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TRANSCRIPT
Calculus I Course Handout
Sijun [email protected]
University of Michiganhttp://www-personal.umich.edu/∼siliu/
Preface
This material is based on my 6 terms teaching of Math 115 (Calculus I) in University of
Michigan. They include:
(1) A problem set for each section of the book. Each problem set contains a summary of
the key definitions/techinques/properties, followed by several problems. Problems are
chosen from web homework, textbook, past exams and the rest are made by myself. Due
to the large volume of problems, I only write the solutions to some of them.
(2) A review problem set for each exam. Most problems come with solutions.
(3) Weekly quizzes. A few quiz samples are provided here, with solution to all problems.
(4) Miscellaneous, including a short tutorial on how to use calculator to find integral,
and the list of derivative rules needed for the gateway test.
Information on the course can be found in the course website:
http://www.math.lsa.umich.edu/courses/115/
You are welcome to use this material if you find it helpful for your teaching. Typos
are always something hard to avoid, so if you find any, please let me know by email
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Contents
1 A library of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.1 Linear function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2 Exponential function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.2.1 Exponential growth . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.2.2 Continuous exponential growth . . . . . . . . . . . . . . . . . . . . 8
1.2.3 Exponential growth v.s Continuous exp growth . . . . . . . . . . . 9
1.2.4 Finding formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3 Logarithm function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3.1 The arithmetic of log functions . . . . . . . . . . . . . . . . . . . . 9
1.3.2 Half life and doubling time . . . . . . . . . . . . . . . . . . . . . . 10
1.4 New function from the old . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.4.1 Composition of functions . . . . . . . . . . . . . . . . . . . . . . . 11
1.4.2 Inverse functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.4.3 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.5 Sinusoidal functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.6 Power function and polynomials . . . . . . . . . . . . . . . . . . . . . . . . 17
1.6.1 Finding power functions . . . . . . . . . . . . . . . . . . . . . . . . 17
1.6.2 Finding polynomials from the graph . . . . . . . . . . . . . . . . . 18
1.6.3 Behavior of power functions and polynomials . . . . . . . . . . . . 18
1.7 Rational functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.8 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.8.1 What means “continuous”? . . . . . . . . . . . . . . . . . . . . . . 19
1.8.2 Property of continuous functions: Intermediate value theorem . . . 20
1.9 Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2
3
1.9.1 What is limit? When it exists? . . . . . . . . . . . . . . . . . . . . 21
1.9.2 How to find limits? . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.9.3 Continuous vs. limit . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2 Key concept: the derivative . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.1 Average velocity and instantaneous velocity . . . . . . . . . . . . . . . . . 25
2.2 Deriviatves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.2.1 Definition of derivatives . . . . . . . . . . . . . . . . . . . . . . . . 26
2.2.2 Estimation of the derivative . . . . . . . . . . . . . . . . . . . . . 26
2.2.3 More problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.3 Properties of derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.4 Interpretation of derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.5 Second derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.6 Derivative, second derivative and concavity . . . . . . . . . . . . . . . . . 32
2.7 Concavity and approximation . . . . . . . . . . . . . . . . . . . . . . . . 34
2.8 Differentibality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2.8.1 What function is not differentiable? . . . . . . . . . . . . . . . . . 35
2.9 Graph the derivative function . . . . . . . . . . . . . . . . . . . . . . . . . 36
2.10 Complicated graph problem . . . . . . . . . . . . . . . . . . . . . . . . . . 37
2.10.1 Build Auto Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
2.10.2 Build an Auto using parts . . . . . . . . . . . . . . . . . . . . . . . 37
3 Short-cuts to differentiation . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.1 Simple derivative rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.2 Product and quotient rules . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.3 Graph problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.4 Increasing/decreasing and concavity . . . . . . . . . . . . . . . . . . . . . 41
3.5 Simple derivative rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.6 Chain rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.7 Inverse function rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.8 Implicit functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.8.1 Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4
3.8.2 Tangent line and approximation . . . . . . . . . . . . . . . . . . . 45
3.9 Tangent line approximation review . . . . . . . . . . . . . . . . . . . . . . 46
3.10 A few more problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4 Using the derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.1 Critical point & Local extrema (local max/min) . . . . . . . . . . . . . . . 49
4.1.1 Critical point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.1.2 Critical points and local extrema (local max/min) . . . . . . . . . 49
4.2 Inflection point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.3 Finding global max/min . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
4.3.1 General method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
4.3.2 An important special case . . . . . . . . . . . . . . . . . . . . . . . 52
4.3.3 Functions involving two related variables . . . . . . . . . . . . . . 53
4.4 Finding the constants in a function . . . . . . . . . . . . . . . . . . . . . . 53
4.5 Optimization problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4.6 Marginal cost/revenue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
4.7 Rates and related rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
4.7.1 Rate is derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
4.7.2 Related rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
5 Key concept: the definite integral . . . . . . . . . . . . . . . . . . . . . 61
5.1 Distance/position/velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
5.2 Estimate distance from t = a to t = b . . . . . . . . . . . . . . . . . . . . . 62
5.3 Definite integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
5.4 Estimate definite integral using left/right sum . . . . . . . . . . . . . . . . 64
5.5 Fundamental theorem of calculus . . . . . . . . . . . . . . . . . . . . . . . 65
5.6 Interpretation of integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
6 Constructing antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . 71
6.1 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
6.1.1 Definition and formulas . . . . . . . . . . . . . . . . . . . . . . . . 71
6.1.2 Use antiderivatives to compute definite integrals . . . . . . . . . . 72
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6.2 Graph antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
7 Exam review materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
7.1 Exam 1 review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
7.2 Exam 2 review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
7.3 Exam 3 review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
8 Solution to selected problem sets . . . . . . . . . . . . . . . . . . . . . 99
8.1 Solution to section 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
8.2 Solution to section 4.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
8.3 Solution to exam 3 review . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
9 Sample quizzes with solutions . . . . . . . . . . . . . . . . . . . . . . . 115
9.1 Quiz 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
9.2 Quiz 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
9.3 Quiz 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
9.4 Quiz 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
9.5 Quiz 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
10 Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
10.1 Using calculator to find integral . . . . . . . . . . . . . . . . . . . . . . . . 127
10.2 Derivative rules for the gateway test . . . . . . . . . . . . . . . . . . . . . 129
Chapter 1
A library of functions
1.1 Linear function
How to find a line:
(1) Know a point (x0, y0) and slope m, use point-slope form: y − y0 = m(x− x0)
(2) know two points: (x1, y1) and (x2, y2), then the slope is m = y2−y1x2−x1
, now use the
point-slope form.
Problem 1. Find the formula of a line which has slope 2 and passes through (1, 1).
Problem 2. Find the formula of a line which passes through (1, 1) and (−2, 2).
Problem 3. Find the formula of a line which has x-intercept 2 and y-intercept 3.
Problem 4. The demand function for a certain product, q = D(p), is linear, where p
is the price per item in dollars and q is the quantity demanded. If p increases by $5,
market research shows that q drops by 2 items. In addtion, 100 items are purchases if
the price is $550. Find the formula.
If line 1 has slope m1, line 2 has slope m2. Then
(1) they are parallel ⇔ m1 = m2
(2) they are perpendicular ⇔ m1m2 = −1. (Caution: it’s −1, not 1 !)
Problem 5. (1) Find a line which is parallel to y = 2x− 1 and passes through (1, 3).
(2) Find a line which is perpendicular to y = 2x− 1 and passes through (1, 3).
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1.2 Exponential function
1.2.1 Exponential growth
(1) If a quantity P increases by r per year/hour/... (r is a percentage), with initial value
(when t = 0) P0, then after t years/hours/... , P = P0(1 + r)t.
(1) If a quantity P decreases by r per year/hour/... (r is a percentage), with initial value
(when t = 0) P0, then after t years/hours/... , P = P0(1− r)t.
In both cases, r is called percent rate.
Rewrite the function as P = P0at, then a is called growth factor (a = 1 + r in case
(1) and a = 1− r in case (2))
Problem 6. (Population model) Population in Ann Arbor in the year 2000 is 2 million,
and the population grows by 1.1% per year. Find the population P in millions, t years
after 2000.
Problem 7. (Radioactive substance) C14 decays by 3% per year. If now we have 30g of
C14, find the amount P in grams, t years after.
Problem 8. A town has a population of 1000 people at time t = 0. In each of the
following cases, write a formula for the population, P , of the town as a function of year
t.
(1) The population increases by 50 people a year.
(2) The population increases by 5% a year.
Problem 9. A patient take 300 mg drug at t = 0, and we know the drug in the body
is eliminated by 20% per hour. Find the amount of drug left t hours after the drug is
taken.
1.2.2 Continuous exponential growth
(1) If P increases by k per year/hour/...continuously (r is a percentage) with initial
value (when t = 0) P0, then after t years/hours/... , P = P0ekt.
(1) If P decreases by k per year/hour/...continuously (r is a percentage) with initial
value (when t = 0) P0, then after t years/hours/... , P = P0e−kt.
In both cases, k is called continuous percent rate.
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Rewrite the function as P = P0at, then a is called growth factor (a = ek in case (1)
and a = e−k in case (2)).
Problem 10. Population in Ann Arbor in the year 2000 is 2 million, and the population
grows by 1.1% per year continuously. Find the population P in millions, t years after
2000.
Problem 11. C14 decays by 3% per year continuously. If now we have 30g of C14,
find the amount P in grams, t years after.
1.2.3 Exponential growth v.s Continuous exp growth
(1) Growing model, P = P0at = P0(1 + r)t = P0e
kt, here a > 1, a = 1 + r = ek
(2) Decaying model, P = P0at = P0(1− r)t = P0e
−kt, here a > 1, a = 1− r = e−k.
Problem 12. Find growth rate a, initial value P0, percent rate r, continuous percent
rate k.
(1, 2) P = 3 · 1.2t, P = 3 · 0.8t
(3) P = 100e−0.45t
1.2.4 Finding formula
set up P = P0at, plug in two points, then we have two equations, divide one by the
other to cancel out P0, then solve for a.
Problem 13. Q = f(t) is exponential function, f(2) = 18, f(−1) = 2/3. Find the
formula.
Problem 14. Q = f(t) is exponential function, f(2) = 1, f(0) = 4. Find the formula.
1.3 Logarithm function
1.3.1 The arithmetic of log functions
Properties of log
(1) log(AB) = log(A) + log(B), log(A/B) = log(A)− log(B), log(At) = t log(A)
10
(2) 10log x = x, log(10x) = x. (log and base 10 cancel out).
Properties of ln
(1) ln(AB) = ln(A) + ln(B), ln(A/B) = ln(A)− ln(B), ln(At) = t ln(A)
(2) elnx = x, ln(ex) = x. (ln and base e cancel out).
Problem 15. Use the properties to simplify.
(1) log(1) (2) log(1/10) (3) log(√103)
(4) 10−2 log(t) (5) ln(t2(t− 1)3) (6) ln(aet/10t)
(7) ln(2 sin(t))− ln(sin(t)) (8) e−2 lnA+lnB
Problem 16. Find exact solutions to the following equations (example: x2 = 2, the
exact solution is ±√2, and ±1.414 is not exact).
(a) 2 = 4 · (1.1)x
(b) 5 = 10 · e−10x
1.3.2 Half life and doubling time
Half life (resp. doubling time) is the time needed for a quantity to halve (resp. double).
Review 1 (exponential growth). Suppose a quantity P has initial value P0, and P is
(1) increasing by r per day/hour/..., then P = P0(1 + r)t
(2) decreasing by r per day/hour/..., then P = P0(1− r)t
Review 2 (continuous exponential growth). Suppose a quantity P has initial value P0,
and P is
(1) increasing by r per day/hour/...continuously, then P = P0(1 + r)t
(2) decreasing by r per day/hour/...continuously, then P = P0(1− r)t
Problem 17. The price of a certain product is increasing by 40% per year. Find the
doubling time of the price.
Problem 18. C14 is decreasing by 2% per year continuously. Find the half life.
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Problem 19. The half life of C14 is 1 million years. How many years does it take until
9% of the original amount is left?
Problem 20. Population in Ann Arbor is increasing exponentially. If the doubling time
for the population is 20 years, find the tripling time.
Problem 21. The water lily population in a pond doubles every five days. The popula-
tion of water lilies after t days is given by P (t). Assume the initial population of water
lilies is P0.
(1) Find a formula for P (t).
(2) A water lily population initially covers 0.1% of Walden Pond. The fraction of the
pond covered by lilies at time t is given by A(t). The pond is completely covered
by lilies when A(t) equals 1. It takes exactly T days for the population of lilies to
entirely cover the pond.
Complete the formula for A(t). You may use P (t) in the expression.
A(t) =
0 ≤ t ≤ T
1 t ≥ T
(1.1)
(3) Find T .
1.4 New function from the old
1.4.1 Composition of functions
Problem 22. Find f(g(x)), g(f(x)), f(f(1)) if
(1) f(x) = x2 + 1, g(x) = sin(x)
(2) f(x) = 2ex, g(x) = x3
Problem 23. Find possible f(x), g(x) such that f(g(x)) =
(1) cos(ln(x+ 1)) (2) ex2+1 (3)
√x3 + 1
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1.4.2 Inverse functions
(1) y = f(x) has inverse function if and only if each y value has a unique x value
(2) horizontal line test: y = f(x) has inverse function if and only if every horizontal
line intersects the graph at most one point
(3) f(x) and f−1(x) cancel out. f(f−1(x)) = f−1(f(x)) = x
(4) f(x) and f−1(x) switch input and output. f(frog) = prince ⇔ f−1(prince) =
frog
Problem 24. Are the following functions invertible?
(1) f(n) is the number of students whose birthday is on the nth day of the year.
(2) f(w) is the cost of buying w lbs of apples.
Problem 25. Let u(t) be the number of Facebook users (in millions) at the beginning
of the year t. What’s the meaning of
(1) u(2009) = 100 (2) u−1(200) = 2001
Problem 26. Let N = f(t) be the total number of cans of cola Sean has consumed by
age t in years. Interpret the following.
(1) f(14) = 400 (2) f−1(50) = 6.
1.4.3 Transformations
Question: What will the function be if we shift/stretch/reflect the graph of f(x)?
Rule 1: shift
new functions transformations on graph of y = f(x)
y = f(x) + k (add k to function) shift up by k
y = f(x)− k (subtract k from function) shift down by k
y = f(x+ h) (replace x with x+ h) shift to the left by k
y = f(x− h) (replace x with x− h) shift to the right by k
Rule 2: vertical reflection and scaling
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new functions transformations on graph of y = f(x)
y = −f(x) (multiply function by −1) reflect graph over x-axis or vertically
y = kf(x) (multiply function by k) k > 0, then just vertically scale graph by k
k < 0, vertically scale graph by |k|, then
reflect graph vertically
Rule 3: horizontal reflection and scaling
new functions transformations on graph of y = f(x)
y = f(−x) (replace x with −x) reflect graph over y-axis or horizontally
y = f(kx) (replace x with kx) k > 0, then just horizontally scale graph by 1/k
k < 0, horizontally scale graph by |1/k|, then
reflect graph horizontally
Rule 4: If y = f(x) has a horizontal/vertical asymptote, do the same thing to the aymp-
tote as what we do to the graph. In addtion, horizontal (resp. vertical) shift/reflection/scaling
doesn’t change the horizontal (resp. vertical) asmptote.
Problem 27. Below (the thick one, not the dashed) is the graph of y = f(x)
-2 -1 1
-1
1
2
(1) Graph f(x+ 1)− 2. (2) Graph −12f(x).
(3) Find the formula of the dashed graph.
Problem 28. How can you get the graph ln(2x) from the graph of ln(x)? (Check all
possible answers)
(a) horizontal scaling (b) vertical scaling (c) horizontal shift (d) vertical shift
14
Problem 29. A(t) is the percentage of the Math 115 midterm exam that the student
Wolv E.Rine will have completed t minutes after the start time of the exam under normal
circumstances, i.e., assume he arrives on time, etc. What will the function be in the
following situations?
(1) Wolv E.Rine arrives p minutes late for the exam.
(2) The instructor accidentally hands Wolv E. Rine a copy of the exam that has the
first p% of the solutions already filled in.
Problem 30. (Winter 07, exam 1, #3) See the back.
15
4
3. (12 points) The graph of y = f(x) is given by the figure below.
−2 2 5
3
−5
y
x
y = f(x)
The graphs of the following functions are related to the graph of f . Determine a formula for eachgraph in terms of the function f .
−2 2 5
−2
2
−5
y
x
y = k(x)
−2 2 5
4
−5
y
x
y = g(x)
k(x) = g(x) =
6
−3 3 6
y
x
y = j(x) 3
−3 3
y
x
y = h(x)
j(x) = h(x) =
16
1.5 Sinusoidal functions
1. Values of sin, cos and tan for special angles.
x 0 π/6 π/4 π/3 π/2
sin(x) 0 1/2√2/2
√3/2 1
cos(x) 1√3/2
√2/2 1/2 0
tan(x) 0√3/3 1
√3 undefined
2. Graph of sinusoidal functions: y = A sin(Bt) + k and y = A cos(Bx) + k. ( B > 0) 3.
Midline y = k.
Amplitude |A| (Caution: not A !)
Period 2π/B
Table 1.1: Properties
Given graph of sinusoidal functions, find the formula.
Tools: look at the y-intercept.
(1) y-intercept is midline, and graph goes up after y-intercept.
Set y = A sin(Bt) + k where A > 0.
(2) y-intercept is midline, and graph goes down after y-intercept.
Set y = A sin(Bt) + k where A < 0.
(3) y-intercept is max.
Set y = A cos(Bt) + k where A > 0.
(4) y-intercept is min.
Set y = A cos(Bt) + k where A < 0.
Then read the midline, amplitude, period from the graph. Compare them with Table
1.1 to find A, k, B.
Problem 31. Find formula for the following graphs.(see next page)
Problem 32. The population of Baffulo in Yellowstone national park oscillates sinu-
soidally between a low of 400 on Jan 1 and a high of 1000 on July 1. Graph the population
P against the time t, in months since the start of the year. (So t = 0 on Jan 1, and
t = 6 on July 1) .Then find the formula of P .
17
[]
Π
-1
3
[] 1
2
4
[] 1
1
[] 3 Π
1
Problem 33. The average weight of a squirrel in Ann Arbor oscillates sinusoidally
between a low of 5 pounds on January 1 and a high of 9 pounds on July 1, and a low of
5 pounds on January 1 next year again. Suppose that the function P (t) gives the average
weight in pounds of an Ann Arbor squirrel t months after January 1. Find a formula
for P (t).
Answer: P (t) = −2 cos(π6 t) + 7
Problem 34. The Awkward Turtle is riding a mini ferris wheel! The wheel has radius
1.5 meters but is lifted off the ground, so that even when he is at the lowest point of the
ride, the Awkward Turtle is still 0.5 meters above the ground, which is, needless to say,
distinctly awkward. The wheel turns at a constant rate of 1 revolution every 90 seconds.
Suppose that precisely at noon, the Awkward Turtle is 2 meters above the ground and
moving toward the ground. Let H(t) denote the height (in meters) of the Awkward Turtle
above the ground, t minutes after noon.
Draw the graph and find the formula of H(t).
1.6 Power function and polynomials
1.6.1 Finding power functions
Problem 35. Poiseuille’s Law gives the rate of flow, R, of a gas through a cylindrical
pipe in terms of the radius of the pipe,r, for a fixed drop in pressure between the two
18
ends of the pipe.
(a) Find a formula for Poiseuille’s Law, given that the rate of flow is proportional to
the fourth power of the radius.
(b) If R=410cm3/s in a pipe of radius 4 cm for a certain gas, find a formula for the
rate of flow of that gas through a pipe of radius r cm.
1.6.2 Finding polynomials from the graph
Let a1, . . . , ak be all the zeros (roots) of a polynomial f(x), then f(x) = c · (x−
a1)n1 · · · (x− ak)
nk , where c is some nonzero constant that can be determined by an
additional point. The value of the powers ni could be 1, 2, 3, depending on what the
graph looks like around ai, more precisely, 3 for seat shape, 2 for U shape, and 1 for
all other shapes.
