
≈
≈
≈
raf
t
f (9.1)
f (9.01)
f (9.001)
f (8.9)
f (8.99)
f (8.999)
√
x−3
lim
x→9 x − 9
√
x−3
x−9
0.1662
0.1666
0.1667
0.1671
0.1667
0.1667
≈ 0.1667
9. Evaluate the limit numerically or state that the limit doesn’t exist.12
2x2 − 18
x→−3 x + 3
Solution:
Solution:
st
D
lim
f (x) =
Fir
f (−2.9)
f (−2.99)
f (−2.999)
f (−3.1)
f (−3.01)
f (−3.001)
2x2 − 18
lim
x→−3 x + 3
≈
≈
≈
≈
≈
≈
2x2 − 18
x+3
−11.800
−11.980
−11.998
−12.200
−12.020
−12.002
≈ −12.000
10. Evaluate the limit numerically or state that the limit doesn’t exist.13
12
13
x3 − 6x2 − 49
x→7 x2 − 8x + 7
lim
You may use ∞, −∞, DNE, or round your answer to three decimal places.
You may use ∞, −∞, DNE, or round your answer to three decimal places.
Page 24 of 70
Calculus I
MTH-121
Essex County College
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Course Overview
Solution:
f (x) =
≈
≈
≈
≈
≈
≈
raf
t
f (7.1)
f (7.01)
f (7.001)
f (6.9)
f (6.99)
f (6.99)
3
x − 6x2 − 49
lim 2
x→7 x − 8x + 7
x3 − 6x2 − 49
x2 − 8x + 7
10.575
10.508
10.501
10.425
10.493
10.499
≈ 10.500
11. Evaluate the limit numerically or state that the limit doesn’t exist.14
sin 10x
x→0
x
st
D
lim
Solution: Once again, make sure your calculator is in radian mode.
f (x) =
Fir
f (0.1)
f (0.01)
f (0.001)
f (−0.1)
f (−0.01)
f (−0.001)
sin 10x
lim
x→0
x
≈
≈
≈
≈
≈
≈
sin 10x
x
8.415
9.983
10.000
8.415
9.983
10.000
≈ 10.000
12. Evaluate the limit numerically or state that the limit doesn’t exist.15
14
15
lim
θ→0
3 cos θ − 3
2θ
You may use ∞, −∞, DNE, or round your answer to three decimal places.
You may use ∞, −∞, DNE, or round your answer to three decimal places.
Page 25 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
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Course Overview
Solution: Once again, make sure your calculator is in radian mode.
f (x) =
≈
≈
≈
≈
≈
≈
raf
t
f (0.1)
f (0.01)
f (0.001)
f (−0.1)
f (−0.01)
f (−0.001)
3 cos θ − 3
lim
θ→0
2θ
3 cos θ − 3
2θ
−0.075
−0.007
−0.001
0.075
0.007
0.001
≈ 0.000
13. Evaluate the limit numerically or state that the limit doesn’t exist.16
x→7
Solution:
2−x
x−7
st
D
lim−
2−x
x−7
f (6.9) = 49.000
f (6.99) = 499.000
f (6.999) = 4999.000
2−x
lim−
= ∞
x→7 x − 7
f (x) =
Fir
14. Evaluate the limit numerically or state that the limit doesn’t exist.17
16
17
7h − 1
h→0
h
lim
You may use ∞, −∞, DNE, or round your answer to three decimal places.
You may use ∞, −∞, DNE, or round your answer to three decimal places.
Page 26 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
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Course Overview
Solution:
f (x) =
≈
≈
≈
≈
≈
≈
raf
t
f (0.1)
f (0.01)
f (0.001)
f (−0.1)
f (−0.01)
f (−0.001)
7h − 1
lim
h→0
h
7h − 1
h
2.148
1.965
1.948
1.768
1.927
1.944
≈ 1.946
15. Evaluate18 the limit numerically or state that the limit doesn’t exist.19
st
D
sec−1 x
lim+ √
x→1
x5 − 1
Solution: Once again, make sure your calculator is in radian mode.
Fir
sec−1 x
f (x) = √
x5 − 1
f (1.1) ≈ 0.550
f (1.01) ≈ 0.624
f (1.001) ≈ 0.632
sec−1 x
lim+ √
≈ 0.632
x→1
x5 − 1
16. Estimate the limit numerically or state that the limit does not exist.20
lim (1 + r)3/r
r→0
Hint: sec−1 x = arccos x−1
19
You may use ∞, −∞, DNE, or round your answer to three decimal places.
20
You may use ∞, −∞, DNE, or round your answer to three decimal places.
18
Page 27 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
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Course Overview
Solution:
=
≈
≈
≈
≈
≈
≈
(1 + r)3/r
17.449
19.788
20.055
23.590
20.391
20.116
raf
t
f (r)
f (0.1)
f (0.01)
f (0.001)
f (−0.1)
f (−0.01)
f (−0.001)
lim (1 + r)3/r ≈ 20.086
r→0
17. Determine the one-sided limit of the function y (x) in the figure (Figure 6, page 28) at
the given point.21
lim− y (x)
-sided limits at c = 1, 2, and 4 of the function g(x) shown in Figure 11, and
st
D
x→2
y
3
2
Fir
1
1
2
3
4
5
x
FIGURE
11of y (x)
Figure
6: Partial graph
Solution:
-hand limit is lim g(x) = 3, whereas the right-hand limit is lim g(x)
x→1−
lim y (x) = 2
x→2−
x→1+
oes not exist at c = 1.
exist, state DNE.
-hand limitIfisan answer
limdoes not
g(x)
= 2, whereas the right-hand limit is lim g(x)
21
x→2−
x→2+
Page 28 of 70
oes not exist at c = 2.
-hand limit is lim g(x) = 2, whereas the right-hand limit is lim g(x)
5
x
Calculus I
MTH-121
Essex 1County
2
3College
4
Division of Mathematics
Assessments
Course Overview
−5
18. Determine the one-sided limit
−10of the function y (x) in the figure (Figure 7, page 29) at
22
the given point.
lim+ y (x)
x→5
termine the one-sided limits of the function f (x) in Figure 14, at the points c = 1, 3, 5, 6.
5
4
3
2
1
−1
2
3
4
−2
−3
5
6
7
8
x
st
D
−4
FIGURE 14 Graph of f (x)
Figure 7: Partial graph of y (x)
ion
im f (x) = lim
f (x) = 3
Solution:
→1−
1
raf
t
y
x→1+
im f (x) = −∞
→3−
im f (x) = 4
→3+
lim y (x) = −3
x→5+
im f (x) = 19.
