Mathematics for agricultural engineers

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Mathematics

For Agricultural Engineers

Practice notes

By

Norbert Bogya, Tam´as Szab´o

July 10, 2019

INVESTING IN YOUR FUTURE European Social

Fund

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Reviewer: F¨ul¨op Vanda

This teaching material has been made at the University of Szeged, and supported by the European Union. Project identity number: EFOP-3.4.3-16-2016-00014.

INVESTING IN YOUR FUTURE European Social

Fund

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Description of the subject 1

1 Set Theory 5

Practice problems . . . 5 Supplementary exercises . . . 8 Quizzes . . . 11

2 Functions 14

Practice problems . . . 14 Supplementery exercises . . . 20 Quizzes . . . 23

3 Limit 27

Practice problems . . . 27 Supplementery exercises . . . 32 Quizzes . . . 35

4 Limits at Infinity, Tangent Line, Derivative 38

Practice problems . . . 38 Supplementary Exercises . . . 43 Quizzes . . . 45

5 Differentiation 48

6 Derivative Tests 59

Midterm #1 72

Version A . . . 73 Version B . . . 77

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7 Concavity; Curve Sketching 81

Practice problems . . . 81

Supplementary exercises . . . 86

Quizzes . . . 89

8 Applications of the Derivative 93 Practice problems . . . 93

Quizzes . . . 102

9 Sequences; Series 106 Practice problems . . . 106

Quizzes . . . 116

10 Integral 119 Practice problems . . . 119

Supplementary Exercises . . . 125

Quizzes . . . 127

11 Matrix Algebra 130 Practice problems . . . 130

Quizzes . . . 144

12 Linear Programming 147 Practice problems . . . 147

Supplementary Exercises . . . 155

Quizzes . . . 157

Midterm #2 160 Version A . . . 161

Version B . . . 166

A Preliminaries 170 Exponents and Roots . . . 170

Intervals . . . 171

Algebraic Fractions . . . 171

Linear and Quadratic Equations . . . 171

Linear Functions . . . 172

Quadratic Functions . . . 173

Absolute Value Function . . . 173

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Title of the subject: Mathematics for Agricultural Engineers Credits: 3 Type of the subject: compulsory

The ratio of the theoretical and practical character of the subject: 10-90 (credit%) The type of the course: practice

The total number of the contact hours: 2 per week Language: English

Type of the evaluation: 2 written tests

Other methods for evaluation of the student’s competence: 9 quizzes The term of the course: I. semester

Prerequisite of the subject: None

The aim of the subject: The courses are directed to students starting their university stud- ies. It provides an overview of techniques to calculate velocity and acceleration, to estimate the rate of spread of a disease, to set levels of production so as to maximize efficiency, to find the best dimensions of a cylindrical can, to find the age of a prehistoric artifact, and for many other applications. Participants will study Algebra which makes them able to find the values that will maximize or minimize a function subject to restrictions.

Course description:

1. Basic Set Theory 2. Real Valued functions 3. Limit and Continuity 4. Differentiation

5. Sequences and Series 6. Integration

7. Linear Programming

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Selected bibliography:

• Bogya N. Szab´o T., Mathematics for Agricultural Engineers, SZTE Repository of Educational Resources (this practice note)

• Herbert B. Enderton, Elements of Set Theory, ISBN-13: 978-0122384400, Academic Press, San Diego, 1977

• Saul I. Gass, Linear Programming, ISBN 0-486-43284-X04, Dover Publications, New York, 1958

• George B. Thomas, Maurice D. Weir, Joel Hass, Frank R. Giordano: Thomas‘ Cal- culus, ISBN-13: 978-0-321-87896-0, Pearson, 2008

General competence promoted by the subject a) Knowledge

• Familiar with the tools and methods of set theory.

• Know the terminology and computational methods of calculus.

• Understand the differences between formal and informal discussions.

• Understand a new or difficult concept, appreciate its full mathematical precision and outcome.

b) Skills

• Able to interpret and the present the results.

• Able to apply the tools and techniques of mathematics.

• Able to recognize the coherency among the different areas of mathematics.

• Able to participate in mathematical projects under competent supervision.

c) Attitude

• Open to cooperate with her/his classmates.

• Ready to understand the concept of infinity.

• Interested in new results, techniques and methods.

• Aspires to use the abstract termology.

d) Autonomy and responsibility

• Able to solve complex problems independently.

• Provides and requires clear explanation.

• Helps her/his classmates in the completion of their projects.

• Able to create mathematical models consciously.

• Able to do autonomous application of theoritical results.

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Special competence promoted by the subject

Knowledge Skills Attitude Autonomy and re-

sponsibilty Know the way to

determine the union and intersection of two or more sets.

Aware of the mean- ing of ”at least”, ”at most” among the real numbers.

Able to carry out the formalization of set operators.

Willing to understand that a paradox is a statement that, despite apparently valid reason- ing from true premises, leads to an apparently-self- contradictory or logically unaccept- able conclusion.

Know the definition of a function, familiar with the elementary functions

Able to determine the domain, the ze- roes of a function.

Independently able to draw and explain the linear transform of a function.

Acquire techniques to calculate different types of limits.

By factoring out the common factors able to solve ”0/0”

type limits.

Undestand the difference among the types of limits, and the difference between the limit and continuity.

Know the definition of differentiability, know the derivative of the elementary functions.

Able to calculate the derivative of a composite function.

Independently able to show that a differen- tiable function is con- tinuous.

Know the definition of the sequences and the series.

Able to determine the limit of a serquence, able to calculate the sum of a series.

Ready to explain the difference and the connection between the concept of the sequence and the series.

Recall the definition of the definite and the indefinite integral as well.

Able to calculate the definite and the indefinite integral.

Aspire to solve practical problems.

Discover the im- proper integral independently.

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Knowledge Skills Attitude Autonomy and re- sponsibilty

Familiar with the properties of matri- ces. Understand that linear programming is a technique for the optimization of a linear objective function, subject to linear equality and linear inequality constraints.