Problem 36. Find the formula for the graph, given that it is a polynomial, that all zeros
of the polynomial are shown in figure 1, that the exponents of each of the zeros are the
least possible, and that it passes through the point (−1,−64).
1.6.3 Behavior of power functions and polynomials
(P1) when x → ∞ or −∞, kxp → ∞ or −∞ when p > 0, and kxp → 0 when p < 0.
(P2) when x → ∞ or −∞, a polynomial has the same behavior as its leading term
axn.
Problem 37. When x → ∞, what about f(x) =
(1) −100x2 (2) πx−2 (3) −ex3 (4) −1010x2 + x3 − 1
(P3) A polynomial of degree n has n many zeros (roots), and they may not be all
different.
19
[Figure 1] [Figure 2]
Problem 38. A cubic polynomial with negative leading coefficient is shown for −10 ≤
x ≤ 10 and −10 ≤ y ≤ 10 in figure 2. In total, how many zeros does this function have?
and which intervals are they in?
1.7 Rational functions
(1) Definition: f(x) = p(x)/q(x) where p(x), q(x) are both polynomials.
(2) Graph of f(x) may have zeros, horizontal asymptote and vertical asymptote.
(2.1) Zeros: x0 such that p(x0) = 0.
(2.2) Vertical asymptote: the line x = x0 where q(x0) = 0.
(2.3) Horizontal asymptote: horizontal line y = L, if f(x) → L when x → ∞ or −∞.
(3) When x → ∞ or −∞, f(x) has the same behavior asleading term of p(x)
leading term of q(x)
Problem 39. R(x) = −2(x−1)(x+4)x2+3x+2
. Find the zeros, vertical asymptotes and horizontal
asymptote.
Problem 40. R(x) = −2(x+1)(x+4)2
x2−3x+2. Find the zeros, vertical asymptotes and horizontal
asymptote.
1.8 Continuity
1.8.1 What means “continuous”?
A function f(x) is called continuous at x = x0 when the graph doesn’t break at when
x = x0.
20
A function f(x) is called continuous on a interval if the graph doesn’t break on that
interval.
A function f(x) is called continuous if the graph doesn’t have break anywhere.
When f(x) is NOT continuous at x = x0?
(1) f(x0) is undefined. E.g. 1/x is not continuous at x = 0. E.g. p(x)/q(x) where
q(x) = 0.
(2) For piecewise defined function, f(x0) has different values using the formulas in
different pieces. E.g. f(x) = x + 1 for x < 1, and x + 2 for x ≥ 1. This f(x)
is not continuous at x = 1.
What function is continuous?
If a function is defined everywhere, and it’s not a piecewise defined function, then
it’s continuous. e.g. sin, cos, linear, exponential, polynomials, kxp with p > 0.
Problem 41. Which function is continuous?
(1) x−1x2−1
(2) 2x+ x1.5 (3) 2x+ x−1 on [−1, 1] (4) 3x−1x2+1
(5) ex/(ex − 1)
(6) f(x) = x for x ≤ 1 and x2 for x > 1 (7) f(x) = x/|x| for x ̸= 0, and 0 for x = 0.
Problem 42. Find k such that f(x) is continuous.
(1) f(x) = kx for 0 ≤ x ≤ 2, and 3x2 for 2 ≤ x
(2) f(x) = 2(x2 − 2)/(x2 − 3x+ 2) when x ̸= 1, and k when x = 1
1.8.2 Property of continuous functions: Intermediate value theorem
Intermediate value theorem: Suppose f is continuous on a closed interval [a, b]. If k is
any number between f(a) and f(b), then there is at least a number c in [a, b] such that
f(c) = k.
Problem 43. Consider the function f(x) = 3x3 + 3x2 + 12, and let c be a number in
the interval [0, 1]. For what values of k is there a c in this interval such that f(c) = k?
21
1.9 Limit
1.9.1 What is limit? When it exists?
If f(x) → L, when x approaches c in all possible ways, then we say the limit exists, and
write it as limx→c
f(x) = L.
Remark. When c is a finite number (i.e. c ̸= ±∞), there are two ways for x to approach
c: from the left side and from the right side. If f(x) approaches different values in the
two ways, then the limit limx→c
f(x) doesn’t exist.
However, we can still define the left side limit and the right side limit.
(1) if f(x) → L when x → c from the left side, then left side limit is L, write as
limx→c−
f(x) = L
(1) if f(x) → L when x → c from the right side, then right side limit is L, write as
limx→c+
f(x) = L
(3) limx→c
f(x) exists if and only if limx→c−
f(x) = limx→c+
f(x), and in this case all limits are
equal.
Table 1.2: Common limits
f(x) limx→∞
f(x) limx→−∞
f(x)
exp function ax (a > 1) ∞ 0
ax (0 < a < 1) 0 ∞e−x 0 ∞
power function xp (p > 0) ∞ −∞power function xp (p < 0) 0 0
Problem 44. Let f(x) = x/|x| when x ̸= 0, and 0 when x = 0.
Find limx→0−
f(x) and limx→0+
f(x). Does limx→0
f(x) exist?
1.9.2 How to find limits?
(1) if f(x) is continuous and f(c) is defined, then limx→c
f(x) = limx→c−
f(x) =
limx→c+
f(x) = f(c). e.g. limx→1
(ex + x2) = limx→1−
(ex + x2) = limx→1+
(ex + x2) = e+ 1.
(2) if f(x) is a polynomial with leading term axn, then limx→∞
f(x) = limx→∞
axn (also
for x → −∞)
22
(3) if f(x) = p(x)/q(x) is a rational function, then limx→∞
f(x) =
limx→∞
leading term of p(x)
leading term of q(x)
(also true for x → −∞)
(4) use known limits: limx→0
sin(kx)/x = k (Note that the unit of x is radians!)
Problem 45. Find the following limits.
(1) limx→1−
(e−x + cos(x)) (2) limx→2
(ln(x) + x) (3) limx→0
sin(2x)/x where x is in degrees.
Problem 46. (Power function and polynomials) Find the following limits.
(1) limx→∞
−2x2 (2) limx→−∞
4x3 (3) limx→−∞
(−2x2+4x+100) (4) limx→∞
(4x3−100x+150)
(5) limx→2
(x3 − x)
Problem 47 (Rational functions). Find the following limits.
(1) limx→∞
−2x2 + x3
−100 + 10x2 − 2x3(2) lim
x→−∞
3x+ 1− 2x3
12− 3x2(3) lim
x→−∞
3x+ 1− 2x3
12− 3x4(4)
limx→1
2x2 − 2
x− 1
Two functions both → ±∞, which one counts?
Consider x13 and x3, as the size of x grows, for example, x → −∞, both functions
→ −∞.
However, as x → −∞, x13 becomes significantly smaller (use symbol ≪) than x3 in
magnitude. e.g. , when x = −106, x13 = −100 ≪ x3 = −1018.
This tells us limx→−∞
(x3+x13 ) = lim
x→−∞x3 since x
13 is too small in magnitude to count
compared to x3.
This suggests, when we take limx→∞ or limx→−∞, we can ignore the significantly
small parts, since they are too small to influence the limit.
Remark. If a log function, a power function, an exponential function all go to ±∞, the
following order shows which is ≪ which.
constant/bounded functions ≪ log(x) or ln(x) ≪ power function ≪
power function with larger power ≪ exp function
23
Problem 48. Find limx→∞
(sin(ex) + e−x + 100x2 − 2x3 + 3ex)
Solution. We have five parts, sin(ex), e−x, 100x2, −2x3, 3ex. Here sin(ex) is bounded
(−1 ≤ sin ≤ 1), e−x becomes a constant 0 as x → ∞. The rest three all → ±∞, −2x3
has a larger power than 100x2, and 3ex is exp function. So we have the following order.
sin(ex) and e−x ≪ 100x2 ≪ −2x3 ≪ 3ex
thus we just keep 3ex since all others are significantly smaller, so
limx→∞
(sin(ex) + e−x + 100x2 − 2x3 + 3ex) = limx→∞
3ex = ∞
Problem 49. Find limx→∞
(ln(x) + 100x13 − 2x3 + 3e−x)
Problem 50. Find limx→∞
e−x + 3x2 + 4x
sin(ex) + x2
If f ≪ g, then f/g → 0, and g/f → ∞ or −∞.
Problem 51. Find limx→∞
x2
exand lim
x→∞
ex
−x2.
Problem 52. Find limx→∞
ex + 3x2 + 4x
sin(ex)− x2
1.9.3 Continuous vs. limit
f(x) is continuous at x = c if and only if limx→c f(x) = f(c).
24
Chapter 2
Key concept: the derivative
2.1 Average velocity and instantaneous velocity
1. Let S(t) be the position of an object, then the average velocity from t = a to t = b is
S(b)−S(a)b−a . Graphically, it’s the slope of the line linking (a, S(a)) and (b, S(b)).
2. The instantaneous velocity (or velocity) v(a) at t = a is
limh→0
S(a+ h)− S(a)
h
The reason is: S(a+h)−S(a)h is the average velocity a short period, h, after t = a, as the
period h gets shorter, the average velocity is getting closer to the velocity at t = a, so
the limit will be the velocity at t = a.
Graphically, v(a) is the slope of the tangent line of S(t) at t = a.
Problem 53. A particle is moving along x-axis. Let S(t) = t2 meters be the position of
the particle on x-axis t seconds after.
(1) Find the average velocity from t = 2 to t = 2.001.
(2) Find the average velocity from t = 2 to t = 2.0001.
(3) Find the average velocity from t = 2 to t = 2.00001.
(4) Find the velocity at t = 2.
Problem 54. The following graph is the distance of a car travelled t hours after. What
can you tell about the speed of the car?
25
26
2.2 Deriviatves
2.2.1 Definition of derivatives
1. The average rate of change of f(x) from x = a to x = b is f(b)−f(a)b−a
2. The instaneous rate of change (derivative) of f(x) is
f ′(x) = limh→0
f(x+ h)− f(x)
h
f ′(a) = limh→0
f(a+ h)− f(a)
h
f ′(a) is the instaneous rate of change of f(x) at x = a, it tells us how fast f(x)
changes at x = a. Graphically, it’s the slope of the tangent line of f(x) at x = a.
3. Let S(t) be the position function, then S′(t) = v(t), the velocity function.
4. Unit of f ′(x) is the unit of f(x) divided by the unit of x.
Problem 55. f(x) = x2. What’s the definition of f ′(1)?
Problem 56. f(x) = 13xsin(x2 + 1). What’s the definition of f ′(2)?
2.2.2 Estimation of the derivative
1. Estimate derivative f ′(a).
(1) f ′(a) ≈ f(a+h)−f(a)h for a small h close to 0. (2) f ′(a) ≈ f(b)−f(a)
b−a for b close to a.
2. The tangent line of f(x) at x = a is: y = f(a) + f ′(a)(x− a).
27
3. f(x) ≈ f(a) + f ′(a)(x − a) for x close to a, where f(a) + f ′(a)(x − a) is called the
tangent line approximation.
Problem 57. Suppose f(x) = xx. (a) Write down the definition of f ′(2). (b) Estimate
f′(2).
Problem 58. f(x) = 3x. Estimate f′(2).
Problem 59. f(13) = 7, f′(13) = 9, estimate f(13.5) and f(12.5).
Problem 60. Suppose the tangent line of f(x) at (1, 2) is y = 3x− 1.
(a) Find f ′(1). (b) Estimate f(1.01).
Problem 61. We have the following table for position function S(t).
t in hours = 0 2 4 6 8 10
S(t) in miles = 18 21 22 21 18 14
(a) Estimate v(2) and v(0). (b) when does the velocity appear to be positive?
(c) If S(t) is continuous, is it invertible?
2.2.3 More problems
Formula: (a+ b)2 = a2 + 2ab+ b2, (a− b)2 = a2 − 2ab+ b2.
Formula: (a+ b)3 = a3 + 3a2b+ 3ab2 + b3, (a− b)3 = a3 − 3a2b+ 3ab2 − b3.
Problem 62. If f(x) = x2 + 2x, write down the definition of f ′(2) and calculate f ′(2).
Problem 63. (1) Find the limit limh→0
(3 + h)3 − 27
husing algebra.
(2) If x3 has derivative 3x2, can you find the limit directly?
Problem 64. If (sin(x))′ = cos(x), calculate limh→0
sin(2 + h)− sin(2)
hand lim
h→0
sin(h)
h.
Problem 65. Look at the figure on the right side.For each of the following pairs of numbers,
decide which is larger.
A. f(3)−f(1)3−1 and f(5)−f(3)
5−3 .
B. f(3)− f(1) and f(5)− f(3).
C. f ′(1) and f ′(3).1 3 5
28
2.3 Properties of derivatives
(1) f ′(x) > 0 ⇔ f(x) is increasing
(2) f ′(x) < 0 ⇔ f(x) is decreasing
(3) if f(x) has a local max/min at x = a, then f ′(a) = 0.
(4) if f ′(x) = m on [a, b], then on [a, b], f(x) = mx+ k, a line segment of slope m.
In particular, if f ′(x) = 0 on [a, b], then f(x) is a constant on [a, b].
(5) f ′(x) is increasing (resp. decreasing) ⇔ f(x) is concave up (resp. down).
Problem 66. Suppose f(x) is given by the table below.
x 0 2 4 6 8 10 12
f(x) 18 16 15 16 20 23 25
(1) Estimate f ′(2) and f ′(0).
(2) For what values of x does f ′(x) appear to be positive?
(3) For what values of x does f ′(x) appear to be negative?
Problem 67. Figure 2.1 is the graph of f(x). When does f ′(x) appear to be postive?
negative? zero?
-1 1 2 3 4 5
3
2
1
-1
-2
-3
Figure 2.1
1 2 3 4 5
3
2
1
-1
-2
Figure 2.2
Problem 68. Figure 2.2 is the graph of f(x) and f ′(x). Tell which is which.
Problem 69. A child inflates a ballon, admires it for a while and then lets the air out
at a constant rate. If V (t) gives the volume of the ballon at time t, then the figure below
29
shows V′(t) as a function of t. At what time does the child
(a) begin to inflate the ballon?
(b) finish inflating the ballon?
(c) begin to let air out?
(d) What would the graph of V ′(t) look
like if the child had alternated between
pinching and releasing the open end of
the balloon, instead of letting the air
out at a constant rate?
3 4 910 1415 18 20
1
-1
-2
Problem 70. The graph of a function f is shown on the right. At which of the labeled x-
values, is
(a) f(x) least?
(b) f(x) greatest?
(c) f ′(x) least?
(d) f ′(x) greatest?
2.4 Interpretation of derivatives
How do we interpret f ′(a) in words?
f ′(a) tells us how fast f(x) is changing at x = a, but how fast? See the formula
f(a+ h)− f(a) ≈ f ′(a)h
From x = a to x = a+ h, f(x) changes by approximately f ′(a)h. This means, at x = a,
increase x by 1 will result in a change of approximately f ′(a) in f(x), in other words,
f(a+ 1) is approximately f ′(a) more than f(a).
This observation provides us 2 ways to interpret f ′(a).
(1) f(a+ 1) is approximately f ′(a) more than f(a).
(2) at x = a, f(x) is changing by approximately f ′(a) per (unit of x).
Then translate everything into words.
To interpret (f−1)′(a), simply replace f with f−1.
30
(1) f−1(a+ 1) is approximately (f−1)′(a) more than f−1(a).
(2) at x = a, f−1(x) is changing by approximately (f−1)′(a) per (unit of x).
Then translate everything into words.
Problem 71. The cost C in thousand dollars of building a house A square feet in area
is given by C = f(A).
(a) Interpret f′(100) = 200.
(b) If 100 square feet house costs 30000 dollars. Estimate the cost of building a house of
100.5 square feet.
Problem 72. P = f(t) is the population in Mexico in millions, where t is the number
of years since 1980. Interpret:
(a) f′(6) = 2
(b) f−1(95.5) = 16
(c) (f−1)′(95.5) = 0.46
Problem 73. (Winter 10, exam 1, #9)Suppose that W (h) is an invertible function
which tells us how many gallons of water an oak tree of height h feet uses on a hot
summer day. Interpret:
(a) W (50)
(b) W−1(40)
(c) W′(5) = 3
31
Problem 74. Let N = f(t) be the total number of cans of cola Sean has consumed by
age t in years. Interpret the following in practical terms, paying close attention to units.
(a) f(14) = 400 (b) f−1(50) = 6
(c) f′(12) = 50
(d) (f−1)′(450) = 1/70
Problem 75. (page 96, #11) An economist is interested in how the price of a certain
item affects its sales. At a price of $p, a quantity, q, of the iteem is sold. If q = f(p),
interpret:
(a) f(150) = 2000
(b) f′(150) = −25
Problem 76. g(v) is the fuel efciency, in miles per gallon, of a car going at a speed of
v miles per hour. Interpret g′(55) = 0.54
Problem 77. (page 96, #19) Water is flowing into a tank; the depth, in feet, of the
water at time t in hours is h(t). Interpret, with units, the following statements.
(a) h(5) = 3
(b) h′(5) = 0.7
(c) h−1(5) = 7
(d) (h−1)′(5) = 1.2
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2.5 Second derivatives
1. For a function f(x), f ′(x) is the derivative of f(x). The derivative of f ′(x) is called
the second derivative of f ′′(x), because it’s what we get by taking derivative of f(x)
twice.
f ′′(x) = limh→0
f ′(x+ h)− f ′(x)
h
2. let S(t) be the position function, then S′(t) is the velocity v(t), S′′(t) = v′(t) is the
acceleration a(t). If the unit of S(t) is meters and unit of t is seconds, then unit of v(t)
is m/s, unit of a(t) is m/s2.
3. Estimate the second derivative: f ′′(x) ≈ f ′(x+h)−f ′(x)h for small h close to 0.
Problem 78. Let S(t) = 2t2 be the position in feet of a particle moving along the x-axis
t seconds after. Use definition to find velcocity v(t) and acceleration a(t).
Problem 79. The position of a car is given as below.
t ( in s) 0 1 2 3 4 5
S(t) ( in ft) 0 32 56 74 87 96
(1) estimate the velocity at t = 0,t = 1 and t = 2.
(2) estimate the average acceleration over the first 2 seconds.
(3) estimate the acceleration at t = 1.
(4) estimate the velocity at t = 0.1.
2.6 Derivative, second derivative and concavity
1. f(x) is concave up ⇔ f ′(x) increasing ⇔ f ′′(x) > 0
2. f(x) is concave down ⇔ f ′(x) decreasing ⇔ f ′′(x) < 0
3. f ′′(x) = 0 where concavity changes. These points are called inflection points.
33
Problem 80. What can you tell about f ′(x) and f ′′(x) if
(1) f(x) is increasing faster and faster.
(2) f(x) is increasing slower and slower.
(3) f(x) is decreasing faster and faster.
(4) f(x) is decreasing slower and slower.
Problem 81. Let S(t) be the distance travelled by a car t hours after noon. If the
driver keeps slowing down in the first hour, keeps speeding up in the second hour, then
gets pulled over by the police due to speeding in the next 30 minutes, and in the rest time
he drives at 65m/h. What does the graph of S(t) look like?
Problem 82. Look at the graph of y = f(x), at which of the marked x values can the
following statements be true?A. f(x) > 0
B. f ′(x) > 0
C. f(x) is increasing.
D. f ′(x) is increasing.
E. The slope of f(x) is negative
F. The slope of f ′(x) is negative.
1 2 3 4 5
Problem 83. Look at the graph of y = f ′(x) (not f(x)), At which of the marked values
of x isA. f(x) greatest?
B. f(x) least?
C. f ′(x) greatest?
D. f ′(x) least?
E. f ′′(x) greatest?
F. f ′′(x) least?
1 2 3 4 5 6
Problem 84. A particle is moving along a straight line. Its distance, s, to the right of
a fixed point is given by the graph below. Estimate:
(a) when the particle is moving to the right and when it’s moving to the left.