2 Plot the function and use the graph to estimate the value of the limit.23
→5−
im f (x) = −3
→5+
sin 8x
x→0 sin 4x
lim
Fir
im f (x) = lim f (x) = ∞
→6−
x→6+
Solution: Your graph (Figure 8, page 30) should like similar to mine.
sin 8x
≈ 2.00
x→0 sin 4x
lim
April 5, 22
2011
If an answer does not exist, state DNE.
23
Round your answer to two decimal places. If an answer does not exist, state DNE.
Page 29 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
2
1
0
-1
-2
1
raf
t
-1
Figure 8: Partial graph of y (x) =
sin 8x
sin 4x
20. Plot the function and use the graph to estimate the value of the limit.24
st
D
7x − 1
lim
x→0 4x − 1
Solution: Your graph (Figure 9, page 31) should like similar to mine.
7x − 1
≈ 1.40
x→0 4x − 1
Fir
lim
24
Round your answer to two decimal places. If an answer does not exist, state DNE.
Page 30 of 70
Essex County College
Division of Mathematics
Assessments
Course Overview
raf
t
Calculus I
MTH-121
4
st
D
3
2
1
-1
0
1
Fir
7x − 1
Figure 9: Partial graph of y (x) = x
4 −1
Page 31 of 70
Calculus I
MTH-121
0.4
Essex County College
Division of Mathematics
Assessments
Course Overview
mth.121.02.03
1. Evaluate the limit using the Limit Laws.
lim 2t−1
t→3
raf
t
Solution: I am using the limit laws, but not explicitly stating them.
lim 2t−1 =
t→3
2. Evaluate the limit using the Limit Laws.
2
3
lim 8x4 − 4x3 + 7x
x→−1
st
D
Solution: I am using the limit laws, but not explicitly stating them.
lim 8x4 − 4x3 + 7x = 5
x→−1
3. Evaluate the limit using the Limit Laws.
lim (15x + 1) (20x − 1)
x→1/5
Solution: I am using the limit laws, but not explicitly stating them.
lim (15x + 1) (20x − 1) = 12
Fir
x→1/5
4. Evaluate the limit using the Limit Laws.
5 − 7t
t→6 t + 6
lim
Solution: I am using the limit laws, but not explicitly stating them.
5 − 7t
37
=−
t→6 t + 6
12
lim
Page 32 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
5. Evaluate the limit using the Limit Laws.
√
x
x→25 x − 9
lim
raf
t
Solution: I am using the limit laws, but not explicitly stating them.
√
5
x
=
lim
x→25 x − 9
16
6. Evaluate the limit assuming that lim f (x) = 6, and lim g (x) = 3.
x→−4
x→−4
lim f (x) g (x) =
x→−4
st
D
Solution: I am using the limit laws, but not explicitly stating them.
lim f (x) g (x) = 18
x→−4
7. Evaluate the limit assuming that lim f (x) = 5, and lim g (x) = 9.
x→−4
x→−4
f (x) + 1
=
x→−4 3 g (x) − 9
lim
Fir
Solution: I am using the limit laws, but not explicitly stating them.
lim
x→−4
f (x) + 1
1
=
3 g (x) − 9
3
8. Can the Quotient Law be applied to evaluate
sin 3x
?
x→0
x
lim
Page 33 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution: No, it’s a quotient and the limit of the bottom is zero.
raf
t
9. Can the Product Law be applied to evaluate
π
lim x −
· tan 8x?
x→π/16
16
Fir
st
D
Solution: No, it’s a product and the limit of tan 8x as x → π/16 does not exist.
Page 34 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
e 14, state0.5
whether
f (x) is left- or right-continuous (or neither) at each point
mth.121.02.04
ble discontinuities?
1. Consider the following graph (Figure 11, page 47) .
5
4
3
2
1
2
3
4
5
6
x
st
D
1
raf
t
y
FIGUREFigure
1410:Graph
of yof f=. f (x)
Partial graph
State whether the function shown in the figure (Figure 11, page 47) is left-continuous
only, right-continuous only, continuous, or neither at the following points:
(a) xat
= 1x = 1; it is right-continuous there.
discontinuous
only
discontinuous Solution:
at x = right-continuous
3; it is neither
left-continuous nor right-continuous th
(b) x = 3
discontinuous Solution:
at x = neither
5; it is left-continuous there.
Fir
(c) x = 5
tinuities are not
removable.
Solution: left-continuous only
he function
g(x)
Figure
2. Let
f (x) in
be the
function15.
f (x) =
4 for x < 3
−1 for x > 3
(a) Determine the value of f (3) if f is left-continuous at x = 3.
y
Solution: If f is left-continuous at x = 3, then f (3) = limx→3− f (x) = 4.
5
(b) Determine the value of f (3) if f is right-continuous at x = 3.
4
3
2
1
Page 35 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution: If f is right-continuous at x = 3, then f (3) = limx→3+ f (x) = −1.
raf
t
3. Use the Laws of Continuity to determine whether the function is continuous or discontinuous.
f (x) = 3x + 5 cos x
Solution: Since x and cos x are continuous, so are 3x and 5 cos x, thus 3x + 5 cos x is
continuous by the Continuity Laws.
4. Use the Laws of Continuity to determine whether the function is continuous or discontinuous.
3
f (x) = − 2
3x + 4
st
D
Solution: Continuous everywhere.
5. Use the Laws of Continuity to determine whether the function is continuous or discontinuous.
3x
f (x) = − x
4 −3
Solution: The function is continuous on its domain. The function has a infinite discontinuity at x = ln 3/ ln 4. Students may need to review MTH 119, especially if they
do not know how to solve 4x − 3 = 0.
6. Determine the points at which the function is discontinuous.
Fir
f (x) =
1
x−3
Classify these as removable, jump, or infinite discontinuities.
Solution: The function has an infinite discontinuity at x = 3.
7. Determine the points at which the function is discontinuous
f (t) =
t2
2
− 3t
Classify these as removable, jump, or infinite discontinuities.
Page 36 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution: The function has an infinite discontinuity at t = 3 and t = 0.
8. Is the following statement true or false? The function f (x) = 2x3/2 − 7x3 is rightcontinuous at x = 0.
raf
t
Solution: True. The function is defined at x = 0 and the limx→0+ f (x) = f (0).