Able to use the array methods to solve simultaneous equations.

Open to study new techniques and methods.

Can independently give a pictorial repre- sentation of a simple linear program with two variables and six inequalities.

Instructor of the course: Dr. Szabó Tamás Zoltán, associate professor

Teachers: Dr. Fülöp Vanda, senior lecturer; Dr. Szabó Tamás, associate professor, Bogya Norbert, assistant research fellow

This practice note was constructed based on the experience and material of the previous years’ Mathematics course for Agricultural Engineers.

There are 12 numbered sections according to the weekly schedule of the practice. In each section there are 3 subsections – Practice problems, Supplementary exercises and Quizzes – in which we explain, deepen and test the material of the actual practice. We start with exercises with well-explained solutions, then we provide several exercises with only final results (without explanations), just for practice. At the end of the section we show some possible quizzes with solutions and scoring, so the students may test their knowledge and score themselves. Moreover, two possible midterms are included among the homework sections to prepare them for the exam.

We are sure that if students attend the classes and use this note, then they will be able to characterise mathematical problems, memorise the types of exercises, recall the possible methods to solve and use them. With their improved mathematical skills, they will be able to explain their own solution clearly and think in a more abstract way.

The Authors

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Set Theory

Practice problems

Elements, subset, power set

Exercise 1. Give the elements of the following sets.

(a) A={x: xis positive integer and x² <20} (b) B ={1−n² |n is even and 3≤n≤6}

(c) The setC of even prime numbers greater than 10.

Solution.

(a) Positive integers are 1,2,3,4, . . . and such a number also has to satisfy the following property: its square should be smaller then 20. It is correct for only the numbers 1,2,3,4, so A={1,2,3,4}.

(b) There are two restrictions ton: it has to be even and between 3 and 6. Only the numbers 4 and 6 (equality is allowed) satisfy these conditions, hence B = {1− 4²,1−6²} = {−15,−35}.

(c) Set C is given by a description. We look for the even primes greater than 10. There is only one even prime number, that is 2. But 2 is not greater then 10. This set has no element, so it isthe empty set,C =∅.

Exercise 2. Ifa, b, c, dare distinct objects, determine which of the five sets{a, b, c},{b, c, a, b}, {c, a, c, b}, {b, c, b, a}, {a, b, c, d} are equal.

Solution. A set is irrelevant to multiplicity and ordering. So we can ignore the multiple appearances of an object and their order:

{a, b, c}={b, c, a, b}={c, a, c, b}={b, c, b, a}.

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Exercise 3. LetA={a, b, c},B ={a, b},C ={a, b, d}, D={a}and E ={b, c}, where a, b, c are distinct objects. State whether each of the following statements is true or false:

(a) B ⊆A, (b) E =C, (c) D⊆B, (d) D⊆A, (e) A=B.

Solution. Notation A⊆B means that each element of set A is also a member of set B.

(a) True, becauseB has two elements: a, b and they are elements of A, too.

(b) True, becauseE contains c, butC does not, so their elements are not the same.

(c) False, since the only element of D isa and it is contained in B as well.

(d) True, the only element of D isa and it is an element ofA, too.

(e) False, sincecis contained in only one of the sets.

Exercise 4. Which of the following sets is the empty set?

(a) H ={x: x is an odd integer and x² = 4} (b) K ={x: xis an integer and x+ 8 = 8}

(c) L={x: x is a positive integer and x <1} Solution.

(a) There are only two numbers whose square is 4, they are the number 2 and−2. But they are even, not odd, soH does not contain any element, H =∅.

(b) The equation of the restriction has exactly one solution which is 0. Zero is an integer, hence K ={0} =∅.

(c) The smallest positive integer is 1, but it is not smaller then 1, as the restriction requires.

Hence, L=∅.

Exercise 5. Give the elements of the following power sets.

(a) P({1,2}) (b) P({a,1,5}) (c) P(∅) (d) P(P(∅))

Solution. Power set of a set A is the set of all subsets of A. So we have to determine the subsets of the given sets. Empty set is always contained in any power set.

(a) Denote{1,2}withA. The empty set is a subset of any set, and that is the only subset with no elements. The subsets ofAwith only one element are{1}and{2}. Furthermore,Ahas only one subset with two elements, that is the whole A. Hence, P(A) ={∅,{1},{2}, A}. (b) Let B ={a,1,5}. Again, the empty set is the only subset of B with zero elements. The subsets of B with one element are {a}, {1} and {5}. But now, there are several subsets with two elements: {a,1}, {a,5} and {1,5}. Of course, the whole set B is a subset of itself. We have P(B) = {∅,{a},{1},{5},{a,1},{a,5},{1,5}, B}.

(c) The empty set has no element, so it has only one subset, itself: P(∅) ={∅}.

(d) From the previous part we know that P(∅) ={∅}. This set has one element, so it has a subset with one element, that is {∅}. Hence,P(P(∅)) ={∅,{∅}}.

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Operations between sets

Exercise 6. Let A = {1,2,3,4,5} and B = {2,4,6,8,9} be two subsets of the universal set U ={1,2, . . . ,10}. Give the elements of the setsA∪B,A∩B,A\B,B \A,A and AB.

Solution. We illustrate the sets with a Venn diagram.

1 3 5 6 8 9

7 10

A B

2 4

• Intersection means the common part of the sets, so A∩B ={2,4}.

• In the union we collect all elements from all of the sets: A∪B ={1,2,3,4,5,6,8,9}.

• Set A\B contains objects that are in A but not in B, hence A\B ={1,3,5}. Similarly, B\A={6,8,9}.

• The complement of a set contains elements that are not in the set: A={6,8,9,7,10}.

• Symmetric difference of A and B is the set of elements, which occur exactly once in the sets A and B, so AB ={1,3,5,6,8,9}.

Exercise 7. Illustrate the following sets on the number line.