(b) when the acceleration of the particle is zero, when it’s negative, and when it’s positive.
34
1 2 3 4
2.7 Concavity and approximation
1. The tangent line at x = a is y = f(a) + f ′(a)(x− a).
2. When x ≈ a, the point on the graph of f(x) (i.e. (x, f(x))) is very close to the
point on the tangent line (i.e. (x, f(a) + f ′(a)(x− a))), so we have
f(x) ≈ f(a) + f ′(a)(x− a)
3. f(a) + f ′(a)(x− a) is called tangent line approximation of f(x) at x = a.
4. If f(x) is concave up, then the tangent line is below the graph of f(x), so the
estimate is an underestimate, i.e. f(a) + f ′(a)(x− a) < f(x).
If f(x) is concave down, then the tangent line is above the graph of f(x), so the
estimate is an overestimate, i.e. f(a) + f ′(a)(x− a) > f(x).
Problem 85. A function f has f(5) = 20, f ′(5) = 2, and f ′′(x) < 0 for x > 5. Which
of the following are possible values for f(7) and which are impossible?
(a) 26 (b) 24 (c) 22
Problem 86. A function f has a tangent line approximation L(x) = −2x+5 at x = 3.
(1) What are f(3) and f ′(3)?
(2) Estimate f(2.8) and f(3.2).
(3) If f ′′(x) < 0 for x < 3, and f ′′(x) > 0 for x > 3. Are the estimates in part (2)
underestimate or overestimate?
35
2.8 Differentibality
1. f(x) is called differentiable at x = a if f ′(a) exists.
2. f(x) is called differentiable if f ′(x) exists for all x.
2.8.1 What function is not differentiable?
f(x) is NOT differentiable at x = a if it satisfies one of the following
(1) f(x) is not continuous at x = a, i.e, there is a break at x = a.
(2) f(x) has a corner point at x = a. e.g. f(x) = |x| is not differentiale at x = 0
because of the corner point at x = 0.
(3) calculate f ′(x) by formulas and f ′(x) is undefined at x = a. e.g. f(x) =√x,
f ′(x) = 12√xis undefined at x = 0.
Result: differentiable function must be continuous, but continuous function may not
be differentiable (may have corner point).
Problem 87. f(x) = |x+ 7| − 2. For which x is f(x) not continuous? differentiable?
Problem 88. Is the following statement true or false?
(1) If f(x) is a differentiable periodic function, then f ′(x) is also periodic.
(2) If f(x) is differentiable on (a, b), then it’s continuous on (a, b).
(3) If f(x) = |x+ 7| − 2 is differentiable at x = 1.
For piecewise defined function
f(x) =
f1(x) : for x ≤ a
f2(x) : for x ≥ a
f(x) is differentiable if and only if f1(a) = f2(a) (continuous) and f ′1(a) = f ′
2(a)
(avoid corner point at x = a).
Problem 89. For what values of a and b is f(x) differentiable if f(x) =
bx+ 8, x > 1
ax2 + x, x ≤ 1
36
Problem 90. For what values of a and b is f(x) differentiable if f(x) =
x2 + a, x > 1
bx− 2, x ≤ 1
2.9 Graph the derivative function
Step 1. Find where f ′(x) = 0, constant, or undefined.
(1). If f(x) has a local max/min at x = a and f ′(a) exists, then f ′(a) = 0.
(2). f ′(x) = m on one interval, if f(x) is a line of slope m on that interval.
(3). f ′ is undefined where f is not continuous or f has a corner point.
Step 2. Find where f ′(x) > 0 or < 0.
(1) f ′ > 0 ⇔ f is increasing; f ′ < 0 ⇔ f is decreasing.
Step 3. Find where f ′(x) is increasing or decreasing.
(1) f ′ is increasing (resp. decreasing) ⇔ f is concave up (resp. down) ⇔ f ′′(x) > 0
(resp. f ′′(x) < 0)
Problem 91. Given the graph of f(x), carefully graph f ′(x) on the same axis. SECTION 4.3 THE FUNDAMENTAL THEOREM OF CALCULUS 311
If is the function whose graph is shown in Figure 2 and, find the values of , , , , , and . Then sketch a
rough graph of .
SOLUTION First we notice that . From Figure 3 we see that is thearea of a triangle:
To find we add to the area of a rectangle:
We estimate that the area under from 2 to 3 is about 1.3, so
For , is negative and so we start subtracting areas:
We use these values to sketch the graph of in Figure 4. Notice that, because ispositive for , we keep adding area for and so is increasing up to ,where it attains a maximum value. For , decreases because is negative.
If we take and , then, using Exercise 27 in Section 4.2, we have
Notice that , that is, . In other words, if is defined as the integral of byEquation 1, then turns out to be an antiderivative of , at least in this case. And if wesketch the derivative of the function shown in Figure 4 by estimating slopes of tangents,we get a graph like that of in Figure 2. So we suspect that in Example 1 too.
a ! 0f !t" ! t
f !t"tx ! 3x ! 3tt " 3t " 3
f !t"t
t!5" ! t!4" # y5
4f !t" dt # 3 # !$1.3" ! 1.7
t!4" ! t!3" # y4
3f !t" dt # 4.3 # !$1.3" ! 3.0
f !t"t ! 3
FIGURE 3
g(1)=1
t0
1
1
22y
t0
1
1
22
2
y
g(2)=3
t0
1
1
22
2
y
3
g(3)Å4.3
t0
1
1
22
42
y
g(4)Å3
t0
1
1
22
42
y
g(5)Å1.7
t!3" ! t!2" # y3
2f !t" dt # 3 # 1.3 ! 4.3
f
t!2" ! y2
0f !t" dt ! y1
0f !t" dt # y2
1f !t" dt ! 1 # !1 ! 2" ! 3
t!1"t!2"
t!5"t!4"t!3"t!2"t!1"t!0"t!x" ! xx0 f !t" dt
t!1" ! y1
0f !t" dt ! 1
2 !1 ! 2" ! 1
t!1"t!0" ! x00 f !t" dt ! 0
t
t!x" ! y x0t dt !
x 2
2
ftt% ! ft%!x" ! xft
tt%! ff
fEXAMPLE 1v
t0
1
1
22
42
y
y=f(t)
FIGURE 2
FIGURE 4
©=j f(t) dta
x
x0
1
1
2
42
y
3
4
53
g
97817_04_ch04_p304-313.qk_97817_04_ch04_p304-313 11/2/10 3:53 PM Page 311
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
37
2.10 Complicated graph problem
2.10.1 Build Auto Parts
If f(x) has a “seat” shape or “U” shape at x = a, and f ′(a) exists, then f ′(a) = 0.
If f(x) has a corner at x = a, then f ′(a) is undefined.
If f ′(x) = m on one interval, then f is a line of slope m on the interval.
Problem 92. Graph f(x) such that it’s both increasing/decreasing and concave up/down?
Problem 93. Graph f(x) such that (1) f is increasing (2) f ′(3) = 0.
Problem 94. Sketch a graph of a continuous function f(x) with the given properties:
(a) f ′′(x) > 0 for x < 2 and for x > 2, and f ′(2) is undefined.
(b) f ′′(x) > 0 for x < 2 and f ′′(x) < 0 for x > 2, and f ′(2) is undefined.
(c) f ′′(x) > 0 for x < 2 and for x > 2, and f ′(2) is undefined, limx→∞
f(x) = 4
Problem 95. Draw a possible graph of f(x) such that:
(a) f(x) is decreasing and concave up. (b) f ′(19) = −1 (c) f(19) = −3.
Answer the following questions.
(1) what do you know about f ′(1)
(2) what happens to f(x) as x → −∞.
(3) where is the zero of f(x)?
2.10.2 Build an Auto using parts
Problem 96. f(x) is a continuous function with all of the following properties:
(1) f(0) = 5 (2) f ′(−3) = 0
(3) f ′(x) ≥ 0 for x < 0 (4) f ′(x) < 0 for x > 3
(5) f ′(x) is decreasing for x < −3 (6) f ′′(x) < 0 for x > 3
(7) f(x) has a constant rate of change of −2 for 0 < x < 3
Draw a possible graph of f(x).
38
Problem 97 (Fall 2010, exam 1, #2). A continuous (but not necessarily differentiable)
function, f , defined for all real numbers has the following properties:
(1) f(2) = 1 (2) f ′(2) = 0
(3) limx→∞
f(x) = 2 (4) f ′′(x) > 0 for x > 5
(5) f ′(x) = 1 for x < −1
(7) f is concave up for −1 < x < 3
Draw a possible graph of f(x).
Problem 98 (Fall 2010, exam 1, #2). A continuous (but not necessarily differentiable)
function, g, defined for all real numbers has the following properties:
a. g′(x) = 2 for 1 < x < 2
b. g′(x) = −2 for 2 < x < 3
c. g(0) = −1
d. g(1) = 0
e. g is decreasing for x > 3
f. g′′(x) < 0 for x > 3
g. g is concave down for x < 1.
Draw a possible graph of f(x).
Chapter 3
Short-cuts to differentiation
3.1 Simple derivative rules
(1) (f(x)+g(x))′ = f ′(x)+g′(x) (2) (c ·f(x))′ = c ·f ′(x) where c is a constant.
(3) c′ = 0 where c is a constant.
(4) power rule: (xn)′ = nxn−1
(5) exponential rule: (ax)′ = ax ln(a), speical case: (ex)′ = ex.
(6) sinusodial function: (sin(x)) = cos(x), (cos(x))′ = − sin(x)
Problem 99. Find derivatives of the following functions.
(1) f(x) = 1√x
(2) f(x) = 2x3 + 5x2 − 9x + 1 (3)
f(x) = 1x − 1
x2
(4) y = t2+1t (5) f(t) = t2−
√t+1
2t
(6) f(x) = 5x3 + 6x + 4 (7) f(q) = eq + qe
(8) f(x) = 2x, find f ′′(x) (9) f(x) = (ln 2)x
Problem 100. Find equation of the tangent line of f(x) at (1, 6), where f(x) = 2x2 +
3x+ 1.
Problem 101. f(x) = a · xn, f ′(3) = 12, f ′(9) = 108. Find a, n.
Problem 102. g(x) = ax2 + bx + c, f(x) = 2x. If f(0) = g(0), f ′(0) = g′(0), f ′′(0) =
g′′(0), find g(x).
39
40
3.2 Product and quotient rules
product rule: (f(x)g(x))′ = f ′(x)g(x) + f(x)g′(x)
quotient rule: (f(x)g(x) )′ = f ′(x)g(x)−f(x)g′(x)
(g(x))2
Problem 103. Find derivatives of the following functions. (Hint: simplify the function
first, if possible)(1) f(x) = x · 15x
(2) f(x) = (√x− 1)(
√x+ 1)
(3) f(x) = x7
4x
(4) f(r) = 5r6r+1
(5) f(z) = z2+1√z
(6) f(x) = 14ex
5x
3.3 Graph problems
(1) f ′(x) is undefined where the graph of f(x) has a corner point.
(2) if f(x) is a linear function, then f ′(x) is the slope.
(3) If something in one expression doesn’t exist, then the whole expression doesn’t
exist. For example, let h(x) = f(x)g(x), we want to find h′(3). Since h′(x) =
(f(x)g(x))′ = f ′(x)g(x) + f(x)g′(x), we have h′(3) = f ′(3)g(3) + f(3)g′(3), then if
f ′(3) doesn’t exist, the whole thing h′(3) doesn’t exist.
Problem 104. Graph (a) gives the graph of f(x) and g(x), where the dashed one is
g(x). Let h(x) = f(x) · g(x), j(x) = f(x)g(x) . Find
(1) h′(1) (2) h′(2)
(3) h′(3) (4) j′(1)
(5) j′(2) (6) j′(3)
Problem 105. Use graph (b), do problem 104 again.
[] 1 2 3 4
1
2
3
4
[] 1 2 3 4
1
2
3
41
3.4 Increasing/decreasing and concavity
(1) Solve f ′(x) = 0 and label the solutions on x-axis, which divide the x-axis into
several intervals. Determine the sign of f ′(x) on each interval by plugging a value
and seeing the sign. Then we know when f ′ > 0 and f ′ < 0, thus when f is
increasing/decreasing.
(2) Solve f ′′(x) = 0 and label the solutions on x-axis, which divide the x-axis into
several intervals. Determine the sign of f ′′(x) on each interval by plugging a value
and seeing the sign.Then we know when f ′′ > 0 and f ′′ < 0, thus the concavity.
Problem 106. On which interval is f(x) = x4 − 4x3 both increasing and concave up?
Problem 107. On which interval is f(x) = x · e−x increasing/decreasing/concave
up/concave down?
3.5 Simple derivative rules
(1) (sin(x))′ = cos(x), (cos(x))′ = sin(x), (tan(x))′ = 1cos2(x)
(Note that we write (cos(x))2 as cos2(x), similarly for sin2(x).)
(2) (ln(x)) = 1x , (log(x))
′ = 1x ln(10)
(3) (arcsin(x))′ = 1√1−x2
, (arccos(x))′ = − 1√1−x2
, (arctan(x))′ = 11+x2
Notation: for y = f(x), we can write the derivative as f ′(x), or dydx . Here the numerator
dy means y is the function, and the denominator dxmeans x is the variable. For example,
s = f(u) can be written as f ′(u) or dsdu .
For y = f(x), the second derivative can be written as f ′′(x) or d2ydx2 , still the symbols
d2y and dx2 tell you what the output and input are.
3.6 Chain rules
The chain rule below enable us to differentiate the composition of two functions.
42
(f(g(x)))′ = f ′(g(x))g′(x)
In practice, to apply chain rule, we can always circle something in the function into
a box � (as your g(x)), and calculate the following two derivatives:
(1) treat the function as a function of the box, and take derivative with respect to the
box; (this is in fact f ′(g(x)))
(2) take the derivative of whatever function you circled in the box; (this is in fact g′(x))
then multiply these two derivatives will give you the answer.
Example 1. we want to take the derivative of sin(x2), we circle x2 into the box �, now
(1) derivative of sin(�) is cos(�) (treat � as the variable);
(2) the function in � is x2, it’s derivative is 2x
so (sin(x2))′ = cos(�) · 2x = cos(x2) · 2x
Problem 108. Find the derivatives (basic examples).(1) e2x
(2) (x+ 10)128
(3)√2− x3
(4) ln(−2x+ a)
(5) cos(sin(x2))
(6) pcos(x) (where p is a constant)
Problem 109. Find the derivatives (advanced examples, simplification may be helpful).
(1)√
(x2 · 7x)7
(2) ( ba+x2 )
2
(3)√
e−5t2 + 5
(4) a sin(x3
b − c) where a, b, c are constants
(5) k(a) = sin7(a) cos7(a)
(6) a(t) = ln((1−sin(t)1+sin(t))
9)
Problem 110. Gateway test level problems.(1) R(s) = (5s+ ln(2)) cos(es)
(2) y = 2(x6 − 3)7 + π
(3) y = (7− ln(2x))π
(4) C(t) = 6(sin(t))2 − e2
(5) f(x) = ln(ln(t)) + ln(ln(2))
(6) z = y3+1cos(y)−1
Practice gateway test: http://instruct.math.lsa.umich.edu/courses/115/gw/
Calculator and notecard are NOT allowed in gateway test. The gateway test
only covers the following rules (however, the exams/quizzes will cover all the rules):
43
(1) Linear: (mx+ b)′ = m, c′ = 0
(2) Power: (xn)′ = nxn−1
(3) Exponential: (ax)′ = ax ln(a), (ex)′ =
ex
(4) Sinusoidal: (sin(x))′ = cos(x),
(cos(x))′ = − sin(x), (tan(x))′ = 1cos2(x)
(5) ln: (ln(x))′ = 1/x
Arithmetic, product/quotient/chain rules
(1) (f(x) + g(x))′ = f ′(x) + g′(x)
(2) (cf(x))′ = cf ′(x)
(3) (f(x)g(x))′ = f ′(x)g(x) + f(x)g′(x)
(4) (f(x)/g(x))′ = f ′(x)g(x)−f(x)g′(x)(g(x))2
(5) (f(g(x))′ = f ′(g(x))g′(x)
Problem 111 (Derivatives regarding arctan, arcsin, arccos). Find f ′(x)
(1) f(x) = arcsin(2x).
(2) f(x) = cos(arctan(3x))
(3) f(x) = sin(arcsin(x+ 1))
Problem 112. Let g(x) = (f(x))2, find g′(x) and g′′(x).
Problem 113. x = x(t) is a function of t, suppose dxdt = x′(t) = x sin(x), find d2x
dt2=
x′′(t).
Problem 114. f(2) = 1, f ′(2) = 5, f(4) = 3, f ′(4) = 7 and g(4) = 2, g′(4) = 6, g(3) =
4, g′(3) = 8. Find:
(a) h′(4) if h(x) = f(g(x)).
(b) h′(4) if h(x) = g(f(x)).
(c) h′(4) if h(x) = f(x)/g(x).
Problem 115. If k(2) = 1, k′(2) = 3
(a) f(x) = k(2x), find f ′(1)
(b) f(x) = k(x+ 1), find f ′(1)
(c) f(x) =√
k(x), find f ′(2)
(d) f(x) = k(√x), find f ′(4)
Problem 116 (Winter 07, exam 2, #2). Suppose f and g are differentiable with the
values given by the table. Calculate the following derivatives:
44
x f(x) g(x) f ′(x) g′(x)
1 2 9 -3 7
3 4 11 15 -19
(a) If h(x) = f(x)g(x), find h′(3).
(b) If j(x) = (g(x))3
f(x) , find j′(1).
(c) If d(x) = x ln(ef(x)), find d′(3).
(d) If t(x) = cos(g(x)), find t′(1).
Problem 117. Given below is a graph of a function f(x) and a table for an invertible
function g(x).Give answers for the following or write “Does not exist”. No partial credit
will be given.
0 1 2 3 4 5
1
2
3
4f HxL
x 1 2 3 4 5
g(x) 0.5 1 4 5 6
g′(x) 1 3 5 4 2
(1) h(x) = f(g(x)), find h′(2)
(2) t(x) = ln(√ef(x)), find t′(2.5)
(3) k(x) = f(x)g(x), find k′(3)
(4) p(x) = 2f(x)− 3g(x), find p′(4)
(5) q(x) = f(x)g(x) , find q′(4)
Problem 118. With t in years, the population of a herd of deer is represented by
P (t) = 4000 + 500 sin(2πt)
(a) Graph P (t) over one year.
(b) When the population is a maximum/minimum? What is that maximum/minimum?
(c) When is the population growing fastest? When is it decreasing fastest?
(d) How fast is the population changing on July 1?
3.7 Inverse function rule
(f−1)′(x) = 1/f ′(f−1(x))
45
Problem 119. If f(x) = x3 + 1.(a) h(x) = (f(x))5, find h′(1).
(b) n(x) = f(x−1), find n′(1).
(c) g(x) = f−1(x), find g′(3).
(d) k(x) = (f(x))−1, find k′(1).
3.8 Implicit functions
3.8.1 Derivative
Given f(x, y) = 0, we can treat y = y(x) as a function of x. To find y′(x), we
take the derivative to f(x, y) = 0 with respect to x, and solve for y′(x). Keep in
mind y is a function, we need to use chain rules sometimes. e.g. (y2)′ = 2y · y′
instead of 2y.
Problem 120. Find dydx (in other words, y′) for the following implicit functions.
(a) x2 + y2 =√7
(b) x2 + xy − y3 = xy2
(c) xy + x+ y = 5
(d) sin(xy) = 2x+ 5
(e)√x = 5
√y
(f) cos2(y) + sin2(y) = y + 2
3.8.2 Tangent line and approximation
Let (x0, y0) be a point on f(x, y) = 0, then
(1) at x = x0, i.e. at (x0, y0), the slope of the tangent line is y′(x0), and the line is
y = y0 + y′(x0)(x− x0)
(2) for x ≈ x0, y ≈ y0 + y′(x0)(x− x0).