9. Determine the points at which the function f (x) is discontinuous.

 x − 3 for x 6= 3
|x − 3|
f (x) =

−9
for x = 3
State the type of discontinuity.
st
D
Solution: x = 3, jump discontinuity. You should look at the limits, both left and
right, and the function’s value at x = 3 to see this. A graph may also prove helpful.
10. Determine the points at which the function f (x) is discontinuous.

1
 sin
for x 6= 4
x−4
f (x) =

0
for x = 4
Classify these as removable, jump, infinite or oscillatory discontinuities.
Solution: Oscillatory discontinuity at x = 4. A graph should prove helpful.
Fir
11. Determine the points at which the function is discontinuous.
f (x) =
e2x
4
− e−x+3
Classify these as removable, jump, or infinite discontinuities.
Solution: Infinite discontinuity at x = 1. Students may need to review MTH 119,
especially if they do not know how to solve e2x − e−x+3 = 0.
12. Determine the domain of the function f (x) =
√
Page 37 of 70
3x2 − 27.
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution: You’ll need to solve 3x2 − 27 ≥ 0. Some students may need to review nonlinear inequalities that were covered in MTH 119. The domain is (−∞, −3] ∪ [3, ∞).
13. Determine the domain of the function f (x) =
4x3
.
5x − x1/2
14. Let f (x) be the function
(
f (x) =
raf
t
Solution: You’ll need to solve 5x − x1/2 = 0, and you should be well aware that we
can not use negative numbers. Some students may need to review interval notation as
it relates to set intersection, and yes, this was covered in MTH 119. The domain is
(0, 1/25) ∪ (1/25, ∞).
3x2
for x ≤ 2
4 − x for x > 2
st
D
Compute the right-hand and left-hand limits at x = 2.
(a) lim+ f (x) =
x→2
Solution:
lim f (x) = lim+ 4 − x = 2
x→2+
x→2
(b) lim− f (x) =
x→2
Solution:
lim f (x) = lim− 3x2 = 12
x→2−
x→2
Fir
Determine whether f (x) is right-continuous, left-continuous, continuous, or neither at
the point where x = 2.
Solution: Left-continuous.
15. Let f (x) be the function
(
x2 − c
for x < 4
f (x) =
6x + 4c for x ≥ 4
Find the value of c that makes the function continuous everywhere.
Page 38 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution: You’ll need to get the sides to agree at x = 4.
Now, if c = −8/5, we have
raf
t
x2 − c = 6x + 4c
16 − c = 24 + 4c
−8 = 5c
8
c = −
5
lim f (x) = lim+ f (x) = f (4)
x→4−
x→4
16. Evaluate the limit using the substitution method.
lim 6 arctan ex
Solution:
st
D
x→0
lim 6 arctan ex =
x→0
3π
2
17. Which of the following quantities would be represented by continuous functions of time
and which would have one or more discontinuities?
(a) Velocity of an airplane during a flight
(b) Temperature in a room under ordinary conditions
(c) Value of a bank account with interest paid yearly
Fir
(d) The salary of a teacher
(e) The population of the world
Solution:
(a) The velocity of an airplane during a flight from Boston to Chicago is a continuous
function of time.
(b) The temperature of a room under ordinary conditions is a continuous function of time.
(c) The value of a bank account with interest paid yearly is not a continuous function of
time. It has discontinuities when deposits or withdrawals are made and when interest
is paid.
Page 39 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
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Course Overview
(d) The salary of a teacher is not a continuous function of time. It has discontinuities
whenever the teacher gets a raise (or whenever his or her salary is lowered).
Fir
st
D
raf
t
(e) The population of the world is not a continuous function of time since it changes by a
discrete amount with each birth or death. Since it takes on such large numbers (many
billions), it is often treated as a continuous function for the purposes of mathematical
modeling.
Page 40 of 70
Calculus I
MTH-121
0.6
Essex County College
Division of Mathematics
Assessments
Course Overview
mth.121.02.05
1. Evaluate the limit or state that it does not exist.25
x2 − 36
x→6 x − 6
raf
t
lim
Solution:
x2 − 36
= lim (x + 6) = 12
x→6 x − 6
x→6
lim
2. Evaluate the limit or state that it does not exist.26
x2 + 17x + 72
x→−8
x+8
Solution:
st
D
lim
x2 + 17x + 72
=
x→−8
x+8
lim
lim (x + 9) = 1
x→−8
3. Evaluate the limit or state that it does not exist.27
x2 − 25
x→5 x − 7
lim
Solution:
x2 − 25
=0
x→5 x − 7
Fir
lim
4. Evaluate the limit or state that it does not exist.28
25
If
If
27
If
28
If
26
an
an
an
an
answer
answer
answer
answer
does
does
does
does
not
not
not
not
exist,
exist,
exist,
exist,
state
state
state
state
2x2 + 8x − 42
lim
x→3
x2 − 9
DNE.
DNE.
DNE.
DNE.
Page 41 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution:
2x2 + 8x − 42
2 (x + 7)
10
=
lim
=
x→3
x→3 x + 3
x2 − 9
3
lim
raf
t
5. Evaluate the limit or state that it does not exist.29
(2 + 3x)3 − 8
x→0
x
lim
Solution:
(2 + 3x)3 − 8
= lim 36 + 54x + 27x2 = 36
x→0
x→0
x
lim
st
D
6. Evaluate the limit or state that it does not exist.30
x2 − x
lim
x→5 x2 − 25
Solution:
x2 − x
x (x − 1)
= lim
= DNE
2
x→5 x − 25
x→5 (x + 5) (x − 5)
lim
7. Evaluate the limit or state that it does not exist.31
62x − 1
x→0 6x − 1
Fir
lim
Solution:
(6x − 1) (6x + 1)
62x − 1
=
lim
x→0 6x − 1
x→0
(6x − 1)
x
= lim (6 + 1) = 2
lim
x→0
29
If an answer does not exist, state DNE.
If an answer does not exist, state DNE.
31
If an answer does not exist, state DNE.