(a) [0,2]∩(1,3) (b) [0,3)∪(−1,2] (c) (0,5]\[1,2]

Solution.

(a) We sketch the intervals with colour blue, their common part is coloured by red. Number 1 is not contained in the intersection, because it is not in the interval (1,3). But number 2 is an element of both of the sets, so it is in the intersection as well.

0 2

1 3

1 2

(b) Now we have to calculate a union. We collect the points of the axis that are covered by at least one of the blue intervals.

0 3

−1 2

−1 3

(c) We delete the common parts from the first set. We have to care about the solid and empty circles: [1,2] is a closed interval, so we delete its endpoints (as well) from (0,5].

0 5

1 2

0 1 2 5

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Supplementary exercises

Elements, subset, power set

Exercise 1. Let A = {∅,{∅},{∅,{∅}}}. Determine whether each of the following statements is true or false.

(a) ∅ ∈A (b) ∅ ⊆A

(c) {∅} ∈A (d) {∅} ⊆A

(e) {{∅}} ∈A (f ) {{∅}} ⊆A

(g) {∅,{∅}} ∈A (h) {∅,{∅}} ⊆A Solution.

(a) True (b) True

(c) True (d) True

(e) False (f ) True

(g) True (h) True Exercise 2. Determine whether each of the following statements is true or false.

(a) a∈ {1, a, b,2} (b) a⊆ {1, a, b,2}

(c) {a} ∈ {1, a, b,2} (d) {a} ⊆ {1, a, b,2}

(e) {a,2} ∈ {1, a, b,2} (f ) {a,2} ⊆ {1, a, b,2} (g) {a,2,3} ∈ {1, a, b,2} (h) {a,2,3} ⊆ {1, a, b,2} Solution.

(a) True (b) False (c) False (d) True

(e) False (f ) True (g) False (h) False Exercise 3. Give the elements of the set P(P(P(∅))).

Solution. P(P(P(∅))) ={∅,{∅},{{∅}},{∅,{∅}}}

Exercise 4. Give the power sets of the following sets.

A={5} B ={2, a} C ={1,0,∅} D={1,2, c, d} Solution.

P(A) =

∅,{5} P(B) =

∅,{2},{a},{2, a} P(C) =

∅,{1},{0},{∅},{1,0},{1,∅},{0,∅},{1,0,∅}

P(D) =

∅,{1},{2},{c},{d},{1,2},{1, c},{1, d},{2, c},

{2, d},{c, d},{1,2, c},{1,2, d},{1, c, d},{2, c, d},{1,2, c, d}

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Operations between sets

Exercise 5. LetA ={a, b, c, d}, B ={b, d, f, h},C ={c, d, e, f}. Find (a) A∩B, A∩C, B∩C,

(b) A∪B, A∪C, B∪C,

(c) A\B, B\A, A\C, C\A,C\B, B\C.

Solution.

(a) A∩B ={b, d}, A∩C ={c, d}, B∩C ={d, f}

(b) A∪B ={a, b, c, d, f, h}, A∪C ={a, b, c, d, e, f},B ∪C={b, c, d, e, f, h}

(c) A\B ={a, c},B\A={f, h},A\C ={a, b},C\A ={e, f},C\B ={c, e},B\C ={b, h} Exercise 6. Let Rbe the set of real numbers, A={x∈R: 1≤x≤3}and B ={x∈R: 2≤ x≤4}. Find the following sets.

(a) A∪B (b) A∩B

(c) (R\A)∩B (d) (R\B)∩A

(e) (R\A)∩(R\B)

(f) (R\B)∩(R\A) (g) (R\A)∪(R\B) (h) B ∪[A∩(R\B)]

(i) [(R\A)∩B]∪[(R\B)∩A]

Solution.

(a) {x∈R: 1≤x≤4} (b) {x∈R: 2≤x≤3} (c) {x∈R: 3< x≤4} (d) {x∈R: 1≤x <2}

(e) {x∈R: x <1 or 4< x}

(f) {x∈R: x <1 or 4< x} (g) {x∈R: x <2 or 3< x} (h) {x∈R: 1≤x≤4},

(i) {x∈R: 1≤x <2 or 3< x≤4}

Exercise 7. Let A ={x ∈ Z: x is a multiple of 10} and B = {x ∈ Z: xis a multiple of 15}. What is A∩B? Can you generalize this result?

Solution. A∩B ={x∈Z: x is a multiple of 30}.

In general, if A = {x ∈Z: x is a multiple of m} and B ={x ∈ Z: x is a multiple of n}, then A∩B ={x∈Z: xis a multiple of k} where k is the least common multiple of m and n.

Exercise 8. Let A = {1,2,3,8,9}, B = {2,4,6,8}, C = {3,6,9}. Determine A\B, C\A, (A\B)∩C, (B ∪C)\(A\C), P(C), P(A\B).

Solution.

• A\B ={1,3,9}

• C\A={6}

• (A\B)∩C ={3,9}

• (B∪C)\(A\C) = {3,4,6,9}

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• P(C) ={∅,{3},{6},{9},{3,6},{3,9},{6,9}, C}

• P(A\B) = {∅,{1},{3},{9},{1,3},{1,9},{3,9},{1,3,9}}

Exercise 9. Let U = {a, b, c, d, e} be the universal set, A = {a, b, c, d}, B = {d, e} and C ={a, b, e}. Determine the elements of the following sets:

A∪B, A∩B, B, A\B, AB, (AC)\B.

Solution.

• A∪B =U

• A∩B ={d}

• B ={a, b, c}

• A\B ={a, b, c}

• AB ={a, b, c, e}

• (AC)\B =∅

Exercise 10. LetA =P({a, b}) and B =P({b, c}). Determine the elements of the following sets:

A∪B, A∩B, A\B, B\A, AB.

Solution.