Problem 121. Consider the curve y3 + xy = 6.
(a) Find dydx .
(b) Find the tangent line at (5, 1).
(c) Estimate y when x = 4.98.
Problem 122. Consider the curve y2 = x2
xy−4 .
(a) Find dydx . (Hint: Simplify first, y2 = x2
xy−4 , so y2(xy−4) = x2, so xy3−4y2−x2 = 0)
46
(b) Find the tangent line at (4, 2).
(c) Estimate the y value when x = 4.01.
Problem 123. Consider the curve given by xy + y2 − 16x = −5. Find all points at
which the curve has a horizontal tangent line. It’s possible that none exists.
Problem 124. Consider the curve given by x2 + y2 − 2y = 3. Find all points at which
the curve has horzontal/vertical tangent line. It’s possible that none exists.
3.9 Tangent line approximation review
Given a function f(x), at the point x = a (or at the point (a, f(a)))
1. We call L(x) = f(a) + f ′(a)(x − a) the tangent line approximation (or local
linearization) at x = a.
2. The tangent line at x = a is y = L(x).
3. For x ≈ a, f(x) ≈ L(x).
4. If the graph of f(x) is concave up, then the tangent line is below the graph,
and L(x) > f(x), the estimate is an overestimate.
If the graph of f(x) is concave down, then the tangent line is above the graph,
and L(x) < f(x), the estimate is an underestimate.
Problem 125. Let f(x) =√x+ 8.
(1) Find the tangent line approximation at x = 1.
(2) Find the tangent line at x = 1.
(3) Estimate f(1.03). Is it an underestimate or overestimate? Why?
Problem 126. The function y = f(x) has a tangent line approximation at x = 3 given
by
L(x) = −2x+ 5
(a) What are the values of f(3) and f ′(3)?
47
(b) Estimate f(2.8).
(c) Suppose that f ′′(3) = 0 and that f ′′(x) < 0 for x < 3. Is your estimate in (b.) an
overestimate or an underestimate?
3.10 A few more problems
Problem 127. True or false question.
(1) f(x) is continuous on (a, b), then f(x) is differentiable on (a, b).
(2) f(x) is periodic function, then f(x+ 1) + 2 is also a periodic function.
(3) f(x) is concave up, then −f(x) + 1 is also concave up.
Problem 128. Give an example.
(1) A function which is continuous, but not differentiable at x = 1 and x = 2.
(2) A function such that limx→1
f(x) ̸= f(1). Is the function continuous?
(3) A function which is contiunous, concave up, not differentiable at x = 1, and limx→∞
f(x) =
3
Problem 129. Suppose g(1) = 2, g(2) = 1, g′(1) = 3, g′(2) = 5. Find (g−1)′(2),
g′(g(2)), h′(1) if h(x) = g(g(x)), t′(1) if t(x) = ln(e2g(x)).
Problem 130. Given below is the graph of f ′(x) where 0 ≤ x ≤ 6.
1 2 3 4 5 6
(1) when is f(x) increasing/decreasing? concave up/down?
(2) when is f ′(x) largest/smallest? positive/negative?
(3) when is f ′′(x) largest/smallest? positive/negative?
(4) when is f ′(x) increasing/decreasing?
48
(5) suppose a particle is moving on x-axis, and the velocity is positive if it moves to
the right. If the graph is the graph of the velocity, when does the particle move to the
left/right? When is the acceleration positive/negative?
(6) if this graph is the graph of the acceleration, and the velocity is 0 at t = 0.5, how does
the particle move in the first 5 seconds? what can you tell about the velocity in the first
5 seconds?
Chapter 4
Using the derivative
4.1 Critical point & Local extrema (local max/min)
4.1.1 Critical point
If f ′(a) = 0 or f ′(a) is undefined, then we say f(x) has a critical point at x = a or at
the point (a, f(a)).
Problem 131. Find critical points of f(x) = x3 − 3x2 − 9x− 1.
Problem 132. Find critical points of f(x) = xe−2x.
4.1.2 Critical points and local extrema (local max/min)
(1) local extrema (local max/min) must be a critical point.
(2) critical point may not be a local extrema.
At a critical point, the graph could be like “∩” (local max), “
∪” (local min), or chair
shape (not local extrema).
Two rules to classify the critical points as a local max/min or neither.
1st derivative test:
if f ′(x) changes sign from “+” to “−” at x = a, then this critical point is a local max;
if f ′(x) changes sign from “−” to “+” at x = a, then this critical point is a local min;
if f ′(x) doesn’t changes sign at x = a, this critical point is neither local max nor local min.
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50
When we use the 1st derivative test, firstly find all the critical points (solve f ′(x) = 0),
then draw the domain of f(x) and label all the critical points. The critical points will
divide the domain into several intervals, for each interval, pick any point inside, the sign
of f ′(x) at that point will be the sign of f ′(x) on that particular interval.
2nd derivative test:
if f ′′(a) < 0, then this critical point is a local max;
if f ′′(a) > 0, then this critical point is a local min;
if f ′′(a) = 0, then this test tells us nothing, use 1st derivative test instead.
Problem 133. Find the critical points of f(x) = x4
4 − x3
3 and classify all the critical
points.
Problem 134. Find the critical points of f(x) = x4(x− 1)5 and classify all the critical
points.
Problem 135. f(x) = 2x3 − 3ax2 where a is a positive constant. Find critical points
in [−a, 32a] and classify them.
Problem 136. Find and classify the critical points of f(x) = e−(x−a)2/b.
4.2 Inflection point
Inflection point is where the concavity changes. At an inflection point, f ′′ equals 0, but
he converse is not true.
We can find the inflection points in the following way.
(1) solve for f ′′(x) = 0. these are candidates for inflection points.
(2) A solution x = a is an inflection point if and only if f ′′(x) changes sign at x = a.
(Reason: f(x) concave up (down) is same as f ′′(x) > 0 (< 0), so the change of concavity
is the same as the sign change of f ′′.)
Problem 137. True or false questions.
(1) m′′(a) = 0, then m(x) has an inflection point at x = a.
(2) There exists a continuous function f(x) which is not differentiable at x = 0 with a
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local maximum at (0, 5).
(3) If j′(x) is continuous everywhere and changes from negative to positive at x = a,
then j has a local minimum at x = a.
(4) If a continuous function H has exactly one local maximum and two local minima,
then there are exactly three distinct values of x sucht that H ′(x) = 0.
Problem 138. Find inflection points of f(x) = x4
12 − x2
2 .
Problem 139 (Winter 10, exam 2, #3). Find all local minima, local maxima, and
inflection points of the function f(x) defined below. Your answer may involve the positive
constant B.
f(x) = e−18x2+B
4.3 Finding global max/min
To distinguish from the local max/min, we call the max/min of a function global
max/min.
For a function defined on an interval
(1) If the interval is not closed ([a, b]), the global max/min may not exist, e.g. f(x) =
1/x on (0, 1]; e.g. g(x) = x2 on (1, 3).
(2) If the interval is closed, the global max/min both exist.
(3) Global max/min can only occur at the critical points and the boundary of the
interval.
4.3.1 General method
To find the global max/min, we find all the critical points in the interval, and compare the
value of f(x) at these critical points and at the boundary points. The largest (smallest)
one is the global max (min). However, if the largest (smallest) value happens at the
boundary, and the boundary is not in the interval, then the global max (min) doesn’t
exist, since this largest (smallest) value cannot be reached.
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Example 2. f(x) = x2 on [−2, 2]. Critical point: x = 0; boundary: x = −2, x = 2.
f(0) = 0 is the smallest, f(2) = f(−2) = 4 is the largest, so global max is 4 at x = ±2,
and global min is 0 at x = 0.
Example 3. f(x) = x2 on (−3, 2]. Critical point: x = 0; boundary: x = −3, x = 2.
f(0) = 0 is the smallest, so it’s global min; f(−3) = 9 is the largest, but since −3 is not
in (−3, 2], the value 9 cannot be reached, so there is no global max.
Example 4. f(t) = 2te−t on (0,∞). f ′(t) = 2e−t(1 − t), so critical point is t = 1;
boundary is x = 0, x = ∞. f(1) = 2e−1, f(0) = 0, f(∞) = limt→∞
f(t) = 0. f(1) is the
largest, so it’s global max. f(0) = f(∞) = 0 is the smallest, but it cannot be reached by
f , so no global min. However, f(t) on [0,∞) has global min, since this time t = 0 is in
[0,∞).
Example 5. f(x) = x2−2x+3 on (−∞,∞). Critical point: x = 1; boundary: x = ±∞.
f(1) = 2, f(∞) = f(−∞) = ∞, so f has global min 2 at x = 1, but no global max.
Problem 140. Find global max/min of f(x) = x− 2 ln(x+ 1) on [0, 2].
Problem 141. Find global max/min of f(x) = x3 − 9x2 − 48x + 52 on the following
intervals.
(a) [−5, 14) (b) [−5,∞) (c) (−∞,∞)
Problem 142. Find global max/min of f(x) = t1+t2
on (−∞,∞).
Problem 143. Find global max/min of f(x) = 2x3 − 3ax2 (here a > 0) on interval
[−a, 1.5a].
Problem 144 (Win 09, exam 2, #2). Suppose a is a positive constant, and consider
the function f(x) = 13x
3− 4a2x. Determine all maxima and minima of f in the interval
[−3a, 5a]. For each, specify whether it is global or local.
4.3.2 An important special case
Most times f has only one critical point on the interval. In this case, if this critical
point is a local max (min), it must be the global max (min) as well, so we can easily
find the global max (min). The first/second derivative test can help us to determine if
the critical point is a local max/min.
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Problem 145. Find global max of f(x) = 6x− x2 + 10 on (−∞,∞).
Problem 146. Find global max of f(x) = 2te−t if t > 0.
Problem 147. Find global max of f(x) = e−18x2+B where B is a constant.
Problem 148. The potential energy, U , of a particle moving along the x-axis is given
by
U = b(a2
x2− a
x)
where a and b are constants and x > 0. What value of x miminizes the potential energy?
4.3.3 Functions involving two related variables
Problem 149. If xy = 150 and x > 0, find the global minimum of 2x+ 3y.
(Hint: it’s same to find global min of f(x) = 2x+ 450x with x > 0)
Problem 150. If 2x+ 4y = 100 and x ≥ 0, y ≥ 0, find the global maximum of 100xy.
(Hint: it’s same to find global max of f(x) = 100x(25− 0.5x) with 0 ≤ x ≤ 50)
Problem 151. If 2x+ 4y = 30, find the global max of e−x2−y2.
4.4 Finding the constants in a function
Problem 152. f(x) = x3 + bx2 + cx+ d has local extrema at x = 1 and x = −3.
(1) Find b, c, d and classify the local extrema as local max or local min.
(2) Find inflection points of f(x).
Problem 153. p(x) = ax4 + bx3 + cx2 + dx + e is symmetric about y−axis, has y−
intercept 1, and global mimima at (1, 0) and (−1, 0). Find a, b, c, d.
Problem 154 ((Bell-shape curve). f(x) = e−(x−a)2
b where b > 0. If f(x) has a local
extrema at x = 1 and an inflection point at x = 2. Find a, b.
Problem 155. f(x) = ae−x + bx has a global minimum at (1, 2). Find a, b.
Problem 156. Consider f(x) = e−ax2+bx+c where a, b and c are constants. Suppose
f(x) has a local extremum at (1, 1) and the y-intercept of f(x) is e. Find a, b and c.
54
Problem 157. Let f(x) = e5x − kx where k > 0.
(1) f(x) has a local minimum. Find it.
(2) Which k maximizes the y coordinate of the local minimum in part (1).
4.5 Optimization problems
In this section, you will be given a word problem and you need to maximize/minimize
some quantity. Procedure is as follows:
(1) set up two variables.
(2) the problem will give you an equality involving the two variables, find it and
solve one variable in terms of the other.
(3) find the function (in two variables) you need to maximize/minimize, then use (2)
to reduce to one variable.
(4) there is only one critical point, and this critical point gives the global max/min
you want. Use derivative tests to classify it and then use the fact that local max/min
implies global max/min.
Problem 158. Finding the dimensions giving the minimum surface area, given that the
volume is 8 cm3.
(1) A closed rectangular box, with a square base x by x cm and height h cm.
(2) A closed cylinder with radius r cm and height h cm.
Problem 159 (Fall 10, exam 2, #8). Farmer Fred is designing a fence next to his barn
for his grass-fed herd of cattle. The fence will be rectangular in shape with wooden fence
on three sides and a chain link fence on the side closest to his barn. The wooden fence
costs $6 per foot and the chain link fence costs $3 per foot. If he wants the fenced area
to be 40,000 square feet, what should the dimensions of his fence be in order to minimize
his total cost?
Problem 160 (Win 07, Final, #4). The zoo has decided to make the new octopus tank
spectacular. It will be cylindrical with a round base and top. The sides will be made of
55
Plexiglas which costs $65.00 per square meter, and the materials for the top and bottom
of the tank cost $50.00 per square meter. If the tank must hold 45 cubic meters of water,
what dimensions will minimize the cost, and what is the minimum cost?
Problem 161 (page 209, #46). On the same side of a straight river are two towns, and
the town people want to build a pumping station. The pumping station is to be at the
river’s edge with pipes extending straight to the two towns. Where should the pumping
station be located to minimize the total length of pipe? (Graph in on the book)
Problem 162. A rectangle has one side on the x-axis and two vertices on the curve
y = 41+x2 . Find the vertices of the rectangle with maximum area.
Problem 163. A piece of wire of length L cm is cut into two pieces. One piece, of
length x cm (x may be equal to 0 or L), is made into a circle. The rest is made into a
square.
(a) Verify that A(x), the sum of the areas of the circle and the square, is given by
A(x) = (1
16+
1
4π)x2 − L
8x+
L2
16
show your work.
(b) Find the critical point(s) of A(x).
(c) Find the values of x that minimize and maximize A(x). Justify.
Problem 164. A box has a bottom with one edge 3 times as long as the other. If the
box has no top and the volume is fixed at V , what dimensions minimize the surface area?
Problem 165. Consider a window the shape of which is a rectangle of height h sur-
mounted by a triangle having a height T that is 1.3 times the width w of the rectangle
(as shown in the figure below).
If the cross-sectional area is A, determine the dimensions of the window which min-
imize the perimeter.
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4.6 Marginal cost/revenue
Let q be the number of items produced. Say the cost of producing q items is C(q),
the revenue is R(q), then the profit is π(q) = R(q)− C(q).
C(0) is called the fixed cost, which is the cost needed (for example, machinary cost)
to begin the production, i.e, it’s the cost before the production.
(1) We call MC = C ′(q) the marginal cost. When the production level is q, i.e,
when q items are produced, the cost of an additional item is approximately C ′(q).
(2) We call MR = R′(q) the marginal revenue. When the production level is q, the
revenue of an additional item is approximately R′(q).
(3) π′(q) = R′(q) − C ′(q) = MR − MC. When the production level is q, the cost
of one more item is about MC, the revenue of one more item is about MR, so the
profit of one more item is about MR−MC.
If MR > MC, we will make money by producing one more item.
If MR < MC, we will lose money by producing one more item.
When the profit π(q) has a local max/min, π′(q) = MR−MC = 0, so MR = MC.
Problem 166. (a) When production is 2200, marginal revenue is 3 dollars per unit
and marginal cost is 7.5 dollars per unit. Do you expect maximum profit to occur at a
production level above or below 2200?
(b) If production is increased by 70 units, estimate the change in profit.
Problem 167. (Final, Fall 10, #6) The Green Bag Company (GBC) makes hand bags
out of recycled materials. A table of the company’s marginal cost, MC, and marginal
57
revenue, MR, at various production levels q is given below. The variable q is the number
of hand bags produced, and marginal cost and marginal revenue are measured in dollars
per bag.
q 1000 2000 3000 4000 5000 6000
MC 100 81 75 96 112 123
MR 125 123 114 110 107 106
Assume for this problem that GBC’s cost and revenue functions are twice differentiable
and that MC and MR are either increasing or decreasing on each interval shown in the
table.
1. At which production level from the table is GBC’s profit increasing the fastest?
2. The CEO of the company thinks profit is maximized at 3000 bags, but the CFO
of the company thinks that profit will be maximized at 4500 bags. Who could be
correct?
Problem 168. The figure below shows cost (blue) and revenue (black) functions over
the domain 0 < q < 30 (in thousands of units).
1. For what production levels is the profit function positive? negative?
2. Estimate the production at which profit is maximized.
58
3. If the blue one is the graph of marginal cost, black one is the graph of marginal
revenue, find the local max/min of the profit.
4.7 Rates and related rates
4.7.1 Rate is derivative
(1) d�d△ is the rate of change of � with respect to △.
(2) when we talk about how fast x is changing, we are talking about dxdt .
Problem 169. (page 226, #36, see Figure 4.87) A lighthouse is 2 km from the long,
straight coastline. Find the rate of change of the distance of the spot of light from the
point O with respect to the angle θ.
Problem 170. (Win 10, Final, #8) A train is travelling eastward at a speed of 0.4
miles/minute along a long straight track, and a video camera is stationed 0.3 miles from
the track, as shown in the figure. The camera stays in place, but it rotates to focus on
the train as it moves.
Suppose that t is the number of minutes that have passed since the train was directly
north of the camera; after t minutes, the train has moved x miles to the east, and the
camera has rotated θ radians from its original position.
1. Write an equation that expresses the relationship between x and t.
2. Write an equation that expresses the relationship between x and θ.
3. Suppose that 7 minutes have passed since the train was directly north of thee cam-
era. How far has the train moved in this time, and how much has the camera
rotated?
4. How fast is the camera rotating (in radians/min) when t = 7?
4.7.2 Related rates
Problem 171. Volume of sphere is given by V = 43πr
3, where r is the radius.
1. what’s the relation between dVdt and dr
dt ?
59
2. When radius is 2 meters, the radius is increasing at a rate of 3m/s, then how fast
is the volume changing?
3. When radius is 2 meters, the radius is increasing at a rate of 3m3/s, then how fast
is the radius changing?
How fast is the surface area changing? (Surface area A = 4πr2)
Problem 172. (page 225, #21) Gasoline is pouring into a cylindrical tank of radius 3
feet. When the depth of the gasoline is 4 feet, the depth is increasing at 0.2 ft/s. How
fast is the volume of gasoline changing at that instant? (Volume V = πr2h where h is
the depth).
Problem 173. A ship is sailing out to sea from a dock, moving in a straight line
perpendicular to the coast. At the same time, a person is running along the coast toward
the dock, hoping desperately to jump aboard the departing ship. Let b(t) denote the
distance in feet between the ship and the dock t seconds after its departure, and let p(t)
denote the distance in feet between the person and the dock t seconds after the ship’s
departure. The situation is depicted below for your reference:
Suppose that 10 seconds after the ship’s departure, it is 40 feet from the dock and is
sailing away at a speed of 20 ft/sec. At the same moment, the person is 30 feet from the
dock and running toward it at 14 ft/sec.
1. What is b′(10)? What is p′(10)?
2. Is the distance between the person and the ship increasing or decreasing 10 seconds
after the ship’s departure? How fast is it increasing or decreasing? (Include units
in your answer, and keep in mind that distance is measured along a straight line
joining the person and the ship.)
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Problem 174. A certain type of spherical melon has weight proportional to its volume
as it grows. When the melon weighs 0.2 pounds, it has a volume of 36 cm3 and its
weight is increasing at a rate of 0.1 pounds per day. [Note: The volume of a sphere is
V = 43πr
3].
1. Find dV/dt when the melon weighs 0.2 pounds (t measured in days).
2. Find the rate at which the radius of the melon is increasing when it weighs 0.2
pounds.
Problem 175. A potter forms a piece of clay into a cylinder. As he rolls it, the length,
l, of the cylinder increases and the radius, r, decreases. If the length of the cylinder is
increasing at 0.1 cm per second, find the rate at which the radius is changing when the
radius is 1 cm and the length is 5 cm.
Problem 176. (page 226, #29)
1. A hemisphere bowl of radius 10 cm contains water to a depth of h cm. Find the
radius of the surface of the water as a function of h.