30
Page 42 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
8. Evaluate the limit or state that it does not exist.32
(h + 3)−2 − 9−1
lim
h→0
h
raf
t
Solution:
st
D
(h + 3)−2 − 9−1 9 (h + 3)2
(h + 3)−2 − 9−1
= lim
·
lim
h→0
h→0
h
h
9 (h + 3)2
9 − (h + 3)2
= lim
h→0 9h (h + 3)2
h (h + 6)
= lim −
h→0
9h (h + 3)2
(h + 6)
2
= lim −
=−
2
h→0
27
9 (h + 3)
9. Evaluate the limit or state that it does not exist.33
x−7
√
lim √
x − 14 − x
x→7
Solution:
Fir
√
√
x−7
x−7
x + 14 − x
√
√
√
lim √
= lim √
·√
x→7
x→7
x − 14 − x
x − 14 − x
x + 14 − x
√
√
(x − 7) x + 14 − x
= lim
x→7
2 (x − 7)
√
√
x + 14 − x √
= lim
= 7
x→7
2
10. Evaluate the limit or state that it does not exist.34
√
50 − x − 1
√
lim
x→49
7− x
32
If an answer does not exist, state DNE.
If an answer does not exist, state DNE.
34
If an answer does not exist, state DNE.
33
Page 43 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution:
√
50 − x − 1
√
lim
=
x→49
7− x
√ √
50 − x − 1 7 + x
50 − x + 1
√
√ ·√
lim
·
x→49
7− x
7+ x
50 − x + 1
√
7+ x
49 − x
·√
= lim
x→49 49 − x
50 − x + 1
√
7+ x
=7
= lim √
x→49
50 − x + 1
raf
t
√
11. Evaluate the limit or state that it does not exist.35
1
8
lim √
−
x→16
x − 4 x − 16
Solution:
√
x−4
x→16 x − 16
√
( x − 4)
√
= lim √
x→16 ( x + 4) ( x − 4)
1
1
= lim √
=
x→16
8
x+4
=
lim
st
D
lim
8
1
√
−
x − 4 x − 16
x→16
12. Evaluate the limit or state that it does not exist.36
cot 5x
lim
x→0 csc 5x
Fir
Solution:
cot 5x
= lim cos 5x = 1
x→0 csc 5x
x→0
lim
13. Evaluate the limit or state that it does not exist.37
csc 7θ
limπ
θ→ 2 cot 7θ
35
If an answer does not exist, state DNE.
If an answer does not exist, state DNE.
37
If an answer does not exist, state DNE.
36
Page 44 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution:
limπ
θ→ 2
csc 7θ
1
= limπ
= DNE
θ→ 2 cos 7θ
cot 7θ
Solution:
lim
18 (1 + x) − 36
(1 − x) (1 + x)
18 (1 − x)
= lim −
x→1
(1 − x) (1 + x)
18
= −9
= lim −
x→1
1+x
= lim
x→1
st
D
x→1
18
36
−
1 − x 1 − x2
raf
t
14. Evaluate the limit or state that it does not exist.38
18
36
lim
−
x→1
1 − x 1 − x2
15. Evaluate the limit or state that it does not exist.39
lim (3t − 2at + 4a)
t→−1
Solution:
lim (3t − 2at + 4a) = 6a − 3
Fir
t→−1
16. Evaluate the limit or state that it does not exist.40
3 (x + h)2 − 3x2
lim
x→0
h
38
If an answer does not exist, state DNE.
If an answer does not exist, state DNE.
40
If an answer does not exist, state DNE.
39
Page 45 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution:
3 (x + h)2 − 3x2
lim
= lim (6x + 3h) = 3h
x→0
x→0
h
raf
t
17. Evaluate the limit or state that it does not exist.41
2x−1 − 2a−1
x→a
x−a
lim
Solution:
Fir
st
D
2x−1 − 2a−1 ax
2x−1 − 2a−1
= lim
·
lim
x→a
x→a
x−a
x−a
ax
2a − 2x
= lim
x→a ax (x − a)
2 (x − a)
= lim −
x→a
ax (x − a)
2
2
=− 2
= lim −
x→a
ax
a
41
If an answer does not exist, state DNE.
Page 46 of 70
Calculus I
MTH-121
0.7
Essex County College
Division of Mathematics
Assessments
Course Overview
mth.121.02.06
squeezed 1.byIn the
u(x)
at47)x is=
3?
At xby=
figureand
(Figurel(x)
11, page
f (x)
squeezed
u (x)2?
and l (x) at x = 3?At x = 2?
y
raf
t
u(x)
f(x)
l(x)
1.5
1
2
3
x
4
FIGURE 7
st
D
Figure 11: Partial graph of f , u, and l.
re is an openSolution:
interval
containing x = 3 on which l(x) ≤ f (x) ≤ u(x) an
Because there is an open interval containing x = 3 on which l (x) ≤ f (x) ≤
u (x) and limx→3 l (x) = limx→3 u (x), f (x) is squeezed by u (x) and l (x) at x = 3.
) and l(x) atBecause
x = there
3. Because
therecontaining
is an xopen
is an open interval
= 2 oninterval
which l (x) ≤containing
f (x) ≤ u (x) x = 2 o
but lim
l (x) 6= lim
(x), f (x) is trapped by u (x) and l (x) at x = 2 but not
m u(x), f (x)squeezed.
is trapped
by uu(x)
and l(x) at x = 2 but not squeezed.
2
x→2
x→2
2. What does the Squeeze Theorem say about lim
f (x) if lim
l (x) = lim
u (x) = 6
f
(x)
if
lim
l(x)
=
lim
u(x)
=
6 and f (x),
ze Theorem
say
about
lim
and f (x), u (x), and l (x) are related as in figure (Figure 12, page 48) provided? The
x→7
x→7
x→7
inequality f (x) ≤x→7
u(x) is not satisfiedx→7
for all x. Does thisx→7
affect the validity of your
conclusion?
uality f (x) ≤ u(x) is not satisfied for all x. Does this affect the validity
Fir
Solution: The Squeeze Theorem does not require that the inequalities l (x) ≤ f (x) ≤
y
u (x) hold for all x, only that the inequalities hold on some open interval containing x =
f(x)containing x = 7. Because
c. It is clear that l (x) ≤ f (x) ≤ u (x) on some open interval
limx→7 u (x) = limx→7 (x) = 6, the Squeeze Theorem guarantees that limx→7 f (x) = 6.
3. Plot the graphs of u (x) = 1 + |x − π/2| and l (x) = sin x on the same set of axes.What
can you say about6
lim f (x)
x→π/2
u(x)
if f (x) is squeezed by l(x) and u(x) at x = π/2?
l(x)
7
Page 47 of 70
FIGURE 8
x
heorem say about lim f (x) if lim l(x) = lim u(x) = 6 and f
x→7
x→7
x→7
y f (x) ≤ u(x) is not satisfied for all x. Does this
Course affect
Overview the val
Calculus I
MTH-121
Essex County College
Division of Mathematics
y
Assessments
raf
t
f(x)
6
u(x)
l(x)
x
7
st
D
FIGURE 8
Figure 12: Partial graph of f , u, and l.