• A∪B ={∅,{a},{b},{a, b},{c},{b, c}}

• A∩B ={∅,{b}}

• A\B ={{a},{a, b}}

• B\A={{c},{b, c}}

• AB ={{a},{c},{a, b},{b, c}}

Exercise 11. Sketch the following sets: (−2; 3]\[1; 4], (−2; 3]∪[1; 4], (−2; 3]∩[1; 4].

Solution.

−2 3

1 4

−2 1

−2 4

1 3

(−2; 3]\[1; 4]

(−2; 3]∪[1; 4]

(−2; 3]∩[1; 4]

Exercise 12. Sketch the following sets: (2,6]\(1; 4), (2,6]∩(1; 4), (2,6]∪(1; 4).

Solution.

2 6

1 4

4 6

2 4

1 6

(2,6]\(1; 4) (2,6]∩(1; 4) (2,6]∪(1; 4)

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Quizzes

Version A

Exercise 1. LetA={1,2,3,8,9}, B ={2,4,6,8} andC ={3,6,9}. Give the elements of the following sets.

(a) AB

(b) (A∩B)∪C (c) (C∪B)\(A∩C) (d) P(C)

Exercise 2. Sketch the following sets on the number line.

(a) (−1; 3]∩[1; 2]

(b) (−2; 2)∪[−1; 4]

(c) (−2; 3]\(0; 5) Version B

Exercise 1. Calculate the power set of (A∪B)∩(BC), whereA={1,2,4,7},B ={2,5,6,8}, C ={3,5,6,8} and the universal set is U ={1,2,3,4,5,6,7,8}.

Exercise 2. Give the following sets as intervals.

(a) {x: x∈R, x² <1}

(b) {x: x∈R, x≤2} ∩(−3,5) (c) [−3,4)\[1,5)

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Solution of Quizzes

Version A

Solution of Exercise 1.

(a) AB ={1,3,4,9,6}

(b) (A∩B)∪C ={2,8} ∪ {3,6,9}={2,3,6,8,9}

(c) (C∪B)\(A∩C) = {2,3,4,6,8,9} \ {3,9}={2,4,6,8} (d) P(C) ={∅,{3},{6},{9},{3,6},{3,9},{6,9},{3,6,9}}

(a) (−1; 3]∩[1; 2]

−1 3

1 2

(b) (−2; 2)∪[−1; 4]

−2 2

−1 4

−2 4

(c) (−2; 3]\(0; 5)

−2 3

0 5

−2 0

1 pt

2 pt

3 pt

4 pt

5 pt

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Version B

• B ={1,3,4,7}

• A∪B ={1,2,3,4,7}

• BC ={2,3}

• (A∪B)∩(BC) ={1,2,3,4,7} ∩ {2,3}={2,3}

• P({2,3}) = {∅,{2},{3},{2,3}}

(a)

x² <1 ⇐⇒ |x|<1 ⇐⇒ −1< x <1 {x:x∈R, x² <1}= (−1,1)

(b) We have to delete the numbers greater than 2 from the interval (−3,5). The remaining interval is (−3,2].

(c)

[−3,4)\[1,5) = [−3,1)

−3 4

1 5

−3 1

1 pt

2 pt

3 pt

4 pt

5 pt

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Functions

Practice problems

Functions

Exercise 1. If y = f(x) = 2x³ −3x² + 1, compute the following values and plot the corresponding points in the coordinate system.

(a) f(0) (b) f(1) (c) f(2) (d) f(−1)

Solution. Substituting the given numbers for x into the function we get the y-coordinate of the corresponding points.

(a) f(0) = 2(0)³−3(0)²+ 1 = 1

Thus y = 1 when x = 0, it gives the ordered pairA(0,1).

(b) f(1) = 2(1)³−3(1)²+ 1 = 0

Thus y = 0 when x = 1, it gives the ordered pairB(1,0).

(c) f(2) = 2(2)³−3(2)²+ 1 = 5

Thus y = 5 when x = 2, it gives the ordered pairC(2,5).

(d) f(−1) = 2(−1)³−3(−1)²+ 1 =−4 Thus y =−4 when x =−1, it gives the ordered pairD(−1,−4).

−1 1 2

−4 1 5

A B

C

D

Exercise 2. Find the following expressions, if g(x) = 3x−2.

(a) g(2a−1) (b) g(x+h) (c) g(x+h)−g(x)

h .

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Solution. We substitute the given expressions into the function. After that we make some simplifications.

(a)

g(2a−1) = 3(2a−1)−2 = 6a−3−2 = 6a−5 (b)

g(x+h) = 3(x+h)−2 = 3x+ 3h−2 (c)

g(x+h)−g(x)

h = (3x+ 3h−2)−(3x−2)

h = 3x+ 3h−2−3x+ 2

h = 3h

h = 3 Exercise 3. Identify the inner and outer parts of the following composite functions.

(a) √³

1−x (b) e^−x² (c) log₅(2−x³) (d) (2−10^x)³ Solution. The outer function corresponds to the last operation we take, calculating a specific value of the given function.

(a) (b) (c) (d)

Outer function: √³

x e^x log₅x x³ Inner function: 1−x −x² 2−x³ 2−10^x

Exercise 4. Letf(x) = 2x−3 and g(x) =x²+ 1. Find the following expressions.

(a) f(g(x)) (b) g(f(x))

Solution. We make composite functions.

(a)

f(g(x)) = 2(g(x))−3 = 2(x²+ 1)−3 = 2x²−11 (b)

g(f(x)) = (f(x))²+ 1 = (2x−3)²+ 1 = 4x²−12x+ 10

Exercise 5. Use the vertical line test to determine whether the given relation is a function by inspecting its graph.

(a) (b) (c)

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Solution.

(a) (b) (c)

Yes, it passes the vertical line test.

No, it fails the vertical line test.

Exercise 6. Find the domain of the following functions.

(a) y= 2x³ (b) y =√

3−2x (c) y= 7−1/x² (d) y= ln(5x+ 4) Solution. The domain of a function is assumed to be the largest subset of the real numbers for which the given formula produces a real number.