2. The water level drops at a rate of 0.1 cm per hour. At what is rate is the radius
of the water decreasing when the depth is 5 cm?
Problem 177. (Fall 09, Final, #5) A cone-shaped container has a small opening at
the bottom, so the water is leaking. At 10 am, the depth of the water is 25 cm, and the
water surface radius is 2 cm, and the volume of the water is decreasing at 1.2 cm3/h.
1. What the relation between depth h and radius r of the water surface?
2. Find the formula for volume V of the water in terms of r.
3. At 10 am, how fast is the radius changing? How fast is the depth changing?
Chapter 5
Key concept: the definite integral
5.1 Distance/position/velocity
The distance traveled is the area between the graph of velocity and t-axis.
The change in position is the area above t-axis minus the area below t-axis, since
the area above t-axis is the distance traveled forward, and the area below t-axis is
the distance traveled backward.
Problem 178. The velocity, in ft/s, of a car is given by v(t) = 9−3t. Find the distance
travelled for 0 ≤ t ≤ 3.
Problem 179. The velocity of a particle along x-axis is given by v(t) = 12− 4t cm/s.
Find the change in position, and the distance traveled, of the particle from t = 0 to
t = 5s.
Problem 180. A car initially going 40 ft/s brakes at a constant rate (constant negative
acceleration), coming to a stop in 8 seconds.
(1) Graph the velocity for t = 0 to t = 8. How far does the car travel before stopping?
(2) How far does the car travel before stopping if its initial velocity is doubled, but it
brakes at the same constant rate?
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62
5.2 Estimate distance from t = a to t = b
Given: The formula/table of v(t), estimate the distance traveled from t = a to t = b.
Method: Divide a ≤ t ≤ b into several, say n subdivisions, then the length of each
subdivision is ∆t = b−an . Estimate the area of each subdivision by a rectangle, where
the base of the rectangle is ∆t = b−an , and height can be chosen as maximal/minimal
height or left/right side height. Take the sum of all subdivisions, we get an estimate
of the area (distance), and the estimate is called the upper/lower estimate or left/right
sum, depending on which height you use.
Remark:
(1) If v(t) is increasing, then lowerestimate=left sum≤distance=area≤ right sum=upperestimate.
(2) If v(t) is decreasing, then upperestimate=left sum≤distance=area≤ right sum=lowerestimate.
Problem 181. The following table gives the velocity of a car, which is decreasing.
t (seconds) 0 2 4 6
v (ft/s) 40 20 10 0
Find the upperestimate/lowerestimate and the left/right sum of the distance traveled in
the first 6 seconds.
Problem 182. The following table gives the velocity of a car, which is increasing.
t (seconds) 0 3 6 9 12
v (ft/s) 31 33 36 39 40
Find the upperestimate/lowerestimate and the left/right sum in the first 12 seconds.
Problem 183. v(t) = 47+t . Use ∆t = 0.2, find an over and under estimate of the
distance travelled for 0 ≤ t ≤ 1.
Problem 184. v(t) = e−t2. Use ∆t = 0.2, find an over and under estimate of the
distance travelled for 0 ≤ t ≤ 1.
5.3 Definite integral
Given the graph of f(x), consider the area of the region bounded between f(x) and
x-axis.
63
Definition
(1) When a ≤ b,∫ ba f(x)dx = area above x-axis − area below x-axis.
(2) When a > b, use∫ ba f(x)dx = −
∫ ab f(x)dx to swap the limits
(3) When a ≤ b,∫ ba |f(x)|dx = total area between f(x) and x-axis.
Some properties
(1)∫ ba f(x)dx+
∫ cb f(x)dx =
∫ ca f(x)dx
(2)∫ ba m · f(x)dx = m ·
∫ ba f(x)dx where m is a constant.
(3)∫ ba f(x)dx = −
∫ ab f(x)dx (switch limits, change sign)
(4)∫ ba mdx = m(b− a) where m is a constant.
Problem 185 (page 271, #31). Find values of
(a)∫ ba f(x)dx (b)
∫ cb f(x)dx (c)
∫ ca f(x)dx (d)
∫ ca |f(x)|dx
Problem 186 (page 271, #32). Given∫ 0−2 f(x)dx = 4, find:
(1)∫ 20 f(x)dx (2)
∫ 2−2 f(x)dx (3) total shaded area.
Problem 187. See the following graph of f(x).
(a) Find∫ 0−3 f(x)dx.
(b) If the area of the shaded region is A, find∫ 4−3 f(x)dx and
∫ 4−2 f(x)dx.
-4 -2 2 4
-1.0
-0.5
0.5
1.0
Problem 188. Graph f(x) = x(x+ 4)(x− 3), use definite integral to express the total
area between the graph and x-axis between x = −4 and x = 3.
(1) The unit of∫ ba f(x)dx is the unit of f(x) times the unit of x
(2) If the velocity is v(t), then the change in position from t = a to t = b is∫ ba v(t)dt,
and the distance traveled is∫ ba |v(t)|dt.
(3) If f(x) is an even function (i.e., f(x) = f(−x), the graph is symmetric about
y-axis), then∫ a−a f(x)dx = 2
∫ 0−a f(x)dx = 2
∫ a0 f(x)dx, and
∫ 0−a f(x)dx =
∫ a0 f(x)dx
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(4) If f(x) is an odd function (i.e., f(−x) = −f(x), the graph is symmetric about
the origin), then∫ a−a f(x)dx = 0, and
∫ 0−a f(x)dx = −
∫ a0 f(x)dx
Problem 189. True or false questions.
(1) From t = 1s to t = 4s, a particle of velocity v(t) (in m/s) traveled∫ 41 v(t)dt meters.
(2) If a(t) is the acceleration in m/s2, then the unit of∫ 41 a(t)dt is m/s.
(3) If∫ 10 x2dx = A, then
∫ 0−1 x
2dx = −A.
(4) If∫ 20 f(x)dx = A, then
∫ 20 (f(x) + 4)dx = A+ 4.
5.4 Estimate definite integral using left/right sum
Result:∫ ba f(x)dx = lim
n→∞left sum = lim
n→∞right sum where n is the number of subdi-
visions.
Therefore we can use left/right sum to estimate the integral. Use more subdivisions, the
left/right sum will be closer to the integral, so the estimate is more accurate.
Remark: If f(x) is increasing, then L ≤∫ ba f(x)dx ≤ R, so the left sum (L)/right sum
(R) is an lower/upperestimate. If f(x) is decreasing, then L ≥∫ ba f(x)dx ≥ R, so the
left sum (L)/right sum (R) is an upper/lowerestimate.
Problem 190. Use left/right sum with n = 4 subdivisions to approximate∫ 73 t2dt.
Problem 191. Use n = 5 subdivisions to approximate∫ 73
11+xdx.
Problem 192. Use n = 5 to estimate∫ 150 f(x)dx.
x 0 3 6 9 12 15
f(x) 50 40 20 10 5 0
Problem 193 (Win 09, Final, #2). L is the left sum for∫ 60 f(x)dx using 3 equal
subdivisions, while R is the right sum using 3 subdivisions. Compare L, R,∫ 60 f(x)dx
for the following graphs.
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5.5 Fundamental theorem of calculus
Fundamental theorem:∫ ba f ′(x)dx = f(b)− f(a)
Problem 194.
(a) Find derivative of x ln(x)− x.
66
(b) Find the exact value of∫ 42 ln(x)dx.
Problem 195. The following is the graph of f ′(x). Suppose f(0) = 0, find
a. f(1) b. f(3) c. compare f(4), f(5).
Problem 196. The following is the graph of f ′(x).
(1) compare f(2), f(4), f(6), f(8).
(2) compare A. f(6)−f(2)2 B. f(4)− f(2) C. f(6)− f(4) D. f(8)− f(6)
1 2 3 4 5 6
-1.0
-0.5
0.5
1.0 2 4 6 8
-6
-4
-2
Problem 197. (a). Find derivative of f(x) = ex2. (b). Find
∫ 10 2xex
2dx.
5.6 Interpretation of integrals
1. Interpret integra∫ ba f(x)dx, there are two ways:
(1) Use the template from the fundamental theorem:
∫ b
arate of change of something = change of that thing from t = a to t = b
(2) Use units: unit of∫ ba f(x)dx = unit of f(x) times unit of x
2. The average of f(x) on a ≤ x ≤ b is defined to be∫ ba f(x)dx
b−a .
Problem 198. Find the average value of f(x) = 2x+ 1 over [2, 6].
Problem 199. The population t years after 2000 is given by P = f(t) = 1.01t million.
Write an expression for the average population from 2000 to 2003.
Problem 200. Interpret:
(1)∫ 60 a(t)dt where a(t) is the acceleration of a car in km/hour2.
67
(2)∫ 20042000 f(t)dt where f(t) is the rate at which world’s population (in billion people) is
growing in year t.
(3)∫ 50 s(x)dx where s(x) is the rate of change of salt concentration (in g/liter per meter)
x meters below the sea surface.
(4) Oil leaks out of a tanker at a rate of r = f(t) liters per minute, where t is in minutes.
If f(t) = Ae−kt, write a definite integral expressing the total quantity of oil which leaks
out of the tanker in the first hour.
(5) (page 279, #23) A two day environmental clean up started at 9 am on the first day.
The number of workers fluctuated as shown in the figure. If the workers were paid $10
per hour, how much was the total personnel cost of the clean up?
Problem 201 (Win 10, Final, #9). Suppose that the standard price of a round-trip plane
ticket from Detroit to Paris, purchased t days after April 30, is P (t) dollars. Assume
that P is an invertible function (even though this is not always the case in real life).
Interpret:
(a)∫ 105 P ′(t)dt.
(b) 15
∫ 105 P (t)dt.
Problem 202 (Fall 10, Final, #7). The rate at which a coal plant releases CO2 into the
atmosphere t days after 12:00 am on Jan 1, 2010 is given by the function E(t) measured
in tons per days. Suppose∫ 310 E(t)dt = 223.
(a) Interpret∫ 5931 E(t)dt.
(b) Interpret E(15) = 7.1.
(c) The plant is upgrading to “clean coal” technology which will cause its July 2010 CO2
emissions to be one fourth of its Jan 2010 CO2 emissions. How much CO2 will the coal
plant release into the atmosphere in July?
(d) Using a left sum with 4 subdivisions, write an expression which approximates∫ 5931 E(t)dt.
Problem 203 (Fall 09, Final, #2). Let C(t) be the temperature, in degree Fahrenheit,
of a warm can of soda t minutes after it was put in a refrigerator. Suppose C(10) = 62.
(a) Interpret C−1(45) = 40.
(b) Interpret C ′(10) = −0.4.
(c) Interpret∫ 100 C ′(t)dt = −5.
68
(d) Assuming the statements in part (a)-(c) are true, determine C(0).
(e) Interpret∫ 10 C(t)dt. (Hint:
∫ 10 C(t)dt =
∫ 10 C(t)dt
1−0 )
Problem 204. The rate q(t) at which cars passed through the intersection of Main Street
and Huron after a football game is presented in the table below.
t (in minutes after the game ended) 0 20 40 60 80 100 120
q(t) (in cars per minute) 10 15 19 21 20 17 13
(a) What”s the meaning of∫ 1200 q(t)dt? Use left sum and n = 6, estimate
∫ 1200 q(t)dt.
(b) Write an expression for the average rate at which cars passed through the intersection
for the first two hours after the game ended.
(c) Estimate q′(30).
(d) If Q(t) denotes the total number of cars that have passed through the intersection t
minutes after the game ended, find and interpret Q′(60).
Problem 205. Jack has a leaky faucet for his kitchen sink. The leak drips water into
the faucet at a constant rate of 10 mL per minute. The water is also draining out of the
sink, but the rate the water drains out is not constant because of the pile of dirty dishes
present. Let r(t) be the rate, in mL per minute, that the water is draining out of the
sink t minutes after 7:00pm. The graph of r(t) can be found below. Assume the sink has
5mL of water in it at 7:00.
5 10 15 20t, minutes
5
10
15
rHtL, mL per minutes
(1) How much water drains out of the sink between 7:00 and 7:15?
(2) When is there the most water in the sink? How much water is in the sink at this
69
time?
(3) How much water is in the sink at 7:20?
(4) Suppose that at 7:20, Jack removes the dirty dishes from the sink and water is allowed
to drain freely at a rate of 15mL per minute. How long will it take for the sink to have
no water in it? [Hint: It may help to use part c]
70
Chapter 6
Constructing antiderivatives
6.1 Antiderivatives
6.1.1 Definition and formulas
(1) Antiderivative of f(x) are the functions (Not unique) whose derivative is f(x).
(2) If F ′(x) = f(x), then F (x) is an antiderivative of f(x).
(2) If F (x) is an antiderivative of f(x), then all antiderivatives have the form F (x) + c
where c is a constant.
(4) We use∫
f(x)dx to represent the antiderivatives of f(x),∫
f(x)dx = F (x) + c.
Formula of antiderivatives:
(1)∫
kdx = kx+ c where k is a constant.
(2)∫
xndx = xn+1
n+1 + c when n ̸= 1.∫
x−1dx =∫
1xdx = ln(|x|) + c
(3)∫
axdx = ax
ln a + c, in particular,∫
exdx = ex + c.
(4)∫
sin(x)dx = − cos(x) + c,∫
cos(x)dx = sin(x) + c,∫
tan(x)dx = 1cos2(x)
+ c
(5)∫
(f(x)± g(x))dx =∫
f(x)dx±∫
g(x)dx.
(6)∫
k · f(x)dx = k ·∫
f(x)dx where k is a constant.
Remark: When use the formulas, we always ingore the constant c in the middle
steps and add it to the answer in the last step.
Problem 206. Find the antiderivatives of f(x).
71
72
(1) f(x) = x3
2 + 1 (2) f(x) = 1x7 − 1
x (3) f(x) = sin(x) + 3 cos(x)
(4) f(x) =√x− x−1/3 (5) f(x) = 2ex − 9x (6) f(x) = (x− 1)2
6.1.2 Use antiderivatives to compute definite integrals
Fundamental theorem: If F (x) is an antiderivative of f(x), i.e. F ′(x) = f(x),
then ∫ b
af(x)dx = F (b)− F (a)
Reason:∫ ba f(x)dx =
∫ ba F ′(x)dx = F (b)− F (a).
Method: To compute∫ ba f(x)dx, firstly find all antiderivatives by calculating∫
f(x)dx, then pick any antiderivative to be F (x), then∫ ba f(x)dx = F (b)− F (a),
and it doesn’t matter which antiderivative you use.
Problem 207. Find
(1)∫ 30 (x2 − 1)dx (2)
∫ π20 (cos(x) + sin(x))dx (3)
∫ c0 3exdx
Problem 208. Find the exact area below the curve y = x3(1− x) and above x-axis.
Problem 209. Find the exact bounded by the x-axis and the graph of y = 6− 5x+ x2.
Problem 210. Find the exact area between the graphs of y = 1 − x and y = ex for
0 ≤ x ≤ 1.7.
Problem 211. Find the exact area between y = sin(x) and y = cos(x) for 0 ≤ x ≤ π.
Problem 212. If the average value of f(x) = 2x on the interval [1, c] is 6. Find c if
c > 1.
6.2 Graph antiderivatives
Given the graph of f ′(x), what information do we know about f(x)?
(1) A few points on the graph of f(x), using the fundamental theorem.
∫ b
af ′(x)dx = f(b)− f(a)
73
(2) The critical points of f(x): where f ′(x) = 0 or undefined.
(3) Inflection points of f(x):
where f ′(x) changes from increasing to decreasing, or from decreasing to increas-
ing.
(4) Where f(x) is increasing (or decreasing):
where f ′ > 0 (or f ′ < 0).
(5) Where f(x) is concave up (down):
where f ′ is increasing (decreasing).
Problem 213 (Fall 08, Final, #2). The graph of the derivative of the continuous func-
tion M(x) is given below. Suppose M(−4) = −2, sketch the graph of M(x). Give all the
critical points, inflection points and endpoints of M(x) on the interval [−4, 4].
-4 -2 2 4
2
1
-1
Problem 214 (Fall 10, Final, #1). Given below is a graph of h′(x)
(a) sketch a possible graph of h(x).
(b) give the x-coordinate of the inflection points of h
(c) give the x-coordinate of the global max of h on [−3, 3].
-3 -2 -1 1 2 3
Problem 215 (Win 07, Final, #7). Given below is a graph of h′(x), and we know
74
h(10) = 10. sketch a possible graph of h(x). Indicate the critical points and inflection
points of h and give coordinates of those points.
-5 5 10 15 20 25 30 35 40 45 50-2
2
Problem 216 (Fall 09, Final #1). Given below is the graph of h′(x).
(1) Find the x-coordinates of all critical points of h in (−5, 5), and classify each as a
local max/min or neither.
(2) Find the x-coordinates of any inflection points of h in the interval (−5, 5).
(3) Find the x-coordinates of on [−5, 5] where h attains its global max/min.
(4) If h(1) = 3, find the linear approximation of h(x) at x = 1. Is this approximation
an over/underestimate of h for points near x = 1?
-5 -4 -3 -2 -1 1 2 3 4 5
-1
1
2
3
Problem 217. Given below is the graph of f ′′(x). Assume f(0) = f ′(0) = 0, decide at
which point,
(a) f(x) is largest? smallest?
(b) f ′(x) is largest? smallest?
(c) f ′′(x) is largest? smallest?
75
a b c d e
76
Chapter 7
Exam review materials
77
78
7.1 Exam 1 review
Sinusoidal function.
#1. The average weight of a squirrel in Ann Arbor oscillates sinusoidally between a low
of 5 pounds on January 1 and a high of 9 pounds on July 1, and a low of 5 pounds on
January 1 next year again. Suppose that the function P (t) gives the average weight in
pounds of an Ann Arbor squirrel t months after January 1.
a). What is the amplitude of P (t)? b).
What is the period of P (t)?
c). Find a formula for P (t).
Interpretation.
#2. The cost, C (in dollars) to produce g gallons of ice cream can be expressed as
C = f(g). Assume f is invertible. Interpret
(1) f ′(100) = 2.5
(2) (f−1)′(100) = 2.5
Finding derivatives.
#3. Given below is a graph of a function f(x) and a table for a function g(x).
Given answers for the following or write “Does not exist”. Each problem is worth 2
1 2 3 4
1
2
3
4
x -3 -2 -1 0 1 2 3
g(x) -2 -3 -2 -1 1 3 2
g′(x) -2 0 1 3 2 0.5 -1
points.
79
(a) h(x) = g(x)f(x) . Find h′(−3). (b) k(x) = 3f(x)− g(x). Find k′(2)
(c) q(x) = f(x)g(x). Find q′(−2) (d) a(x) = g(−f(x)). Find a′(1)
(e) l(x) = e2f(x). Find l′(3) (f) p(x) = sin( πf(x)). Find p′(1)
(g) t(x) = ln((eg(x))2). Find t′(−2)
Concavity, f ′′, f ′, and f
#4. “Winning the war on poverty” has been described cynically as slowing thee rate at
which people are slipping below the pverty line. Let N be the number of people below
the poverty line at time t, answer the following questions, and draw a possible graph.
(1) If N is increasing at a faster and faster rate.
Is N ′(t) increasing, decreasing or neither? Draw a possible graph.
(1) If N is decreasing at a slower and slower rate.
Is N ′(t) increasing, decreasing or neither? Draw a possible graph.
#5. A particle is moving along a straight line. Its distance, s, to the right of a fixed
point is given by the figure 7.1. Estimate:
(a) when the particle is moving to the right and when it’s moving to the left.
(b) when the acceleration of the particle is zero, when it’s negative, and when it’s positive.
#6. Figure 7.2 is the graph of y = f ′(x) (not f(x)), At which of the marked values
of x is
A. f(x) greatest? B. f(x) least?
C. f ′(x) greatest? D. f ′(x) least?
E. f ′′(x) greatest? F. f ′′(x) least?
#7. Kenny opens the concert in Ford Field with his recent hit “Living in Fast Forward”
and spends the entire song, 0 ≤ t ≤ F , running along the front of the stage. Beth
contains her excitement enough to record Kenny’s velocity v(t) as a function of time
t, shown in the graph below; when velocity is positive Kenny is running to the right.