Solution:
graph (Figure
page inequalities
49) should look similarl(x)
to mine.≤Because
heorem does
not Your
require
that13,the
f (x) ≤ u(x
there is an open interval containing x = π/2 on which l (x) ≤ f (x) ≤ u (x) and
l (x) = lim
squeezed
by u (x) and
at xclear
= π/2. that l(x) ≤
en intervallimcontaining
xu (x),=f (x)
c.isIn
Figure
8, lit(x) is
1
ecause lim u(x) = lim l(x)lim=f (x)
6,=the
Squeeze Theorem guara
x→π/2
x→π/2
x→π/2
x→7
x→7
4. Use cos
the Squeeze
to evaluate
suming that
x ≤Theorem
f (x)
≤ 1.the limit.
lim x cos
Fir
x→0
5
x
os x = lim 1 = 1, it follows that lim f (x) = 1 by the Squeez
Solution: I strongly recommend that you graph
the inequalities to make sure it holds
x→0
x→0
on some open interval containing x = 0.
5
lity provides sufficient information
− |x| ≤ x cos
≤to
|x| determine lim f (x), an
x
+2
lim − |x| = 0
x→0
lim |x| = 0
x→0
lim x cos
x→0
5
= 0
x
Page 48 of 70
x→1
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
3
2
0
raf
t
1
1
2
-1
Figure 13: Partial graph of u, and l.
5. Use the Squeeze Theorem to evaluate the limit.
st
D
x−3
lim x2 − 9
x→3
|x − 3|
Solution: I strongly recommend that you graph the inequalities to make sure it holds
on some open interval containing x = 3.
x−3
≤ x2 − 9
− x2 − 9 ≤ x2 − 9
|x − 3|
2
lim − x − 9 = 0
x→3
lim x2 − 9 = 0
x→3
Fir
x−3
= 0
lim x2 − 9
x→3
|x − 3|
6. Evaluate the limit using Theorem 242 as necessary.
sin (8x) · sec (7x)
x→0
5x
lim
42
lim
θ→0
sin θ
= 1,
θ
lim
θ→0
1 − cos θ
=0
θ
Page 49 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
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Course Overview
Solution:
sin (8x) · sec (7x)
8 sin (8x)
1
= lim ·
·
x→0
x→0 5
5x
8x
cos (7x)
8
=
5
raf
t
lim
7. Evaluate the limit using Theorem 243 as necessary.
sin2 5t
t→0
t
lim
Solution:
st
D
sin2 5t
sin 5t
lim
= lim 5 sin 5t ·
=0
t→0
t→0
t
5t
8. Evaluate the limit using Theorem 244 as necessary.
x2
x→0 sin2 3x
lim
Solution:
x2
lim
= lim
x→0 sin2 3x
x→0
3x
3x
1
·
·
9 sin 3x sin 3x
Fir
9. Evaluate the limit using Theorem 245 as necessary.
1 − 9 cos x
x→π/2
4x
lim
43
lim
θ→0
sin θ
= 1,
θ
lim
θ→0
1 − cos θ
=0
θ
44
sin θ
= 1,
θ→0 θ
lim
1 − cos θ
=0
θ→0
θ
lim
45
lim
θ→0
sin θ
= 1,
θ
lim
θ→0
1 − cos θ
=0
θ
Page 50 of 70
=
1
9
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution:
1 − 9 cos x
1
=
x→π/2
4x
2π
10. Evaluate the limit.
raf
t
lim
sin 2x
x→0
x
lim
Solution:
st
D
sin 2x
sin x
lim
= lim 2 cos x ·
=2
x→0
x→0
x
x
11. Evaluate the given limit using the hint.46
sin 2h
h→0 sin 7h
lim
Solution:
sin 2h
lim
= lim
h→0 sin 7h
h→0
Fir
12. Evaluate the limit.
7h
2 sin 2h
·
·
7
2h
sin 7h
=
2
7
=
7
2
tan 7x
x→0
2x
lim
Solution:
tan 7x
lim
= lim
x→0
x→0
2x
sin 7x
7
·
7x
2 cos 7x
46
sin Ax
A sin Ax
Bx
=
·
·
sin Bx
B
Ax
sin Bx
Page 51 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
13. Evaluate the limits.
sin 10x
=
|10x|
lim−
sin 10x
=
|10x|
x→0
x→0
raf
t
lim+
sin 10x
=
x→0 |10x|
lim
Solution:
sin 10x
sin 10x
= lim+
=1
x→0
x→0
|10x|
10x
sin 10x
sin 10x
lim−
= lim− −
= −1
x→0
x→0
|10x|
10x
sin 10x
= DNE
lim
x→0 |10x|
st
D
lim+
14. Use the identity
1 − cos x
1
=
2
x→0
x
2
lim
to evaluate the following limit.
cos 2x − 1
x→0
6x2
lim
Fir
Solution: Here I am letting u = 2x which means that 23 u2 = 6x2 .
cos 2x − 1
2 1 − cos u
1
lim
= lim − ·
=−
2
2
x→0
u→0
6x
3
u
3
15. Use the value
1
1 − cos x
=
2
x→0
x
2
lim
Page 52 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
to evaluate the following limits.
lim+
x→0
√
lim−
x→0
√
lim
x→0
Solution:
3 − 3 cos x
=
2x
3 − 3 cos x
=
2x
raf
t
√
3 − 3 cos x
=
2x
Fir
st
D
! √
√ r
√
3 − 3 cos x
3
1 − cos x
6
lim+
= lim+
·
=
2
x→0
x→0
2x
2
x
4
!