(a) There are no restrictions on the numbers substituted forx, so the domain consists of all real numbers, x∈R.

(b) Since √

3−2x is not definied for 3−2x <0 (that is for x >3/2), we conclude that the domain of the function is the interval x≤3/2 or (−∞,3/2].

(c) Since a denominator of a fraction cannot be zero, the domain consists of all real numbers except the zero point of x², that is 0.

(d) The argument of a logarithm is always positive, so the domain is the set of all real numbers satisfying 5x+ 4>0, that means x >−4/5.

Exercise 7. Draw the graphs of the following functions.

(a) f(x) =







−x, x <0

−x², 0≤x <2 x−1, 2≤x

(b) f(x) =







2, x≤1/2

1

x, 1/2< x <3

−3, 3≤x Solution.

(a) (b)

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Exercise 8. Give the equation of the straight lines which correspond the following descriptions.

(a) The line goes through the pointP(2,−1) and its slope is m= 1/2.

(b) The line goes through the pointsA(−1,2) andB(−2,5).

Solution.

(a) We use the equationy =mx+b of a straight line, mis given. So the half-ready equation is

y= 1 2x+b.

After substitution of the given point we get a value forb.

y = 1

2 ·x+b

−1 = 1

2 ·2 +b

−1 = 1 +b

−2 = b So the equation of the line isy = 1

2x+b.

(b) From two points of a line, we can calculate its slope:

m = ∆y

∆x = y2−y1

x2−x1

= 5−2

−2−(−1) =−3.

Then we can follow the method used in the exercise above. We have the slope and a given point of the line. After substituting the pointB into the half-ready equationy=−3x+b, we get b =−1. The final equation of the line isy =−3x−1.

Manipulating Graphs

Exercise 9. Match the equations listed to the graphs in the accompanying figure.

1. The base function is y=x². (a) y= (x−1)²−4

(b) y= (x−2)²+ 2 (c) y= (x+ 2)² + 2 (d) y= (x+ 3)² −2

−3−2 1 2

−4

−2 2

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2. The base function is y= lnx.

(a) y= ln 3x (b) y= ln¹₂x (c) y= 3 lnx (d) y= ¹₂lnx

3. The base function is y=√ x.

(a) y=√

−x (b) y=−√

x

Solution.

1. The graph of the base functiony=x² (a) moves right 1 unit and moves down 4

units, soy= (x−1)²−4 is red;

(b) moves right 2 units and moves up 2 units, soy= (x−2)²+ 2 is orange;

(c) moves left 2 units and moves up 2 units, so y= (x+ 2)²+ 2 is light blue;

(d) moves left 3 units and moves down 2 units, soy= (x+ 3)²−2 is blue.

−3−2 1 2

−4

−2 2

2. The graph of the base functiony= lnx (a) contracts in the horizontal direction, so

y= ln 3x isred;

(b) stretches in the horizontal direction, so y= ln¹₂x isblue;

(c) stretches in the vertical direction, so y= 3 lnx islight blue;

(d) contracts in the vertical direction, so y= ¹₂lnx isorange.

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3. The graph of the base functiony=√ x (a) is reflected through the y-axis, so

y=√

−x is red;

(b) is reflected through the x-axis, so y=−√

x is blue.

Exercise 10. Graph each function, not by plotting points, but by starting with the graph of the function f0(x) =x³ and applying the appropriate transformations.

(a) y= (x−1)³−2 (b) y= (1−x)³+ 2 Solution.

(a) 1. step:

f₁(x)=f₀(x−1) = (x−1)³ Graph off0 moves right 1 unit.

2. step:

f₂(x)=f₁(x)−2 = (x−1)³−2

Graph off1 moves down 2 units. ¹

(b) 1. step:

f1(x)=f0(x+ 1) = (x+ 1)³ Graph off0 moves left 1 unit.

2. step:

f2(x)=f1(−x)= (−x+ 1)³

Graph off1 is reflected to the y-axis.

3. step:

f3(x)=f2(x)+ 2 = (1−x)³+ 2 Graph off2 moves up 2 units.

−1 1

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Supplementery exercises

Functions

Exercise 1. Substitute the following expressions into the functionf(x) = 1−log₂(x²−3) and give their simplest form.

(a) x=−2 (b) x=√

11 (c) x=√

t Solution.

(a) f(−2) = 1 (b) f(√

11) =−2 (c) f(√

t) = 1−log₂(t−3)

Exercise 2. Substitute the following expressions into the function g(x) = x² 2^x·√

x and give their simplest form.

(a) x= 1 (b) x= 4 (c) x=t²

Solution.

(a) g(1) = 1

2 (b) g(4) = 1

2 (c) g(t²) = t³

2^t² Exercise 3. Give the f(g(x)) andg(f(x)) composite functions, where

(a) f(x) = √³

x, g(x) = 2−x²; (b) f(x) = 2x−3, g(x) =e^x;

(c) f(x) = 1−√

x, g(x) = log₃x;

(d) f(x) = x²−1, g(x) = 3/x.

Solution.

(a) f(g(x)) = √³

2−x² g(f(x)) = 2−√3

2−x²2

(b) f(g(x)) = 2e^x−3 g(f(x)) =e^2x−3 (c) f(g(x)) = 1−

log₃x g(f(x)) = log₃(1−√ x) (d) f(g(x)) = 9

x² −1 g(f(x)) = 3 x²−1

Exercise 4. Determine the domain of the following functions.

(a) f(x) = 4(x−3)² (b) f(x) = log₂(1−3x)

(c) f(x) = log₂(1−√ x) (d) f(x) = 3

4−x²

(e) f(x) = √³ 3 + 2x (f) f(x) = √

x−log₂(x−3) (g) f(x) = √

9−x² (h) f(x) = 2

e^1+x

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Solution.

(a) x∈R (b) x <1/3

(c) x∈[0,1) (d) x∈R\ {−2,2}

(e) x∈R (f) x >3

(g) x∈[−3,3]

(h) x∈R Exercise 5. Draw the graph of the following functions.