Further, let s(t) represent Kenny’s position right of the center of the stage at time t.
Answer the following questions using the values of t labelled on the graph. If no intervals
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1 2 3 4
Figure 7.1: Problem 5
1 2 3 4 5 6
Figure 7.2: Problem 6
or labelled points satisfy the condition you must write “NONE” to receive credit. No
explanation is needed.
A B C D E FΠ 5 Π3 3 Π 11 Π3 5 Π 6 Πt
vHtL
(1) During which interval(s) is Kenny’s acceleration is positive?
(2) At which point(s) does Kenny turn around?
(3) During which interval(s) is the graph of s(t) increasing?
(4) During which interval(s) is the graph of s(t) concave down?
(5) At which point is Kenny running the fastest toward the left end of the stage?
(6) During the interval 0 ≤ t ≤ D, at which point is Kenny farthest to the right?
(7) During which interval(s) are both Kenny’s velocity increasing and his acceleration
increasing?
transformations
#8. A(t) is the percentage of the Math 115 Midterml Exam that the student Wolv E.
Rine will have completed t minutes after the ocial start time of the exam under normal
81
circumstances, i.e. assuming he arrives on time, etc. What will the function be in the
following situations?
(1) Wolv E. Rine arrives p minutes late for the exam.
(2) The instructor accidentally hands Wolv E. Rine a copy of the exam that has the first
p% of the solutions already filled in.
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7.2 Exam 2 review
1. Derivative problem
Caution: Firstly find the derivative of the function completely, then plug in the value.
Example: f(1) = 1, f ′(1) = 2, let h(x) = (f(x))2, find h′(1). Correct answer is:
(1) Find the derivative compleltely. h′(x) = [(f(x)2)]′ = 2f(x)f ′(x).
(2) Then plug in 1. h′(1) = 2f(1)f ′(1) = 2 · 1 · 2 = 4.
Note: we should write (f(x))′ as f ′(x).
Wrong solution: h′(1) = [(f(1))2]′ = 2f(1)(f(1))′ = 2 · 1 · 2 = 4.
Mistake 1: We cannot plug in number before we find the derivative of the function, in
fact, [(f(1))2]′ = 0 since (f(1))2 is constant.
Mistake 2: (f(1))′ is not f ′(1), since in fact (f(1))′ = 0 b/c it’s constant. We should
write f ′(1) instead of (f(1))′.
#1. Given below is a graph of a function f(x) and a table for an invertible function
g(x).
1 2 3 4 5
1
2
3
4f �x�
x 1 2 3 4 5
g(x) 0.5 1 4 5 6
g′(x) 1 3 5 4 2
Calculate the following derivatives. If the derivative doesn’t exist, write “DNE”.
(1) [f(f(x))]′ at x = 1.5
(2) [ (f(x))2
g(x) ]′ at x = 1
(3) [π2π − π2f(x)]′ at x = 4
(4) [g(2x− 1) + cos(x2)]′ at x = 1
(5) h(x) = 1g−1(x)
, find h′(4).
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2. Tangent line approximation/linear localization
#1. The graph of the derivative, f ′(x), of a continuous function f is shown below:
-4 -2 2
-2
2
4
f 'HxL
(a) If f(−3) = 5, find the tangent line approximation, L(x), to f(x), near x = -3.
L(x) =
(b) Compare each pair of quantities using one of the signs >,< or =. No explanation is
needed.
i) L(−2.9) f(−2.9)
ii) L(−4) f(−4)
iii) f(0) f(1)
iv) f ′(−3) f ′(3)
v) f ′′(−4) f ′′(2)
#2. Last week, the 2011 World Snail Racing Championships were held in Norfolk,
England. This years winner is Zoomer, a snail from Ann Arbor. Let s(t) be the distance
Zoomer has run (in inches), where t is the time that has passed (in seconds) since the
start of the race, and s(t) is given by
s(t) = et(7− 4t+ t2)− 7
(a) Find the velocity of Zoomer at t = 1.
Solution: s′(1) = e1(3− 2 · 1 + 12) = 2e inch/s
(b) Find the linear localization (in other words, tangent line approximation) of s(t) at
84
t = 1.
Solution: s′(1) = 2e, s(1) = 4e − 7, so the tangent line approximation is L(t) =
4e− 7 + 2e(t− 1) = 2et+ 2e− 7
(c) Use part (b) to approximate the distance Zoomer has travelled in the first 1.02 sec-
onds.
Solution: s(1.02) ≈ 2e · 1.02 + 2e− 7 = 4.04e− 7 inches
(d) Is your estimate in part (c) and overestimate or underestimate? Use concavity to
justify your reasoning.
Solution: s′′(t) = et(1 + t2) > 0, so graph of s(t) is concave up, the estimate must be
an underestimate.
3. Implicit function
#1. Consider the curve given by x2 + xy + y2 = y.
(a) Compute dydx . (In other words, y′)
Solution: take the derivative, we have 2x + y + xy′ + 2yy′ = y′, so xy′ + 2yy′ − y′ =
−(2x + y), factor y′ out on the left side, we have y′(x + 2y − 1) = −(2x + y), so
y′ = − 2x+yx+2y−1 .
(b) Find all points at which the curve has a horizontal tangent line. If none exists, write
“None.”
Solution: y′ = − 2x+yx+2y−1 = 0 at the points where the curve has a horizontal tangent
line.
so 2x+ y = 0, y = −2x. Plug this back into the original equation x2 + xy + y2 = y, we
find 3x2 = −2x, so x = 0 (so y = −2x = 0) or x = −23 (so y = −2x = 4
3).
Therefore there are 2 such points, one is (0, 0), the other is (−23 ,
43).
4. Critical pts and inflection pts, functions with constants
#1. Consider f(x) = e−ax2+bx+c where a, b, and c are constants. Suppose f(x) has a
local extremum at (1, 1) and the y-intercept of f(x) is e. Find a, b, and c.
Solution: We know f(1) = 1, f ′(1) = 0, and f(0) = e.
f(1) = 1 ⇒ 1 = e−a+b+c ⇒ −a+ b+ c = 0
85
f ′(x) = e−ax2+bx+c(−2ax+ b), so f ′(1) = 0 ⇒ 0 = e−a+b(−2a+ b) ⇒ −2a+ b = 0
f(0) = e ⇒ e = ec ⇒ c = 1
Solving the system of equations −a+b = −1 and −2a+b = 0, we get a = −1 and b = −2
#2. Consider f(x) = ax3 + bx2 + cx where a, b, and c are constants. Suppose f(x)
has a local maximum at x = −1 and an inflection point at (1,−10). Find a, b, and c.
Solution: f ′(x) = 3ax2 + 2bx+ c, f ′′(x) = 6ax+ 2b
f(x) has a local maximum at x = −1 implies f ′(−1) = 0, so 3a − 2b + c = 0 (equation
[1])
(1,−10) is on the graph of f(x), so f(1) = a+ b+ c = −10 (equation [2])
f(x) has an inflection point at (1,−10) implies f ′′(1) = 6a+ 2b = 0 (equation [3])
equation [1]-equation [2], we have 2a− 3b = 10 (equation [4])
from equation [3], we have b = −3a, plug this into equation [4]
we find, 11a = 10, so a = 10/11
so b = −3a = −30/11, and use equation [2], we have c = −10− a− b = −90/11
5. Optimization problem
#1. The cost of fuel to propel a boat through the water (in dollars per hour) is pro-
portional to the cube of the speed. A certain ferry boat uses $100 worth of fuel per
hour when cruising at 10 miles per hour. Apart from fuel, the cost of running this ferry
(labor, maintenance, and so on) is $675 per hour. At what speed should it travel so as
to minimize the cost per mile travelled?
Solution: Suppose v is the speed in miles per hour, and t is the time used to travel one
mile, then vt = 1.
On one hand, for the cost of fuel, we know
cost of fuel in dollars per hour = constant · speed3
Since it uses $100 worth of fuel per hour when speed is 10 miles per hour. We know this
constant is 0.1.
86
Therefore we have
cost of fuel in dollars per hour = 0.1v3
Since it takes t hours to run for 1 mile, so 1 mile cost 0.1v3t dollars for fuel.
One the other hand, the cost of running the ferry for one mile is 675t dollars.
Therefore to run one mile, the total cost is 0.1v3t+ 675t
Now because vt = 1, we have t = 1/v, so the cost now is
C = 0.1v3 · 1v+ 675/v = 0.1v2 + 675/v
Take the derivative, C ′ = 0.2v − 675/v2 = 0, we have 0.2v = 675/v2, then 0.2v3 = 675,
so v3 = 3375, so v = 15, there is only one critical point.
C ′′ = 0.2 + 1350/v3, so C ′′(15) > 0, so the critical point is a local min by the second
derivative test.
Now because there is only one critical point and it’s a local min, we know it must be a
global min.
Therefore when the speed v is 15 miles per hour, the cost of running for 1 mile is small-
est, which is C(15) = 0.1 · 152 + 675/15 = 67.5 dollars.
#2. We want to construct a box with a square base and we only have 10 m2 of ma-
terial to use in construction of the box. Assuming that all the material is used in the
construction process determine the maximum volume that the box can have.
#3. A printer need to make a poster that will have a total area of 200 in2 and will
have 1 inch margins on the sides, a 2 inch margin on the top and a 1.5 inch margin on
the bottom. What dimensions will give the largest printed area?
#4. Dirk just finished his last class of the week Friday afternoon at 3pm and he is
very excited about the weekly 2 vs 2 free throw shooting contest scheduled later that
night at 11pm. In order to prepare himself for the competition, he can either practice
87
shooting free throws or get focused by listening to motivational music. His free throw
percentage, P , is a function of how many hours, S, he spends shooting and how many
hours, M , he spend listening to music and is given by
P = (9 + 8S − S2)e0.2M
How should Dirk spend his time in order to maximize his free throw percentage in the
contest, and what will his free throw percentage be if he prepares optimally?
Solution: From 3pm to 11pm, there are 8 hours in total, so S + M = 8, we have
M = 8−S, and we know 0 ≤ S ≤ 8. Substitute M with 8−S, the free throw percentage
is:
P = (9 + 8S − S2)e0.2(8−S)
Let P ′ = 0.2 · e0.2(8−S)(S2− 18S+31) = 0, we have S2− 18S+31 = 0, by the quadratic
formula, we find 2 solutions, S = 5√2 + 9 ≈ 16 or S = −5
√2 + 9 ≈ 2. Only the second
solution is in the domain 0 ≤ S ≤ 8, so P has a unique critical point at S = 2.
Use the first derivative test, pick 1 from interval (0, 2), P ′(1) > 0, so P ′ > 0 on this
interval;
pick 3 from interval (2, 8), P ′(3) < 0, so P ′ < 0 on this interval.
By the first derivative test, we know P has a local max at S = 2, and since it’s the
unique critical point, P must have a global max there.
Therefore, when S = 2, M = 8− S = 6, the max free throw percentage P = (9 + 8 · 2−
22)e0.2(8−2) = 69.7.
#5. Henri is a lazy student, who never studies until the night before the exam. Tomor-
row there are two important exams, calculus and linear algebra. In order to survive in
these two exams, he decides to study the whole night without sleeping. Now there are
only 12 hours left before the exams. According to his experience, if he spends x hours
on calculus, he can study 65e−0.2x calculus problems per hour; if he spends y hours on
linear algebra, he can study 5e−0.2(12−y) linear algebra problems per hour.
(a) In order to maximize the total number of problems he can study, how many hours
88
should he spend on calculus? Show all your steps!
Solution: Suppose he spends x hours on calculus, then he will spend 12− x hours on
linear algebra. The total number of problems he can do would be
P = x · 65e−0.2x + (12− x) · 5e−0.2(12−(12−x)) = 60(1 + x)e−0.2x
P ′ = 12(4− x)e−0.2x, so the unique critical point is at x = 4.
Now compare the values of P at end points and the critical points, P (0) = 60, P (4) =
135, P (12) = 71, so the global max occurs at the critical point, and this global max is
135.
Therefore he should spend 4 hours on calculus, and he can do a max of 135 problems in
total.
(b) Suppose he has to spend at least 8 hours on calculus and 1 hour on linear alge-
bra to pass both exams. And according to his experience, he has to study at least 120
problems in total to get an average GPA A. Is it possible for him to get an average A?
Explain.
Solution: He has to spend at least 8 hours and at most 11 hours on calculus, so
8 ≤ x ≤ 11. P doesn’t have any critical point in [8, 11] (since P has a unique critical
point at x = 4), so we just compare the value of P at end points.
P (8) = 109, P (11) = 80, so the global max is 109, which is smaller than the required
120, therefore he will not be able to get an average GPA A.
#6. Inspired by Math 115, you have decided to try to optimize your budget for the
coming month. You only buy two things: Diet Coke at 1 dollar per liter and Top Ramen
noodles at 4 dollars per kilogram. You have 30 dollars to spend this month. If you buy
C liters of Diet Coke and R kilograms of Top Ramen noodles, you will be malnourished
89
at the end of the coming month with probability P given by
P = e−6√R−
√C
What combination of pop and noodles should you buy to minimize the probability that
you become malnourished? Include units.
Solution: We have 1 · C + 4 · R = 30, so C = 30 − 4R. Substitute, now the prob-
ability we want to minimize is P = e−6√R−
√30−4R.
Since C ≥ 0, we have 30− 4R ≥ 0, so R ≤ 7.5, so the domain is 0 ≤ R ≤ 7.5.
P ′ = −(3√R
− 2√30− 4R
)e−6√R−
√30−4R = 0
We have 3√R= 2√
30−4R, square both sides, we have 9
R = 430−4R , so 9(30− 4R) = 4R, so
R = 6.75.
Now let’s check P has a local minimum at R = 6.75. Pick any value in (0, 6.75) and
(6.75, 7.5), say 1 and 7, we have P ′(1) < 0 and P ′(7) > 0, so P ′ < 0 on (0, 6.75) and
P ′ > 0 on (6.75, 7.5). By the first derivative test, P has a local minimum at R = 6.75,
and since there is only one critical point, P has a global minimum at R = 6.75. Therefore,
R = 6.75 and C = 30− 4 · 6.75 = 3 minimize the probability.
90
7.3 Exam 3 review
1. basics on derivatives
1.1 definition: f ′(x) = limh→0
f(x+ h)− f(x)
h, f ′(a) = lim
h→0
f(a+ h)− f(a)
h
#1. Let f(x) = xx, write down the definition of f ′(2).
1.2 estimation:
(1) f ′(a) ≈ f(a+h)−f(a)h for h close to 0
(2) f(x) ≈ f(a) + f ′(a)(x− a)
Here the right side, L(x) = f(a) + f ′(a)(x− a) is called the tangent line approximation
or local linearization at x = a.
If f(x) is concave up at x = a (f ′′(a) > 0 or f ′(x) is increasing at x = a), then it’s
underestimate, L(x) < f(x).
If f(x) is concave down at x = a (f ′′(a) < 0 or f ′(x) is decreasing at x = a), then it’s
overestimate, L(x) > f(x).
#2. Let f(x) = xx, estimate f ′(2).
#3. Suppose E(t) is the total amount (in tons) of CO2 released into the air during
the first t days in 2011. The following is the table for E′(t) (not E(t)!).
t 3 5 7 9 11 13 15
E′(t) 1.5 2 2.2 3 3.1 4 4.2
(1) Interpret E′(3) = 2.
(2) If E(3) = 5, estimate E(4). Is your estimate an overestimate or underestimate?
Why?
(3) Use integral to represent the amount of CO2 released from Jan 3 to Jan 15.
91
(4) Use n = 3 subdivisions to estimate the integral in part (3). Find an over and
under estimate.
1.3 Interpretation. There are a lot of problems on the handout and past exams.
2. advanced topics on derivatives
2.1 Optimization problems
Keep in mind:
(1) there is always one critical point. If it’s a local max, it’s a global max; if it’s a local
min, it’s a global min.
(2) check it’s a local max/min, and say “since there is only one critical pt, it’s a local
max(min) implies it’s a global max (min)”.
#4. A company is producing and selling chairs. The fixed cost is $100, and the ad-
ditional cost is $5 per chair. If we sell the chair at a price of $10 per chair, 100 chairs
can be sold, and the experience is that if we increase the price by $1, 5 less chairs can
be sold.
(1) If we sell the chair at a price of x dollars, how many chairs can be sold?
(2) use your answer is part (1), find the cost and revenue.
(3) maximize the profit, what’s the price then?
2.2 Related rates problem
#5. page 227, #45.
92
#6. When in space, the force, in Newtons, exerted on a certain satelite is inversely
proportional to the square of the distance between the satelite and the Earth.
(1) Suppose that when the satelite is 90 kiolmeters away from earth, the force exerted on
it is 50 Newtons. Find a formula for the F (d) force (in Newtons) exerted on the satelite
when it is d miles from Earth.
(2) In order to gather more data, the satelite starts moving further from earth at a
rate of 4 km/hr. At what rate is the force exerted on the satelite changing (in new-
tons/hr).
(3) The time, T (measured in minutes), it takes for the satelite to make a full orbit
around Earth can be found by T 2 = 341.8d3. How fast is the time required for one orbit
changing?
3. Integral
3.1 Interpration of integrals, see the handout.
3.2 use left/right sum to estimate integral.
(1) if f(x) is increasing, then left sum≤∫ ba f(x)dx ≤ right sum, left sum is an underes-
timate and right sum is an overestimate.
(2) if f(x) is decreasing, then right sum≤∫ ba f(x)dx ≤ left sum, left sum is an overes-
timate and right sum is an underestimate.
#7. For a super bowl party, Timmy brought a giant bowl filled with 200 ounces of
93
spinach dip for a communal snack. Let f(t) be the rate at which the dip is being eaten,
in ounces per minute, t minutes after the start of the game. Some values of f(t) are in
the table below. Assume f(t) is decreasing.
t 0 10 20 30 40 50 60
f(t) 5 4 3.5 3 2 1 0
(1) Interpret (but do not compute)∫ 4010 f(t)dt. Include units.
(2) Using the table, find a left hand riemann sum ( in other words, left sum ) ap-
proximation of∫ 4010 f(t)dt using n = 3 subintervals ( in other words, subdivisions).
(3) Not all of the dip is eaten during the game. How big of a container does Timmy need
to make sure he can store the leftovers? [Hint : find an underestimation of how much
dip is eaten during the game]
3.3 understand integral from graph.
#8. In the distant future, the University of Michigan will install electrically heated
sidewalks. These heaters cause accumulated snow to melt at a rate of 2 inches per day.
Suppose that p(t) (which is graphed below) gives the rate, also measured in inches per
day, at which snow accumulates during one snowy winter week. Assuming there is no
snow on the ground at time t = 0, answer the following questions.
94
H0, 3L
H3, 9L
1 2 3 4 5 6 7t, days
-2
2
4
6
8
10
pHtL, inches per day
(1) Estimate the time t on [0, 7] when the depth of the snow is the greatest.
(2) Estimate the depth of the snow when t = 5.
(3) how much snow falls per day, on average, in the week for which data is given?
#9. Jack has a leaky faucet for his kitchen sink. The leak drips water into the faucet at
a constant rate of 10 mL per minute. The water is also draining out of the sink, but the
rate the water drains out is not constant because of the pile of dirty dishes present. Let
r(t) be the rate, in mL per minute, that the water is draining out of the sink t minutes
after 7:00pm. The graph of r(t) can be found below. Assume the sink has 5mL of water
in it at 7:00.
5 10 15 20t, minutes
5
10
15
rHtL, mL per minutes
95
(1) How much water drains out of the sink between 7:00 and 7:15?
(2) When is there the most water in the sink? How much water is in the sink at
this time?
(3) How much water is in the sink at 7:20?