√ r
√
√
3 − 3 cos x
3
1 − cos x
6
= lim− −
·
=−
lim−
2
x→0
x→0
2x
2
x
4
√
3 − 3 cos x
lim
= DNE
x→0
2x
Page 53 of 70
Calculus I
MTH-121
0.8
Essex County College
Division of Mathematics
Assessments
Course Overview
mth.121.02.07
1. Evaluate the limit.
4x
x→∞ 9x + 9
Solution:
4x
=
x→∞ 9x + 9
lim
2. Evaluate the limit.
4
4
=
x→∞ 9 + 9/x
9
lim
2x2 + 8x
x→∞ 4x4 + 3x3 − 29
st
D
lim
Solution:
raf
t
lim
2x2 + 8x
=
x→∞ 4x4 + 3x3 − 29
lim
3. Evaluate the limit.
2/x2 + 8/x3
=0
x→∞ 4 + 3/x − 29/x4
lim
4x + 6
x→∞
3
lim
Solution:
lim
Fir
x→∞
4. Evaluate the limit.
4x + 6
= ∞
3
7x2 − 5
x→∞ 6 − 29x
lim
Solution:
7x2 − 5
= −∞
x→∞ 6 − 29x
lim
Page 54 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
raf
t
5. Find the equations of the horizontal asymptotes (there are two).
√
6x2 + 7
f (x) =
9x + 4
Solution:
p
√
√
6 + 7/x2
6
lim
=
note: x2 = x if x > 0
x→∞ 9 + 4/x
9
√
6
y =
9
p
√
√
2
√
6 + 7/x2
6x + 7
6
lim
= lim
=−
note: x2 = −x if x < 0
x→−∞ 9x + 4
x→−∞ −9 − 4/x
9
√
6
y = −
9
st
D
√
6x2 + 7
lim
=
x→∞ 9x + 4
6. Find the equation of the horizontal asymptote.
f (x) =
Solution:
3ex
1 + e−x
3ex
= ∞
x→∞ 1 + e−x
3ex
= 0
lim
x→−∞ 1 + e−x
y = 0
Fir
lim
7. Find the equations of the horizontal asymptotes (there are two).
f (x) =
9x1/3
(64x2 + 10)1/6
Page 55 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
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Course Overview
Solution:
x→∞
9x1/3
(64x2
+ 10)
1/6
=
y =
lim
x→∞
9x1/3
(64x2
+ 10)
1/6
=
lim
x→∞
9
(64 +
lim −
x→∞
9
(64 +
10/x2 )1/6
9
2
8. Evaluate the limit.
note:
√
6
x2 = x1/3 if x > 0
=−
9
2
note:
st
D
9. Evaluate the limit.
x2 = −x1/3 if x < 0
1
1 + 7/x5/3
=
lim p
x→−∞
4
16 + 6/x4
note:
√
x 4 = x2
6x − 4
lim √
x→−∞
4x2 + 5
√
−6 + 4/x
lim p
= −3 note: x2 = −x if x < 0
x→−∞
4 + 5/x2
Fir
6x − 4
lim √
=
x→−∞
4x2 + 5
√
6
x2 + 7x1/3
lim √
x→−∞
16x4 + 6
x2 + 7x1/3
lim √
=
x→−∞
16x4 + 6
Solution:
9
2
9
2
y = −
Solution:
10/x2 )1/6
=
raf
t
lim
10. Evaluate the limit.
lim 6 arctan
x→∞
x
7
Solution:
lim 6 arctan
x→∞
x
π
= 6 · = 3π
7
2
Page 56 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
11. Calculate the limit.
lim
x→∞
9x
1
−
x x+2
Assessments
Course Overview
Solution:
lim
x→∞
1
9x
−
x x+2
12. Calculate the limit.
=
x + 2 − 9x2
= −9
x→∞
x2 + 2x
raf
t
lim
lim [ln (5x + 1) − ln (2x + 1)]
x→∞
Solution:
st
D
5x + 1
lim [ln (5x + 1) − ln (2x + 1)] = lim ln
x→∞
x→∞
2x + 1
5x + 1
= ln lim
x→∞ 2x + 1
5
= ln
2
13. Calculate the limit.
√
lim ln 5x4 + 2 − ln x
x→∞
Solution:
√
5x4 + 2
lim ln
x→∞
x
!
√
5x4 + 2
= ln lim
x→∞
x
= ∞
Fir
√
4
lim ln 5x + 2 − ln x =
x→∞
14. Calculate the limit.
lim arctan
x→∞
x2 + 5
x+5
Page 57 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution:
lim arctan
x→∞
x2 + 5
x+5
15. Calculate the limit.
x2 + 5
= arctan lim
x→∞ x + 5
π
=
2
lim arctan
x→∞
Solution:
lim arctan
x+9
x+3
x+9
= arctan lim
x→∞ x + 3
π
= arctan 1 =
4
Fir
st
D
x→∞
x+9
x+3
raf
t
Page 58 of 70
Calculus I
MTH-121
0.9
Essex County College
Division of Mathematics
Assessments
Course Overview
mth.121.02.08
1. Using IVT, does the function
f (x) =
x2
x7 + 4
raf
t
have the value 0.172 for some x in [0, 1]?
Solution: f (0) √= 0 < 0.172; f (1) = 1/5 > 0.172; f (x) is continuous at all points
x where x 6= − 7 4 and this negative number is not on the given interval, therefore
f (x) = 0.172 for some x between 0 and 1.
2. Using IVT, does the function
f (x) = x2 · tan x
have the value 0.032 for some x in [0, π/8]?
st
D
Solution: f (0) = 0 < 0.032; f (π/8) ≈ 0.0639 > 0.032; f (x) is continuous at all
points x where x ∈ [0, π/8], therefore f (x) = 0.032 for some x between 0 and π/8.
3. Using IVT, does 2x + 7x = 8x have a solution?
Solution: Let f (x) = 2x + 7x − 8x . Observe that f is continuous on [0, 2] with
f (0) = 1 > 0 and f (2) = −11 < 0. Therefore, by the IVT, there is a c ∈ (0, 2) such
that f (c) = 2c + 7c − 8c = 0, or 2c + 7c = 8c
4. Using IVT, does cos x = 4 arccos x have a solution in (0, 1)?
Fir
Solution: Let f (x) = cos x − 4 arccos x. Observe that f is continuous on [0, 1] with
f (0) = 1 − 2π < 0 and f (1) ≈ 0.540 > 0. Therefore, by the IVT, there is a c ∈ (0, 1)
such that f (c) = cos c−4 arccos c = 0; equivalently we could state that cos c = 4 arccos c
for some c ∈ (0, 1).
5. Find an interval of length 0.1 in [1, 2] containing a root of the equation.
x7 + 6x − 10 = 0
Solution: Using a calculator, the solution is roughly 1.16911, so a good interval to
choose is [1.16, 1.26].
Page 59 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
10
-1
0
-5
-10
1
raf
t
5
2
3
Figure 14: Partial graph of f (x) = x3 − 8x − 0.25.
st
D
6. The figure (Figure 14, page 60) shows that f (x) = x3 − 8x − 0.25 has a root on interval
[2.75, 3].
Which of the intervals of length 0.025 contains this root?
Solution: Using a calculator the solution in this interval is roughly 2.8439
(a) [2.825, 2.850]
Solution: YES. This interval has length 0.025 and contains 2.8439.