(a) f(x) =







x², if x <0

−x, if 0≤x≤1 2^x, if x >1

(b) g(x) =







x⁴−1, if x <0

√x, if 0≤x <1 1 + _x¹, if x≥1

Solution. (a) (b)

Exercise 6. Give the equation of the straight lines which satisfy the following properties.

(a) The line goes through the pointP(0,3) and it has a slope m= 2.

(b) The line goes through the pointsA(−1,5) andB(2,2).

(c) The line goes through the pointsA(−1,1) andB(0,−1).

Solution.

(a) y= 2x+ 3 (b) y=−x+ 4 (c) y=−2x−1

Manipulating Graphs

Exercise 7. Using transformation steps draw the graphs of the following functions.

(a) f(x) = (x−2)³+ 1 (b) f(x) = 2√³

x−1 (c) f(x) = 1−1

2log₂x (d) f(x) = log₂(3−x)

(e) f(x) = 2¹⁻^x (f) f(x) =

1− x 2

2

(g) f(x) = √ 2x−3 (h) f(x) = x

x+ 1

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Solution.

(a)

2 1

(b)

−1

(c) (d)

1 2

(e) (f)

2

(g)

3/2

(h) x

x+ 1 = x+ 1−1

x+ 1 = 1− 1 x+ 1

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Quizzes

Version A

Exercise 1. Substitute the following expressions into the function g(x) = log₃(2x−5) and give their simplest form.

(a) x= 3 (b) x= 7 (c) x= 1

Exercise 2. Draw the graph of the function

f(x) =







−2x−1, if x <0,

√x, if 0≤x≤1, 1, if x >1.

Exercise 3. Using transformation steps draw the graph of the functionf(x) =−√

2x+ 4.

Version B

Exercise 1. Give the domain of the functionf(x) = e¹⁻^1/x−log₂(2x+6) as unions of intervals.

Exercise 2. Give the equation of the straight line that passes through the points (2,−1) and (−2,4).

Exercise 3. Using transformation steps draw the graph of the functionf(x) = √³

4−2x.

Version C

Exercise 1. Let f(x) =√

1−x, g(x) = 1−2x. Determine the functionsf(g(x)) and g(f(x)) and calculate the values f(g(2)) and g(f(2)).

Exercise 2. Give the equation of the line which goes through the pointP(1,−2) and it has a slope m=−²₃.

Exercise 3. Use linear transformation steps to sketch the graph of the function f(x) = 1− 2

x−1.

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Solution of Quizzes

Version A

(a) g(3) = log₃(2·3−5) = log₃1 = 0

(b) g(7) = log₃(2·7−5) = log₃9 = log₃3² = 3 (c) The function is not defined at x= 1.

−1/2 1

−1

• f0(x) = √ x

• f0(x+ 4) =√

x+ 4 =f1(x) Graph of f0 moves left 4 units.

• f1(2x) = √

2x+ 4 =f2(x)

Graph of f1 is compressed horizontally.

• −f2(x) = −√

2x+ 4 =f(x)

Graph of f₂ is reflected through the x- axis.

−4 −2 2

−2 2

1 pt

2 pt

3 pt

4 pt

5 pt

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Version B

Solution of Exercise 1. We have two restrictions: x = 0 and 2x+ 6 > 0. The domain consists of numbers greater than −3 except 0, that is (−3,0)∪(0,∞).

The slope of the line is m= 4−(−1)

−2−2 =−5 4. Using y = mx+b, we have y = −5

4x+b. Substituting (2,−1), we have −1 = −5

4 ·2 +b, hence b= 3

2.

The equation of the line is y=−5 4x+ 3

2. Solution of Exercise 3.

• f0(x) = √³ x

• f0(x+ 4) = √³

x+ 4 =f1(x) Graph of f0 moves left 4 units.

• f1(2x) = √³

2x+ 4 =f2(x)

Graph of f1 is compressed horizontally.

• f2(−x) = √³

−2x+ 4 =f(x)

Graph of f2 is reflected through the y- axis.

−4 −2 1 1

1 pt

2 pt

3 pt

4 pt

5 pt

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Version C

• f(g(x)) =

1−(1−2x) = √ 2x

• g(f(x)) = 1−2(√

1−x) = 1−2√ 1−x

• f(g(2)) =√

2·2 =√ 4 = 2

• g(f(x)) is not defined at x= 2

Solution of Exercise 2. By y = mx+b, we have y = −2

3x+b. Substituting P we get

−2 =−2

3 +b, thusb =−4

3. The equation of the line isy=−2 3x− 4

3. Solution of Exercise 3.

• f0(x) = 1 x

• f0(x−1) = 1

x−1 =f1(x) Graph of f0 moves right 1 unit.

• 2f₁(x) = 2

x−1 =f₂(x)

Graph of f2 is stretched vertically.

• −f2(x) = − 2

x−1 =f3(x)

Graph of f2 is reflected through the x- axis.

• f3(x) + 1 = 1− 2

x−1 =f(x) Graph of f3 moves up 1 unit.

1 1

1 pt

2 pt

3 pt

4 pt

5 pt

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Limit

Practice problems

Limits from Graph

Exercise 1. Investigating the graph, determine the following limits.

(a) lim

x→−1⁺f(x) (b) lim

x→0⁻f(x) (c) lim

x→0f(x)

(d) lim

x→1f(x) (e) lim

x→2⁺f(x) (f) lim

x→2f(x)

-1 1 2

1

y =f(x)

x y

Solution.

(a) As x approaches −1 through the values −0.9, −0.99, −0.999..., the function approaches the value 1. Therefore lim

x→−1⁺f(x) = 1.

(b) As x approaches 0 through the values −0.1, −0.01, −0.001..., the function approaches the value 0. Therefore lim

x→0⁻f(x) = 0.