(4) Suppose that at 7:20, Jack removes the dirty dishes from the sink and water is
allowed to drain freely at a rate of 15mL per minute. How long will it take for the sink
to have no water in it? [Hint : It may help to use part c]
3.4 calculate integrals.
(1) use definition for∫ ba f(x)dx, it’s area above x-axis minus area below x-axis. When
use the definition, make sure a ≤ b.
(2) use fundamental theorem for∫ ba f ′(x)dx, which equals f(b)− f(a).
(3) average of f(x) on [a, b] is∫ ba f(x)dx
b−a
Exercise: see 5.4 handout, #7.
#10. f(x) is an even function. The graph of f(x) with x ≥ 0 is given below, which
contains a quarter circle with radius of 2 and two line segments.
96
1 2 3 4 5 6
-2
-1
1
2
Also suppose g(x) is an odd function, with g(1) = 1, g(2) = 2, g(4) = 3,∫ 40 g(x)dx = 5
and∫ 20 g(x)dx = 3. Find the following integrals. (1)
∫ 31 f(x+ 1)dx
(2)∫ 60 (−2f(x) + 1)dx
(3)∫ 41 f ′(g(x))g′(x)dx
(4)∫ 42 g(x)dx
(5)∫ 2−2 (g(x) + f(x))dx
3.5 use integral to calculate area between curves.
#11. Compute the total shaded area between y = x2 − x− 1 and y = −x2 + x+ 3.
y= x2 - x - 1
y=-x2 + x + 3
97
4 Sketch f(x) from f ′(x)
Given graph of f ′(x), find critical/inflection points of f(x), and sketch f(x). There are
a lot of problems on 6.1&6.2 handout, part 2.
98
Chapter 8
Solution to selected problem sets
8.1 Solution to section 2.4
1. The cost C in dollars of building a house A square feet in area is given by C = f(A).
(a) Interpret f′(100) = 200.
Solution: The cost of building a house of area 101 square feet is approximately 200
dollars more than building a house of area 100 square feet.
Or you may interpret like this: For a house of area 100 square feet, it costs approxi-
mately 200 dollars per square feet for extra area.
(b) If 100 square feet house costs 30000 dollars. Estimate the cost of building a house
of 100.5 square feet.
Solution: Use the second interpretation, 0.5 square feet more in area will cost approx-
imately 0.5 · 200 = 100 dollars, so the cost would be approximately 30000+100 = 30100
dollars.
2. P = f(t) is the population in Mexico in millions, where t is the number of years
since 1980. Interpret:
(a) f′(6) = 2
Solution: In 1986, the population of Mexico was increasing by approximately 2 million
people per year.
Or, the population of Mexico in 1987 is approximately 2 million more than the pop-
99
100
ulation in 1986.
(b) f−1(95.5) = 16
Solution: When the population is 95.5 million, the year is 1996.
(c) (f−1)′(95.5) = 0.46
Solution: When the population is 95.5 million, it takes approximately 0.46 years for
the population to increase by 1 million.
Or, when the population is 95.5 million, it takes approximately 0.46 years for the
population to reach 96.5 million.
3. (Winter 10, exam 1, #9)Suppose that W (h) is an invertible function which tells
us how many gallons of water an oak tree of height h feet uses on a hot summer day.
Interpret:
(a) W (50)
Solution: A 50 feet tall Oak tree uses W (50) gallons of water on a hot summer day.
(b) W−1(40)
Solution: W−1(40) is the height of an oak tree in feet, which uses 40 gallons of water
on a hot summer day.
(c) W′(5) = 3
Solution: An oak tree which is 6 feet tall uses approximately 3 more gallons of water
per hot summer day than a 5 feet tall oak tree does.
Or, if a 5 feet tall oak tree grew an extra foot, it uses approximately 3 more gallons
of water per hot summer day.
4. Let N = f(t) be the total number of cans of cola Sean has consumed by age t
in years. Interpret the following in practical terms, paying close attention to units.
(a) f(14) = 400
Solution: By age 14, Sean had consumed 400 cans of cola.
101
(b) f−1(50) = 6
Solution: Sean had consumed 50 cans of cola by age 6.
(c) f′(12) = 50
Solution: At age 12 Sean was consuming approximately 50 cans of cola per year.
(d) (f−1)′(450) = 1/70
Solution: At the time Sean had consumed 450 cans of cola, it took approximately one
year to consume 70 additional cans.
5. (page 96, #11) An economist is interested in how the price of a certain item af-
fects its sales. At a price of $p, a quantity, q, of the item is sold. If q = f(p), interpret:
(a) f(150) = 2000
Solution: When the price is 150 dollars, 2000 items are sold.
(b) f′(150) = −25
Solution: When the price is 150 dollars, approximately 25 less items are sold for 1
dollar increase in price.
6. g(v) is the fuel efficiency, in miles per gallon, of a car going at a speed of v miles per
hour. Interpret g′(55) = 0.54
Solution: If the car speeds up from 55 mph to 56 mph, then the car becomes less fuel
efficient by approximately 0.54 miles per gallon.
7. (page 96, #19) Water is flowing into a tank; the depth, in feet, of the water at
time t in hours is h(t). Interpret, with units, the following statements.
(a) h(5) = 3
Solution: 5 hours after the depth of water in the tank is 3 feet.
(b) h′(5) = 0.7
102
Solution: 5 hours after, the depth of water is increasing by approximately 0.7 feet per
hour.
Or, the depth of water 6 hours after is approximately 0.7 feet more than the depth
5 hours after.
(c) h−1(5) = 7
Solution: It takes 7 hours for the water to be 5 feet deep.
(d) (h−1)′(5) = 1.2
Solution: It takes approximately 1.2 hours for the water to be from 5 feet deep to 6
feet deep.
8.2 Solution to section 4.6
In this section, we study rates and related rates.
(1) If y = f(x), then dydx = f ′(x) tells us how fast y is changing with respect to x.
(2) If y = f(x) and x = g(t), we want to know how fast y is changing with respect to t,
i.e, dydt . We use a new version of chain rule:
dy
dt=
dy
dx· dxdt
Part 1. Rates
1. According to the US Census, the world population P , in billions, was approximately
P = 6.7e0.011t, where t is in years since Jan 1, 2007. At what rate was the world’s
population increasing on that date?
Solution: The rate is dPdt |t=0 = 6.7 · 0.011 · e0.011·0 = 0.0737 billion people per year.
2. With time t, in minutes, the temperature, H, in degrees Celsius, of a bottle of
water put in the refrigerator at t = 0 is given by
H = 4 + 16e−0.02t
103
How fast is the water cooling initially? After 10 minutes?
Solution: H ′(t) = −0.32e−0.02t. So the temperature is changing at a rate of H ′(0) =
−0.32 degree celsius per min initially, and at a rate of H ′(10) = −0.32e−0.2 degree celsius
per min after 10 minutes.
3. (page 226, #36, see Figure 4.87) A lighthouse is 2 km from the long, straight coast-
line. Find the rate of change of the distance of the spot of light from the point O with
respect to the angle θ.
Solution: x = 2tan(θ), so the rate of change is dxdθ = 2
cos2(θ).
4. (Win 10, Final, #8) A train is travelling eastward at a speed of 0.4 miles/minute
along a long straight track, and a video camera is stationed 0.3 miles from the track, as
shown in the figure. The camera stays in place, but it rotates to focus on the train as it
moves.
Suppose that t is the number of minutes that have passed since the train was directly
north of the camera; after t minutes, the train has moved x miles to the east, and the
camera has rotated θ radians from its original position.
(a) Write an equation that expresses the relationship between x and t.
Solution: x = 0.4t
(b) Write an equation that expresses the relationship between x and θ.
Solution: x = 0.3 tan(θ).
(c) Suppose that 7 minutes have passed since the train was directly north of thee camera.
How far has the train moved in this time, and how much has the camera rotated?
Solution: x = 0.4 · 7 = 2.8 miles.
(d) How fast is the camera rotating (in radians/min) when t = 7?
Solution: Since x = 0.3 tan(θ) and x = 0.4t, we have 0.4t = 0.3 tan(θ), so tan(θ) =
104
4t/3, so θ = arctan(4t/3). Take the derivative, we have
dθ
dt=
1
1 + (4t/3)2· 43
So when t = 7, the camera is rotating at a rate of dθdt |t=7 =
11+(4·7/3)2 ·
43 = 0.015 radians
per minute.
Part 2. Related rates
Method: Find formula, take derivative (use chain rule dydt = dy
dx ·dxdt ), then two rates are
known, find the unknown.
1. Volume of sphere is given by V = 43πr
3, where r is the radius.
(1) what’s the relation between dVdt and dr
dt ?
Solution:
dV
dt=
dV
dr· drdt
= 4πr2 · drdt
(2) When radius is 2 meters, the radius is increasing at a rate of 3m/s, then how fast is
the volume changing?
Solution: We know when r = 2, drdt = 3. Therefore, when r = 2, we have
dV
dt= 4π · 22 · 3 = 48πcm3/s
Therefore when r = 2, the volume is increasing at a rate of 48πcm3/s
(3) When radius is 2 meters, the volume is increasing at a rate of 3m3/s, then how
fast is the radius changing?
How fast is the surface area changing? (Surface area A = 4πr2)
Solution: We know when r = 2, dVdt = 3. Therefore, when r = 2, we have
3 =dV
dt= 4π · 22 · dr
dt
105
So when r = 2, the radius is increasing at a rate of drdt = 3/(16π)cm/s.
Now let’s find dAdt . We have dA
dt = dAdr · dr
dt = 8πr · drdt . Therefore when r = 2, the area is
increasing at a rate of
dA
dt= 8π · 2 · 3
16π= 3cm2/s
2. (page 225, #21) Gasoline is pouring into a cylindrical tank of radius 3 feet. When
the depth of the gasoline is 4 feet, the depth is increasing at 0.2 ft/s. How fast is the
volume of gasoline changing at that instant? (Volume V = πr2h where h is the depth).
Solution: Since r = 3, we know the volume of gasoline is V = π · 32 · h = 9πh.
Therefore
dV
dt=
dV
dh· dhdt
= 9π · dhdt
On the other hand, we know when r = 3, dhdt = 0.2, so when r = 3, the volume of gasoline
is changing at a rate of dVdt = 9π · dh
dt = 9π · 0.2 = 1.8π ft3/s.
3. A potter forms a piece of clay into a cylinder. As he rolls it, the length, l, of
the cylinder increases and the radius, r, decreases. If the length of the cylinder is in-
creasing at 0.1 cm per second, find the rate at which the radius is changing when the
radius is 1 cm and the length is 5 cm.
Solution: The volume of the potter is constant, since when the radius is 1 cm, length
is 5 cm, we know the volume V = π · 12 · 5 = 5π. Therefore we have the formula
5π = V = πr2l, where r, l are both functions of t.
Now let’s take the derivative. We have 0 = 2πr drdt l + πr2 dl
dt . When r = 1, l = 5, we
know dldt = 0.1, plug in all these values, we have
0 = 2π · 12 · drdt |r=1,l=5
· 5 + π · 12 · 0.1
We have drdt |r=1,l=5
= −0.01. So the radius is decreasing at 0.01cm/s.
Alternative solution: We found V = 5π = πr2l, so l = 5r2. Now we take the
derivative of both sides, we have dldt =
−10r3
drdt , so
drdt =
−r3
10dldt . Now when r = 1, l = 5, we
know dldt = 0.1, plug in these values, we find the same answer.
106
4. (page 226, #29) (a) A hemisphere bowl of radius 10 cm contains water to a depth of
h cm. Find the radius of the surface of the water as a function of h.
Solution: (10− h)2 + r2 = 102, so r =√102 − (10− h)2 =
√20h− h2.
(b) The water level drops at a rate of 0.1 cm per hour. At what is rate is the ra-
dius of the water decreasing when the depth is 5 cm?
Solution: Since r =√20h− h2, we have
dr
dt=
dr
dh· dhdt
=10− h√20h− h2
· dhdt
When h = 5, we know dhdt = −0.1, so when h = 5,
dr
dt=
10− h√20 · h− h2
· dhdt
=10− 5√20 · 5− 52
· (−0.1) = −0.057cm/hour
5. (Fall 09, Final, #5) A cone-shaped container has a small opening at the bottom, so
the water is leaking. At 10 am, the depth of the water is 25 cm, and the water surface
radius is 2 cm, and the volume of the water is decreasing at 1.2 cm3/h.
(a) What the relation between depth h and radius r of the water surface?
(b) Find the formula for volume V of the water in terms of r.
(c) At 10 am, how fast is the radius changing?
How fast is the depth changing?
Solution: on the course website.
107
8.3 Solution to exam 3 review
#1. Let f(x) = xx, write down the definition of f ′(2).
Solution: f ′(2) = limh→0
f(2 + h)− f(2)
h= lim
h→0
(2 + h)2+h − 22
h
#2. Let f(x) = xx, estimate f ′(2).
Solution: choose a small h, for example, h = 0.001, then
f ′(2) ≈ f(2 + 0.001)− f(2)
0.001≈ 6.779
#3. Suppose E(t) is the total amount (in tons) of CO2 released into the air during the
first t days in 2011. The following is the table for E′(t) (not E(t)!).
t 3 5 7 9 11 13 15
E′(t) 1.5 2 2.2 3 3.1 4 4.2
(1) Interpret E′(3) = 2.
Solution: E(4) − E(3) ≈ E′(3) = 2, here E(4) − E(3) is the amount released on the
4th day, so the answer is:
The amount of CO2 released on Jan 4 is approximately 2 tons.
Or: E′(3) = 2 tells us on Jan 3 (t = 3), the amount is increasing at a rate of 2 tons per
day. Drop the word “rate”, add the word “approximately”, the answer is:
On Jan 3, the amount is increasing by approximately 2 tons per day.
(2) If E(3) = 5, estimate E(4). Is your estimate an overestimate or underestimate?
Why?
Solution: E(4) ≈ E(3) + E′(3)(4− 3) = 5 + 2 · 1 = 7 tons.
It’s an underestimate, since E′(t) is increasing implies E(t) is concave up.
(3) Use integral to represent the amount of CO2 released from Jan 3 to Jan 15.
Solution:∫ 153 E′(t)dt.
(4) Use n = 3 subdivisions to estimate the integral in part (3). Find an over and
108
under estimate.
Solution: since E′(t) is increasing, we know left sum is underestimate and right sum is
overestimate.
the length of one subdivision is (15− 3)/3 = 4, so
left sum= 4 · (E′(3) + E′(7) + E′(11)) = 4 · (1.5 + 2.2 + 3.1),
right sum= 4 · (E′(7) + E′(11) + E′(15)) = 4 · (2.2 + 3.1 + 4.2)
#4. A company is producing and selling chairs. The fixed cost is $100, and the ad-
ditional cost is $5 per chair. If we sell the chair at a price of $10 per chair, 100 chairs
can be sold, and the experience is that if we increase the price by $1, 5 less chairs can
be sold.
(1) If we sell the chair at a price of x dollars, how many chairs can be sold?
Solution: when we increase price from $5 to $ x, the price is increased by x − 5 dol-
lars, so 5(x − 5) chairs can be sold. Therefore the number of chairs can be sold is
100− 5(x− 5) = 125− 2x.
(2) use your answer is part (1), find the cost and revenue.
Solution: cost= 100 + 5x dollars, revenue is x(125− 2x) = −2x2 + 125x dollars.
(3) maximize the profit, what’s the price then?
Solution: the profit is π(x) = (−5x2 + 125x)− (100 + 5x) = −5x2 + 120x− 100. Take
the derivative, π′(x) = −10x + 120 = 0, we have x = 12. Now π
′′(x) = −10 < 0, we
know the critical point is a local max, so must be a global max. The global max is
π(12) = 620 dollars, and the price is 12 dollars.
#5. page 227, #45
Solution:
(1) h(t) = 300− 30t.
(2) tan(θ) = (h(t)−100)/150 = (200−30t)/150, so we have θ = arctan((200−30t)/150)
dθdt = 1
1+((200−30t)/150)2· (−1
5).
109
(3) θ is decreasing at a rate of |dθdt | =15 · 1
1+((200−30t)/150)2.
When ((200− 30t)/150)2 = 0, this rate is 15 .
When ((200 − 30t)/150)2 > 0, the denominator 1 + ((200− 30t)/150)2 > 1, so the rate
will be smaller than 15 . Therefore θ is decreasing fastest when ((200 − 30t)/150)2 = 0,
so t = 20/3 and h(t) = 100.
There is another way to do this problem.
since tan(θ) = (h(t)− 100)/150, we can take the derivative ans use h′(t) = −30.
1
cos2(θ)
dθ
dt=
h′(t)
150= −1/5
so we have dθdt = −1
5 cos2(θ), therefore θ is decreasing at a rate of 1
5 cos2(θ), the largest
rate happens when cos2(θ) = 1, so cos(θ) = 1, θ = 0. Therefore θ is decreasing fastest
when θ = 0.
#6. When in space, the force, in Newtons, exerted on a certain satelite is inversely
proportional to the square of the distance between the satelite and the Earth.
(1) Suppose that when the satelite is 90 kiolmeters away from earth, the force exerted on
it is 50 Newtons. Find a formula for the F (d) force (in Newtons) exerted on the satelite
when it is d miles from Earth.
Solution: Suppose F = kd2, then k = 902 · 50 = 40500, so F = 40500
d2.
(2) In order to gather more data, the satelite starts moving further from earth at a
rate of 4 km/hr. At what rate is the force exerted on the satelite changing (in new-
tons/hr).
Solution: dFdt = −2·40500
d3· dddt = −81000
903· 4 = −4
9 Newtons/hr.
(3) The time, T (measured in minutes), it takes for the satelite to make a full orbit
around Earth can be found by T 2 = 341.8d3. How fast is the time required for one orbit
changing?
110
Solution: T 2 = 341.8d3, so T =√341.8d1.5, so dT
dt =√341.8 · 1.5 · d0.5 · dd
dt =√341.8 · 1.5 ·
√90 · 4 = 1052.35 min/hr
#7. For a super bowl party, Timmy brought a giant bowl filled with 200 ounces of
spinach dip for a communal snack. Let f(t) be the rate at which the dip is being eaten,
in ounces per minute, t minutes after the start of the game. Some values of f(t) are in
the table below. Assume f(t) is decreasing.
t 0 10 20 30 40 50 60
f(t) 5 4 3.5 3 2 1 0
(1) Interpret (but do not compute)∫ 4010 f(t)dt. Include units.
Solution:∫ 4010 f(t)dt is the total number of oz’s of dip eaten between 10 minutes after
and 40 minutes after the start of the game.
(2) Using the table, find a left hand riemann sum ( in other words, left sum ) ap-
proximation of∫ 4010 f(t)dt using n = 3 subintervals ( in other words, subdivisions).
Solution: left sum= (f(10) + f(20) + f(30)) · 10 = (4 + 3.5 + 3) · 10 = 105.
(3) Not all of the dip is eaten during the game. How big of a container does Timmy need
to make sure he can store the leftovers? [Hint : find an underestimation of how much
dip is eaten during the game]
Solution: The right sum will be an underestimate, which is (f(10) + f(20) + f(30) +
f(40)+ f(50)+ f(60)) · 10 = 135. So at least 135 oz’s are eaten, therefore the size of the
container should be at least 200− 135 = 65 oz.
#8. In the distant future, the University of Michigan will install electrically heated
sidewalks. These heaters cause accumulated snow to melt at a rate of 2 inches per day.
Suppose that p(t) (which is graphed below) gives the rate, also measured in inches per
day, at which snow accumulates during one snowy winter week. Assuming there is no
snow on the ground at time t = 0, answer the following questions.
111
H0, 3L
H3, 9L
1 2 3 4 5 6 7t, days
-2
2
4
6
8
10
pHtL, inches per day
(1) Estimate the time t on [0, 7] when the depth of the snow is the greatest.
Solution: From t = 0 to t = 4, the rate of snow fall is larger than the rate of snow
melting, so the depth keeps increasing from t = 0 to t = 4. After t = 4, they have the
same rate, so the depth is unchanged from t = 4 to t = 7. Therefore the snow is deepest
on [4, 7].