(b) [2.850, 2.875]
Solution: NO. This interval has length 0.025 but does not contain 2.8439.
(c) [2.875, 2.900]
Fir
Solution: NO. This interval has length 0.025 but does not contain 2.8439.
(d) [2.900, 2.925]
Solution: NO. This interval has length 0.025 but does not contain 2.8439.
(e) [2.925, 2.950]
Solution: NO. This interval has length 0.025 but does not contain 2.8439.
Page 60 of 70
Calculus I
MTH-121
0.10
Essex County College
Division of Mathematics
Assessments
Course Overview
mth.121.03.01
1. Let f (x) = 3x2 .
(a) Find f (3) =
Solution:
raf
t
f (3) = 27
(b) Find f (3 + h) =
Solution:
f (3 + h) = 3 (3 + h)2
= 27 + 18h + 3h2
f (3 + h) − f (3)
=
h
Solution:
st
D
(c) Find
27 + 18h + 3h2 − 27
f (3 + h) − f (3)
=
h
h
= 18 + 3h, h 6= 0
(d) Compute f 0 (3) by taking the limit as h → 0.
Solution:
f (3 + h) − f (3)
h→0
h
= lim 18 + 3h = 18
f 0 (3) = lim
h→0
Fir
2. Compute f 0 (a) in two ways.
f (x) = 7 + 5x + 2x2 ,
a=5
f (a + h) − f (a)
.
h→0
h
Solution:
(a) f 0 (a) = lim
f (5 + h) − f (5)
h→0
h
82 + 25h + 2h2 − 82
= lim
h→0
h
= lim 25 + 2h = 25
f 0 (5) = lim
h→0
Page 61 of 70
2 + h3 − 8
8 + 12h + 6h
2 ) = 12.
Essex County College
Assessments
=
lim
(12
+
6h
+
h
= lim
Division of Mathematics
Course Overview
h
h→0
h→0
Calculus I
MTH-121
f (x) − f (a)
.
x→a
x−a
Solution:
(b) f 0 (a) = lim
3− f−(5)8
f (x) − f (2)
x
f (x)
0
"
f (a) =
= lim
lim
f (2) = lim
x→5
x−5
x−2
x→2
x→2
7+x
5x −
+ 2x22 − 82
raf
t
= lim
x→5
x−5
(2x + 15) (x − 5)
(x − 2)(x 2 + 2x
+ 4)
= lim
=
(x 2 + 2x + 4) = 12.
= lim
x→5
x − lim
5
x − 2 = x→5
lim 2x + 15 =x→2
25
x→2
3. Find11.
the slope of the secant line (Figure 16, page 63) through (2, f (2)) and (2.5, f (2.5)).
refer to Figure
0
Is it larger or smaller than f (2)?
y
st
D
3.0
2.5
2.0
1.5
1.0
0.5
f (x)
0.5
1.0
1.5
2.0
2.5
3.0
x
FIGURE 11
Fir
Figure 15: Partial graph of f (x) and a line tangent to f at x = 2.
he slope of the
secant line through (2, f (2)) and (2.5, f (2.5)). Is it larger or
Solution:
f 0 (2) = 0.75
f (2.5) − f (2)
msec =
2.5 − 2
2.5 − 2
=
=1
2.5 − 2
msec > f 0 (2)
the graph, it appears that f (2.5) = 2.5 and f (2) = 2. Thus, the slope of the
f (2.5)) is
4.
f (2.5) − f (2)
2.5 − 2
=
= 1.
2.5 − 2
2.5 − 2
Estimate f 0 (1) and f 0 (2) (Figure 16, page 63).
s also clear that the secant line through (2, f (2)) and (2.5, f (2.5)) has a larger s
Page 62 of 70
er words, the slope of the secant line
through (2, f (2)) and (2.5, f (2.5)) is larg
f (2 + h) − f (2)
f (2 + h) − f (2)
which Calculus I
=Essex
0. County College
refer to Figure
11.h
MTH-121
Division of Mathematics
Assessments
Course Overview
y
f (x)
f (2 + h) − f (2)
h
raf
t
3.0
2.5
2.0
1.5
1.0
0.5
ust have f (2 + h) = f (2). Now, f (2) = 2,x and the only othe
= 2. Thus, 2 + h = 0, or 0.5
h =1.0−2.1.5 2.0 2.5 3.0
FIGURE 11
Figure 16: Partial graph of f (x) and a line tangent to f at x = 2.
st
D
slope of12.
the
secant line through (2, f (2)) and (2.5, f (2.5)). Is it larger or
Solution:
oheFigure
f 0 (1) = 0
f 0 (2) = 0.75
the graph, it appears that f (2.5) = 2.5 and f (2) = 2. Thus, the slope of the
f (2.5)) is
5. Determine f 0 (a) for a = 1, 2, 4, 5, 7 (Figure 19, page 65).
f (2.5) − f (2)
2.5 − 2
=
= 1.
y 2.5 − 2
2.5 − 2
5
Fir
s also clear that the secant
4 line through (2, f (2)) and (2.5, f (2.5)) has a larger s
er words, the slope of the
3 secant line through (2, f (2)) and (2.5, f (2.5)) is larg
f (2 + h) − f (2) 2
te
for h = −0.5. What does this quantity represent? Is it la
h
1
x
1 2 from
3 4 the
5 graph
6 7 it8appears
9
h = −0.5, 2 + h = 1.5. Moreover,
that f (1.5) = 1.7 a
(a) f 0 (1) =
Partial
graph of f of
12
1.7 −(x).2 f (x).
fFIGURE
(2 +Figure
h) −17:f
(2)Graph
=
= 0.6.
h
−0.5
sents the slope of the secant line through the points (2, f (2)) and (1.5, f (1.5))
Page(1.5,
63 of 70
t line through the points (2, f (2)) and
f (1.5)) has a smaller slope than the t
=
1,
2,
4,
7.
2 + h) − f (2)
for h = −0.5 is smaller than f " (2).
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution:
f 0 (1) = 0
Solution:
raf
t
(b) f 0 (2) =
f 0 (2) = 0
(c) f 0 (4) =
Solution:
f 0 (4) = 0.5
st
D
(d) f 0 (5) =
Solution:
f 0 (5) = DNE
(e) f 0 (7) =
Solution:
f 0 (7) = 0
6. For which values47 of x is f 0 (x) < 0 (Figure 19, page 65)?
Fir
Solution:
(7, 9)
7. Show that f 0 (3) does not exist (Figure 19, page 65).
47
Use interval notation.