(c) Asxapproaches 0 through either positive or negative values, the function approaches the value 0. Therefore lim

x→0f(x) = 0 even if f is defined at x= 0 as f(0) = 1.

(d) We can see that lim

x→1f(x) does not exist because the left-handed limit, which is clearly 1,

xlim→1⁻f(x) = 1, is different from the right-handed limit, which is 0, lim

x→1⁺f(x) = 0.

(e) Right-handed limit lim

x→2⁺f(x) is not defined because the domain of f is the [−1,2] closed interval, that is the function is not defined for values grater than 2.

(f) Since 2 is the right hand end point of the domain off, we have lim

x→2f(x) = lim

x→2⁻f(x) = 0.

27

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Limits by Substitution

Exercise 2. Find the following limits.

(a) lim

x→5(x−3) (b) lim

x→−3(x²+ 4x+ 3) (c) lim

y→−2

y²+ 5

(d) lim

x→−1(π−1) (e) lim

v→2

v³+ 8 v⁴ −6

(f) lim

v→−2

v³+ 8 v⁴−6 (g) lim

z→3(2^z−1) (h) lim

t→2log₂(4t−7) Solution. To calculate limit of a function, we always start with substitution.

(a) Substituting 5 for x we have

x→5lim(x−3) = 5−3 = 2.

(b) Substituting −3 for x we have

xlim→−3(x²+ 4x+ 3) = (−3)²+ 4·(−3) + 3 = 9−12 + 3 = 0.

(c) Similarly,

ylim→−2

y²+ 5 =

(−2)²+ 5 =√ 9 = 3.

(d) Since there is no x in the expression π−1, for any x value the function value is π−1, that is we have a constant function, so

xlim→−1(π−1) = π−1.

(e)

limv→2

v³ + 8

v⁴−6 = 8 + 8 16−6 = 16

10 (f)

vlim→−2

v³+ 8

v⁴−6 = −8 + 8 16−6 = 0 (g)

zlim→3(2^z−1) = 2³−1 = 7 (h)

limt→2log₂(4t−7) = log₂1 = 0

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Eliminating Zero Denominators Algebraically

(a) lim

x→2

(x−3)(x−2) x−2 (b) lim

x→−2

x²+ 4x+ 4 x+ 2 (c) lim

y→3

y²−9 y²−3y

(d) lim

z→1

z²−2z+ 1 z² +z−2 (e) lim

t→1

√t−1 t−1

(f) lim

h→0

h 1−√

h+ 1 (g) lim

v→4

√v−2 v²−16

Solution. If substitution does not give a valid number, then we have to figure out something else. When after substitution both the numerator and the denominator have a value 0, pro- ducing the meaningless fractional form 0/0, we try to cancel common factors or rationalize the numerator and/or the denominator, and reevaluate the simplified form by direct substitution.

(a) Substituting 2 forxwe have the form 0/0. However, in determining the limit at 2 we are concerned with only the values of the function when x is close to 2, butx = 2. Thus we can divide numerator and denominator by x−2 to obtain

xlim→2

(x−3)(x−2)

x−2 = lim

x→2(x−3).

We can calculate this new limit by direct substitution to have

xlim→2

(x−3)(x−2) x−2 = lim

x→2(x−3) = 2−3 = −1.

(b) By direct substitution we obtain the form 0/0. Therefore both the numerator and the denominator have a common factor of (x+ 2). Since x²+ 4x+ 4 = (x+ 2)², we have

x→−2lim

x²+ 4x+ 4

x+ 2 = lim

x→−2

(x+ 2)²

x+ 2 = lim

x→−2(x+ 2) =−2 + 2 = 0.

(c) Substitution yields form 0/0. Because of y²−9 = (y−3)(y+ 3), y²−3y=y(y−3), we can cancel common factors:

ylim→3

y² −9

y²−3y = lim

y→3

(y−3)(y+ 3)

y(y−3) = lim

y→3

y+ 3

y = 3 + 3 3 = 2.

(d) Substitution gives 0/0.

zlim→1

z²−2z+ 1

z²+z−2 = lim

z→1

(z−1)²

(z+ 2)(z−1) = lim

z→1

z−1 z+ 2 = 0

(e) Form 0/0. In this case, first we rewrite the fraction by rationalizing the numerator.

limt→1

√t−1 t−1 = lim

t→1

√t−1 t−1 ·

√t+ 1

√t+ 1 = lim

t→1

(√

t−1)·(√ t+ 1) (t−1)·(√

t+ 1)

= lim

t→1

t−1 (t−1)·(√

t+ 1) = lim

t→1

√ 1

t+ 1 = 1 2

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(f) Form 0/0. We rewrite the fraction by rationalizing the denominator.

hlim→0

h 1−√

h+ 1 = lim

h→0

h 1−√

h+ 1 · 1 +√ h+ 1 1 +√

h+ 1 = lim

h→0

h(1 +√ h+ 1) 1−(h+ 1)

= lim

h→0

h(1 +√ h+ 1)

−h = lim

h→0

1 +√ h+ 1

−1 = 2

−1 =−2 (g) Form 0/0.

v→4lim

√v−2

v²−16 = lim

v→4

√v−2 v²−16 ·

√v+ 2

√v+ 2 = lim

v→4

v−4 (v²−16)·(√

v+ 2)

= lim

v→4

v−4 (v−4)(v + 4)(√

v+ 2) = lim

v→4

1 (v+ 4)(√

v+ 2) = 1 32

One-sided Limits

(a) lim

x→0

2x−1 x (b) lim

x→2

1−x x(2−x)

(c) lim

x→0

3 x²(x−2) (d) lim

x→−1

x²+ 1 x²−3x−4

Solution. When after substitution only the denominator has a value 0 the limit is either ∞,

−∞ or does not exist. We calculate one-sided limits to decide on that.

(a) By direct substitution we obtain the form −1/0. Hence, we are looking at the one-sided limits. First consider

xlim→0⁺

2x−1 x .