(2) Estimate the depth of the snow when t = 5.
Solution: depth=∫ 50 p(t)dt− 2 · 5 = 15.5 inches.
(3) how much snow falls per day, on average, in the week for which data is given?
Solution: ∫ 70 p(t)dt
7− 0=
29.5
7= 4.2 inch per day
#9. Jack has a leaky faucet for his kitchen sink. The leak drips water into the faucet at
a constant rate of 10 mL per minute. The water is also draining out of the sink, but the
rate the water drains out is not constant because of the pile of dirty dishes present. Let
r(t) be the rate, in mL per minute, that the water is draining out of the sink t minutes
after 7:00pm. The graph of r(t) can be found below. Assume the sink has 5mL of water
in it at 7:00.
112
5 10 15 20t, minutes
5
10
15
rHtL, mL per minutes
(1) How much water drains out of the sink between 7:00 and 7:15?
Solution: it’s the area below r(t), bounded between t = 0 and t = 15, so answer is 100
mL.
(2) When is there the most water in the sink? How much water is in the sink at
this time?
Solution: when t = 17.5.
The amount of water = initial amount + in − out = 5 + 175− 118.75 = 61.25 mL.
(3) How much water is in the sink at 7:20?
Solution: From t = 17.5 to t = 20, the water draining out is 31.25 mL, the water
coming in is 25 mL, so the total amount of water at 7:20 is 61.25− 31.25+ 25 = 55 mL.
(4) Suppose that at 7:20, Jack removes the dirty dishes from the sink and water is
allowed to drain freely at a rate of 15mL per minute. How long will it take for the sink
to have no water in it? [Hint : It may help to use part c]
Solution: the water is out at a constant rate of 15− 10 = 5 mL per minute, so it takes
55/5 = 11 minutes.
#10. f(x) is an even function. The graph of f(x) with x ≥ 0 is given below, which
contains a quarter circle with radius of 2 and two line segments.
113
1 2 3 4 5 6
-2
-1
1
2
Also suppose g(x) is an odd function, with g(1) = 1, g(2) = 2, g(4) = 3,∫ 40 g(x)dx = 5
and∫ 20 g(x)dx = 3. Find the following integrals. (1)
∫ 31 f(x+ 1)dx
Solution:∫ 31 f(x+ 1)dx =
∫ 42 f(x)dx = 0.
(2)∫ 60 (−2f(x) + 1)dx
Solution:∫ 60 (−2f(x) + 1)dx = −2
∫ 60 f(x)dx+
∫ 60 1dx = −2 · 1
4 · π · 22 + 6 = 6− 2π
(3)∫ 41 f ′(g(x))g′(x)dx
Solution:∫ 41 f ′(g(x))g′(x)dx =
∫ 41 (f(g(x)))′dx = f(g(4))− f(g(1)) = f(3)− f(1) =
0− 0 = 0
(4)∫ 42 g(x)dx
Solution:∫ 42 g(x)dx =
∫ 40 g(x)dx−
∫ 20 g(x)dx = 5− 3 = 2
(5)∫ 2−2 (g(x) + f(x))dx
Solution:∫ 2−2 (g(x) + f(x))dx =
∫ 2−2 g(x)dx +
∫ 2−2 f(x))dx = 0 + 2 ·
∫ 20 f(x)dx =
0 + 2 · 0 = 0
#11. Compute the total shaded area between y = x2 − x− 1 and y = −x2 + x+ 3.
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y= x2 - x - 1
y=-x2 + x + 3
Solution: x2 − x− 1 = −x2 + x+3, so 2x2 − 2x− 4 = 2(x− 2)(x+1) = 0, so x = −1
or x = 2.
The shaded area is∫ 2−1 ((−x2 + x + 3) − (x2 − x − 1))dx =
∫ 2−1 (−2x2 + 2x + 4)dx =
(−23x
3 + x2 + 4x+ C)|2−1 = 9
Chapter 9
Sample quizzes with solutions
115
116
9.1 Quiz 1
1. A biologist is growing mold in a Petri dish for an experiment. At noon she starts the
experiment by innoculating (putting mold into) the dish. At 4 pm, she notices that the
mold has grown to cover an area of 5 cm2. At 7pm, she returns and finds the mold now
covers an area of 7 cm2.
(a) Assuming the area of the mold grows linearly over time, find a formula l(t) which
gives the area the mold covers t hours after noon. (3 points).
Solution: let y = l(t) = mt+ b, then l(4) = 5 and l(7) = 7, so slope m = 7−57−4 = 2
3 . Use
the point slope form, we have y − 7 = 23(x− 7), simplify, y = l(t) = 2
3 t+73 .
(b) Assuming the area of the mold grows exponentially over time, find a formula e(t)
which gives the area the mold covers t hours after noon. (3 points).
Solution: Let y = e(t) = P0at, then
e(4) = 5 ⇒ P0a4 = 5
e(7) = 7 ⇒ P0a7 = 7
Divide the second equation by the first and cancel out P0, we have 1.4 = a3, so
a = 1.413 = 1.1187, plug this back into either equation, we have P0 = 3.1925. So
e(t) = 3.1925 · (1.1187)t.
(c) Using your formula e(t) from part (b), how long does it take for the area of the mold
to triple? (3 points).
Solution: 3P0 = P0(1.187)t, cancel out P0, 3 = (1.1187)t, so t = ln(3)/ ln(1.1187) ≈
9.795 years.
(d) Suppose the biologist returns at 9pm to find the mold has grown to 8.8 cm2. Which
of your formulas more accuately predicts this growth? Justify your reasoning. (3 points).
Solution: l(9) = 23 · 9 + 7
3 = 8.33, e(9) = 3.1925 · (1.1187)9 = 8.76, so exponential is
more accurate because it predicts 8.76 cm2 of mold, which is closer than the 8.33 cm2
of mold predicted by l(t).
117
2. Given below is the graph of y = f(x).
-2 -1 1 2
2
1
-1
Find the function for the above graphs in terms of f(x). (No partial credit, 8 points)
-1 1 2 3
2
1
-1
-1 1
1
0.5
-1
-1 1
1
0.5
-1
-2 -1 1 2
1
-1
-2
Write your answer here
(a) shift down by 1 and right by 1, so f(x− 1)− 1.
(b) vertical and horizontal scaling by 1/2, so 12f(2x)
(c) reflect over x-axis and shift up by 1, so −f(x) + 1.
(d) reflect over y-axis and shift down by 0.5, so f(−x)− 0.5.
118
9.2 Quiz 2
1. [6 points] The average weight of a squirrel in Ann Arbor oscillates sinusoidally be-
tween a low of 5 pounds on January 1 and a high of 9 pounds on July 1, and a low of 5
pounds on January 1 next year again. Suppose that the function P (t) gives the average
weight in pounds of an Ann Arbor squirrel t months after January 1.
a). What is the amplitude of P (t)?
Solution: 9−52 = 2
b). What is the period of P (t)?
Solution: 12 (months)
c). Find a formula for P (t).
Solution: Suppose y = A cos(Bt) + k, then A < 0. Midline is y = 5+92 = 7, so k = 7;
amplitude is 2, so |A| = −A = 2, so A = −2; period is 2πB = 12, so B = π
6 . Therefore
the formula is
P (t) = −2 cos(π
6t) + 7
2. [2 points]f(x) = sin(x)x, write down the limit definition of f ′(3). (You don’t need to
calculate it, just write the definition.)
Solution:
f ′(3) = limh→0
f(3 + h)− f(3)
h= lim
h→0
sin(3 + h)3+h − sin(3)3
h
3. [4 points]The cost, C (in dollars) to produce g gallons of ice cream can be expressed
as C = f(g). Assume f is invertible. Interpret
(1) f ′(100) = 2.5
Solution: When 100 gallons of ice cream are produced, it costs approximately 2.5
119
dollars to produce 1 more gallon of ice cream.
Or, It costs approximately 2.5 more dollars to produce 101 gallons of ice cream than
producing 100 gallons.
(2) (f−1)′(100) = 2.5
Solution: When the cost is 100 dollars, approximately 2.5 gallons more ice cream will
be produced if 1 more dollar is spent.
4. [6 points]Consider a particle, whose position, s, is given by the table
t (seconds) 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
S (feet) 0.5 1.4 3.8 6.5 9.6 9.2 8.7 6.1
(1) Estimate the velocity of the particle at t = 0.2. (Don’t forget the unit)
Solution: v(0.2) = S′(0.2) ≈ S(0.4)−S(0.2)0.4−0.2 = 1.4−0.5
0.2 = 4.5 ft/s
(2) Use (1) and the table to estimate the position of the particle at t = 0.17. (Don’t
forget the unit)
Solution: S(0.17) ≈ S(0.2) + S′(0.2)(0.17− 0.2) = 0.5 + 4.5 · (−0.03) = 0.365 ft.
(3) For which t, does the velocity appear to be positive?
Solution: For 0.2 ≤ t ≤ 1.2, S(t) is increasing, so v(t) = S′(t) appear to be positive
for 0.2 ≤ t ≤ 1.2.
5.[2 points] See the following figure, which graph represents the position of an object
that is speeding up and then slowing down?
Answer: (c) Since speed is v(t) = S′(t) = slope of tangent line at t, speed is increasing
then decreasing, so slope of tangent lines is increasing then decreasing.
120
t
distance
t
distance
t
distance
t
distance
121
9.3 Quiz 3
1. [14 points] Given below is a graph of a function f(x) and a table for a function g(x).
Given answers for the following or write “Does not exist”. Each problem is worth 2
-4 -3 -2 -1 1 2 3 4
-3
-2
-1
1
2
3
x -3 -2 -1 0 1 2 3
g(x) -2 -3 -2 -1 1 3 2
g′(x) -2 0 1 3 2 0.5 -1
points.
a) h(x) = g(x)f(x) . Find h′(−3).
Solution: h′(x) = g′(x)f(x)−g(x)f ′(x)(f(x))2
, so
h′(−3) =g′(−3)f(−3)− g(−3)f ′(−3)
(f(−3))2=
−2 · 3− (−2) · 032
= −2/3
b) k(x) = 3f(x)− g(x). Find k′(2)
Solution: k′(x) = 3f ′(x) − g′(x), so k′(2) = 3f ′(2) − g′(2). Since f ′(2) doesn’t
exist, k′(2) doesn’t exist.
c) q(x) = f(x)g(x). Find q′(−2)
Solution: q′(x) = f ′(x)g(x)+f(x)g′(x), so q′(−2) = f ′(−2)g(−2)+f(−2)g′(−2) =
0 · (−3) + 3 · 0 = 0
d) a(x) = g(−f(x)). Find a′(1)
Solution: a′(x) = g′(−f(x))(−f ′(x)), so a′(1) = g′(−f(1))(−f ′(1)) = g′(1) · 2 =
2 · 2 = 4
e) l(x) = e2f(x). Find l′(3)
Solution: l′(x) = e2f(x) · 2f ′(x), so l′(3) = e2f(3) · 2f ′(3) = e2·0 · 2 · 3 = 6
f) p(x) = sin( πf(x)). Find p′(1)
Solution: p′(x) = cos( πf(x)) · (−π · (f(x))−2 · f ′(x)), so p′(1) = cos( π
f(1)) · (−π ·
(f(1))−2 · f ′(1)) = cos(−π) · (−π) · (−1)−2 · (−2) = −2π
122
g) t(x) = ln((eg(x))2). Find t′(−2)
Solution: t(x) = ln((eg(x))2) = 2 ln(eg(x)) = 2g(x), so t′(x) = 2g′(x). t′(−2) =
2g′(−2) = 2 · 0 = 0
2. [6 points]“Winning the war on poverty” has been described cynically as slowing thee
rate at which people are slipping below the pverty line. Let N be the number of people
below the poverty line at time t, answer the following questions.
(1) If N is increasing at a faster and faster rate.
Is N ′(t) increasing, decreasing or neither? Answer: increasing
Which is a possible graph for N? If none is possible, write None. Answer: a
(1) If N is decreasing at a slower and slower rate.
Is N ′(t) increasing, decreasing or neither? Answer: increasing
Which is a possible graph for N? If none is possible, write None. Answer: d
123
9.4 Quiz 4
1. [6 points] Suppose f(x) = ex2.
(1) [2 points] Find the tangent line approximation of f(x) at x = 1.
Solution: f ′(x) = 2xex2, so f ′(1) = 2e, the tangent line approximation L(x) =
f(1) + f ′(1)(x− 1) = e+ 2e(x− 1) = 2ex− e.
(2) [2 points] Use your answer in part (1) to estimate e1.012.
Solution: e1.012= f(1.01) ≈ L(1.01) = 2e · 1.01− e = 2.02e− e.
(3) [2 points] Is your estimation in part (2) an overestimate or underestimate? Use
concavity to explain your answer.
Solution: f ′′(x) = (4x2 + 2)ex2> 0, so f(x) is always concave up, so the estimation is
an underestimate.
2. [6 points]Suppose that x and y satisfy the relation given by the curve
x2 + xy + y2 = 3
(1) [4 points] Find dydx (in other words, y′).
Solution: take derivative directly to both sides, we have 2x + y + xy′ + 2yy′ = 0, so
(x+ 2y)y′ = −(2x+ y), so y′ = −2x+yx+2y .
(2) [2 points]Find all points on the curve at which the tangent line is horizontal.
Solution: The tangent line is horizontal means the slope is 0. So let y′ = −2x+yx+2y = 0,
we have 2x+ y = 0, so y = −2x. Then plug y = −2x back in the original equation, we
have x2 + x · (−2x) + (−2x)2 = 3, so 3x2 = 3, then x = 1 (so y = −2 ) or x = −1 (so
y = 2). We find 2 points, (1,−2) and (−1, 2).
3. [7 points] Let f(x) = x2(x − 1)2. Find all the critical points of f(x) and use ei-
ther the first derivative test or the second derivative test to classify these critical points
as a local max, local min, or neither. Show all your steps!
124
Solution: f(x) = x4 − 2x3 + x2, so f ′(x) = 4x3 − 6x2 + 2x = 2x(2x2 − 3x + 1) =
2x(2x − 1)(x − 1). Let f ′(x) = 0, we have 3 solutions, x = 0, x = 0.5 or x = 1. These
are where the critical points locate.
Let’s use the second derivative test. f ′′(x) = 12x2 − 12x + 2, so f ′′(0) = 2 > 0,
f ′′(0.5) = −1 < 0, f ′′(1) = 2 > 0. By the test, we know f(x) has a local min at x = 0
and x = 1, and a local max at x = 0.5.
We can also use the first derivative test. Now the three critical points divide the x-axis
into 4 intervals, (−∞, 0), (0, 0.5), (0.5, 1) and (1,∞). We pick any value on each interval
to find the sign of f ′ on that interval.
(−∞, 0), e.g, pick −1, f ′(−1) = −12 < 0, so f ′ < 0 on (−∞, 0);
(0, 0.5), e.g, pick 0.25, f ′(0.25) = 0.1875 > 0, so f ′ > 0 on (−∞, 0);
(0.5, 1), e.g, pick 0.75, f ′(0.75) = −0.1875 < 0, so f ′ < 0 on (−∞, 0);
(1,∞), e.g, pick 2, f ′(2) = 12 > 0, so f ′ > 0 on (−∞, 0);
By the first derivative test, we can get the same result.
125
9.5 Quiz 5
1. The following is the graph of f ′(x) ( not f(x)).
-4 -3 -2 -1 1 2 3 4 5
-2
-1
1
2
(1) Find the exact value of∫ 2−4 f ′(x)dx.
Solution: 0.
(2) If∫ 4−2 f ′(x)dx = −0.2, find the area of the shaded region from x = 3 to x = 5.
Solution: Suppose the shaded area is A, then∫ 4−2 f ′(x)dx = 1 − A/2 = −0.2, so
A = 2.4.
(3) If f(−4) = 1, use part (1) and the fundamental theorem to find f(2).
Solution: By the fundamental theorem, 0 =∫ 2−4 f ′(x)dx = f(2) − f(−4), so
f(2) = f(−4) = 1.
2. Oil leaks out of a tanker at a rate of r = f(t) liters per minute, where t is in
minutes.
(1) interpret∫ 600 f(t)dt. (Make sure to include units!)
Solution: During the first hour (or first 60 minutes),∫ 600 f(t)dt liters oil leaks out of
the tanker.
(2) moreover, suppose f(t) = e−t, use n = 5 subdivisions, write the expression of the
left sum of∫ 600 f(t)dt. (Don’t evaulate)
Solution: left sum= e−0 · 12 + e−12 · 12 + e−24 · 12 + e−36 · 12 + e−48 · 12.
(3) Is the left sum an overestimate or underestimate?
Solution: The left sum is an overestimate, since f(t) = e−t is decreasing.
126
3. In the free time, Tom enjoys inflating perfectly spherical balloons. One day, Tom is
inflating a balloon and he notices that when its radius is equal to 3 cm, its volume is
increasing at 4 cm3 per second. We know that a sphere of radius r has volume V = 43πr
3,
and surface area s = 4πr2. Answer the following 2 questions.
(1) At this moment, how fast is the radius of the balloon increasing?
Solution: V = 43πr
3, so dVdt = 4πr2 drdt . Plug in r = 3, dV
dt = 4, we have 4 = 4π · 32 · drdt ,
so the radius is increasing at drdt =
19π cm/s.
(2) At his moment, how fast is the surface area increasing? (hint: use part (1))
Solution: s = 4πr2, so dsdt = 8πr · drdt , plug in r = 3, dr
dt =19π , we know the surface area
is increasing at a rate of dsdt =
83 cm2/s
Chapter 10
Miscellaneous
10.1 Using calculator to find integral
Method: Caluculate∫ ba f(x)dx.
Here the function you want to integrate is f(x), the lower limit is a, upper limit is b.
(1) Graph y = f(x), choose the x-window to be a ≤ x ≤ b (in other words, xmin = a,
xmax = b).
(2) Click 2nd and trace , choose option 7 (which is∫f(x)dx).
step (3) and (4) is to put in the lower and upper limit.
(3) click -4, then enter .
(4) click 4, then enter .
(5) click enter , the calculator will give us the answer.
Example: Calculate∫ 4−4 (16− x2)dx.
Method: here the function is 16− x2, lower limit is −4 and upper limit is 4.
127
128
Procedure: (1) Graph y = 16− x2, choose window −4 ≤ x ≤ 4.
(2) Click 2nd and trace , choose option 7,∫f(x)dx.
(3) put in the lower limit -4, click -4, then enter .
(4) put in the upper limit 4, click 4, then enter .
(5) click enter , we find the answer.
129
10.2 Derivative rules for the gateway test
The following are the rules for the gateway test. You need to keep this rules in mind,
since you are not allowed to bring notecard or calculator for the gateway test. Inverse
function rule and implicit function rule will not show up in the test, but will show up in
the subsequent exams.
Basic arithmetic rules:
• (c · f(x))′ = c · f ′(x) , here c is a constant
• (f(x) + g(x))′ = f ′(x) + g′(x)
• Product rule: (f(x)g(x))′ = f ′(x)g(x) + f(x)g′(x)
• Quotient rule: (f(x)g(x) )′ = f ′(x)g(x)−f(x)g′(x)
(g(x))2
• Chain rule: (f(g(x)))′ = f ′(g(x))g′(x)
Derivative of special functions:
• constant function: (c)′ = 0
• linear function: (mx+ b)′ = m
• power function: (xn)′ = nxn−1
• exponential function: (ax)′ = ax ln(a), (ex)′ = ex
• sinusoidal function: (sin(x))′ = cos(x), (cos(x))′ = − sin(x), (tan(x))′ = 1cos2(x)
• log function: (ln(x))′ = 1x
Test preparation: do as many practice gateway tests as you can, ask me if anything
is confusing. The problems are not hard, but it doesn’t mean you can get the correct
answer; even if you know the correct answer, you may type in the computer a wrong
answer (be careful of the parenthesis!)
website for practice gateway test:
http://instruct.math.lsa.umich.edu/classes/115/gw/