Page 64 of 70
f (212.
+ h) − f (2)
o Figure
which Calculus I
=Essex
0. County College
h
MTH-121
Division of Mathematics
y
4
ust have f (2 + h)
= 2. Thus, 2 + h =
= 1, 2, 4, 7.
3
2= f (2).
10, or h =
Now, f (2) = 2, and the only othe
−2.
1 2 3 4 5 6 7 8 9
x
Figure 18:12
Partial
graph of f of
(x). f (x).
FIGURE
Graph
st
D
o Figure 12.
f (2 + h) − f (2)
h
raf
t
5
Assessments
Course Overview
5
y
Fir
at the value of the derivative
of f at x = a can be interpreted as
4
t x = a. From Figure
3 12, we see that the graph of y = f (x) is a
erval 0 ≤ x ≤ 3. Accordingly,
f ! (1) = f ! (2) = 0. On the inter
2
1 ; thus, f ! (4) = 1 . Finally, the line tangent to the graph of y =
2
21
is f ! (x) < 0?
1 2 3 4 5 6 7 8 9
x
Figure 19:12
Partial
graph of f of
(x). f (x).
FIGURE
Graph
then the slope of the tangent line at x is negative. Graphically, th
! (x) < 0 fo
Pagegraph,
65 of 70
ing
for
increasing
x.
From
the
it
follows
that
f
= 1, 2, 4, 7.
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
Solution:
f (3 + h) − f (3)
= 0
h→0
h
f (3 + h) − f (3)
= 0.5
lim+
h→0
h
f (3 + h) − f (3)
lim
= DNE
h→0
h
raf
t
lim−
8. Use the limit definition to calculate the derivative of the linear function.
f (x) = 9x − 5
Solution:
f (x + h) − f (x)
h→0
h
9x + 9h − 5 − 9x + 5
= lim
h→0
h
9h
= lim
=9
h→0 h
st
D
f 0 (x) = lim
9. What is an equation of the tangent line at x = 3, assuming that y (3) = 5 and y 0 (3) = 2.
Solution:
Fir
y − 5 = 2 (x − 3)
y = 2x − 1
10. Find f (3) and f 0 (3), assuming that the tangent line to y = f (x) at a = 3 has equation
y = 3x + 7.
Solution:
f (3) = 3 · 3 + 7 = 16
f 0 (3) = 3
Page 66 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
11. Suppose that f (x) is a function such that the relationship given is true.
f (5 + h) − f (5) = 6h2 + 3h
(a) What is f 0 (5)?
Solution:
f (5 + h) − f (5)
h→0
h
2
6h + 3h
= lim
h→0
h
= lim 6h + 3 = 3
raf
t
f 0 (5) = lim
h→0
(b) What is the slope of the secant line through (5, f (5)) and (8, f (8))?
Solution:
= 6h2 + 3h
= 0
= 54 + 9 = 63
63
= msec =
= 21
3
st
D
f (5 + h) − f (5)
f (5 + 0) − f (5)
f (5 + 3) − f (5)
f (8) − f (5)
8−5
1
12. Let f (x) = √ . Compute
7x
lim
h→0
Solution:
f (28 + h) − f (28)
h
Fir
f (28 + h) − f (28)
1
1
1
√
lim
= lim
−√
h→0
h→0 h
h
196 + 7h
196
"√
# √
√
√
1
196 − 196 + 7h
196 + 196 + 7h
√ √
√
= lim
·√
h→0 h
196 196 + 7h
196 + 196 + 7h
1
1
−7h
√ √
√
= lim
·√
h→0 h
196 196 + 7h
196 + 196 + 7h
−7
1
√
√
= lim
·
h→0 14 196 + 7h
14 + 196 + 7h
1
= −
784
Page 67 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
13. Use the limit definition to compute f 0 (a).
f (x) = 11 − x2 ,
a = −1
Solution:
raf
t
Find an equation of the tangent line.48
f (x) − f (−1)
x→−1
x+1
11 − x2 − 10
= lim
x→−1
x+1
1 − x2
= lim
x→−1 x + 1
= lim 1 − x = 2
f 0 (−1) =
lim
x→−1
st
D
y − 10 = 2 (x + 1)
y = 2x + 12
14. Use the limit definition to compute the derivative of the function at x = −7.
f (x) =
1
x+9
And find an equation of the tangent line at x = −7.49
Solution:
f 0 (−7) =
Fir
=
f (x) − f (−7)
x→−7
x+7
1/ (x + 9) − 1/2 2 (x + 9)
lim
·
x→−7
x+7
2 (x + 9)
−x − 7
lim
x→−7 2 (x + 7) (x + 9)
1
1
lim −
=−
x→−7
2 (x + 9)
4
1
− (x + 7)
4
1
5
− x−
4
4
lim
=
=
y−
1
=
2
y =
48
49
Write your answer as an equation using the variables y and x.
Write your answer as an equation using the variables y and x.
Page 68 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
15. The limit represents a derivative f 0 (a). Find f (x) and a.
x4 − 16
x→2 x − 2
lim
raf
t
Solution:
f (x) = x4
a = 2
16. Sketch the graph (Figure 20, page 69) of f (x) = x5/2 on [0, 6].
80
70
50
40
30
20
10
0
st
D
60
1
2
3
4
5
6
Figure 20: f (x) = x5/2 on [0, 6]
(a) Use the sketch (Figure 20, page 69) to justify the inequalities for h > 0:
Fir
f (4) − f (4 − h)
f (4 + h) − f (4)
≤ f 0 (4) ≤
h
h
Solution: Visually the slopes are increasing from left to right.
(b) Use (a) to compute f 0 (4) to four decimal places.
Solution: Taking h = 0.00001 we find that
f (4) − f (4 − h)
f (4 + h) − f (4)
≤ f 0 (4) ≤
h
h
19.9999625 ≤ f 0 (4) ≤ 20.0000375
The slope looks to be about 20.0000.
Page 69 of 70
Calculus I
MTH-121
Essex County College
Division of Mathematics
Assessments
Course Overview
(c) Use a graphing utility to plot f (x) and the tangent line at x = 4, using your
estimate for f 0 (4).
Solution: Your graph should roughly look similar (Figure 21, page 70).
80
60
50
40
30
20
10
0
1
2
3
raf
t
70
4
5
Fir
st
D
Figure 21: f (x) = x5/2 on [0, 6]
Page 70 of 70
6