As x approaches 0 from the right, the numerator is eventually a negative quantity approaching−1 and the denominator a positive quantity approaching 0. Consequently, the ratio is a negative quantity that decreases without bound. This means

x→0lim⁺

2x−1

x =−∞. Now consider

xlim→0⁻

2x−1 x .

Asxapproaches 0 from the left, the numerator is a negative quantity approaching−1 and the denominator a negative quantity approaching 0. Consequently, the ratio is a positive quantity that increases without bound. That is,

xlim→0⁺

2x−1 x =∞. The side-limits are not the same, so

x→0lim

2x−1 x does not exist.

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(b) Substitution yields the form −1/0. Let us write the function as 1−x

x(2−x) = 1−x

x · 1

2−x

to separate the factor responsible for the 0 in the denominator. Then, by

xlim→2

1−x x = −1

2 , lim

x→2⁺

1

2−x =−∞, lim

x→2⁻

1

2−x =∞ we have

xlim→2⁺

1−x

x(2−x) =∞, lim

x→2⁻

1−x

x(2−x) =−∞. Therefore

x→2lim

1−x x(2−x) does not exist.

(c) Substitution gives 3/0. By

3

x²(x−2) = 3 x−2 · 1

x² and

xlim→0

3

x−2 = 3

−2, lim

x→0⁺

1

x² =∞, lim

x→0⁻

1 x² =∞ we have

xlim→0⁺

1−x

x(2−x) =−∞, lim

x→0⁻

1−x

x(2−x) =−∞. It means, that

xlim→0

3

x²(x−2) =−∞. (d) Form 2/0. By

x²+ 1

x²−3x−4 = x²+ 1 x−4 · 1

x+ 1 and

xlim→−1

x²+ 1 x−4 = 2

−5, lim

x→−1⁺

1

x+ 1 =∞, lim

x→−1⁻

1

x+ 1 =−∞, we have

x→−lim1⁺

x²+ 1

x²−3x−4 =−∞, lim

x→−1⁻

x²+ 1

x²−3x−4 =∞. Therefore

xlim→−1

x²+ 1 x²−3x−4 does not exist.

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Supplementery exercises

Limits from Graph

Exercise 1. Using the graph, determine the following limits.

(a) lim

x→−1⁺f(x) (b) lim

x→0⁻f(x) (c) lim

x→0f(x)

(d) lim

x→1f(x) (e) lim

x→2⁺f(x) (f) lim

x→2f(x)

-1 1 2

1

y= f(x)

x y

Solution.

(a) 0 (b) 1

(c) Does not exist.

(d) 1

(e) Not defined.

(f) 0

Limits by Substitution

Exercise 2. Find the following limits and simplify your result.

(a) lim

x→−7(2x+ 5) (b) lim

x→2(−x²+ 5x−2) (c) lim

t→6(t−5)(t−7)

(d) lim

x→2

x+ 3 x+ 6 (e) lim

x→5

4 x−7 (f) lim

y→−5

y² 5−y

(g) lim

z→0(2z−8)^1/3 (h) lim

t→−3(5−t)^4/3 (i) lim

h→0

√ 3

3h+ 1 + 1 Solution.

(a) −9 (b) 4

(c) −1

(d) 5/8 (e) −2 (f) 5/2

(g) −2 (h) 16

(i) 3/2

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Eliminating Zero Denominators

(a) lim

x→5

x−5 x²−25 (b) lim

x→−5

x²+ 3x−10 x+ 5 (c) lim

x→−3

x+ 3 x²+ 4x+ 3 (d) lim

t→1

t² +t−2 t² −1

(e) lim

t→−1

t²+ 3t+ 2 t²−t−2 (f) lim

y→0

5y³+ 8y² 3y⁴−16y² (g) lim

u→1

u⁴−1 u³−1 (h) lim

x→9

√x−3 x−9

(i) lim

h→0

√3h+ 1−1 h (j) lim

h→0

√5h+ 4−2 h (k) lim

x→4

4x−x² 2−√

x (l) lim

y→1

y−1

√y+ 3−2 Solution.

(a) ₁₀¹ (b) −7

(c) −¹₂ (d) ³₂

(e) −¹₃ (f) −¹₂ (g) ⁴₃ (h) ¹₆

(i) ³₂ (j) ⁵₄ (k) 16

(l) 4

One-sided Limits

(a) lim

x→2⁻

3 x−2 (b) lim

x→2⁺

3 x−2 (c) lim

x→2

3 x−2

(d) lim

t→3

1 t−3 (e) lim

z→−8⁺

2z z+ 8 (f) lim

z→−8⁻

2z z+ 8

(g) lim

y→7

4 (y−7)² (h) lim

p→−2⁺

2−p p²+p−2 (i) lim

q→1⁻

2 +q q²−3q+ 2 Solution.

(a) −∞

(b) ∞

(c) Does not exist.

(d) Does not exist.

(e) −∞

(f) ∞

(g) ∞ (h) −∞

(i) ∞

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Mixed exercises

(a) lim

u→2

u−2 u²−4 (b) lim

u→−2

u−2 u²−4

(c) lim

v→3

v+ 3 v² −6v+ 9 (d) lim

v→3

v−3 v² −6v+ 9

(e) lim

h→0

(x+h)²−x² h (f) lim

x→0

(x+h)²−x² h Solution.

(a) ¹₄

(b) Does not exist.

(c) ∞

(d) Does not exist.

(e) 2x (f) h

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Quizzes

Version A

Exercise. Find the following limits.

(a) lim

x→−2

x+ 2 x²+ 4 (b) lim

x→5

x²−3x−10 x−5 (c) lim

y→4

√y−2 y²−4y (d) lim

t→3

t−2 t² −t−6 Version B

Exercise. Find the following limits.

(a) lim

p→3

p−3 p²+ 3p (b) lim

x→−2

x²+ 2x x²+x−2 (c) lim

x→2

√x+ 2−2 x²−3x+ 2 (d) lim

t→3

2t−2 t² −5t+ 6