Calculus Super Review

By Editors of REA

Get all you need to know with Super Reviews! Each Super Review is packed with in-depth, student-friendly topic reviews that fully explain everything about the subject. The Calculus I Super Review includes a review of functions, limits, basic derivatives, the definite integral, combinations, and permutations. Take the Super Review quizzes to see how much you've learned - and where you need more study. Makes an excellent study aid and textbook companion. Great for self-study! DETAILS - From cover to cover, each in-depth topic review is easy-to-follow and easy-to-grasp - Perfect when preparing for homework, quizzes, and exams! - Review questions after each topic that highlight and reinforce key areas and concepts - Student-friendly language for easy reading and comprehension - Includes quizzes that test your understanding of the subject

• Language: English
• Publisher: Research & Education Association
• Release date: Jan 1, 2013
• ISBN: 9780738665672

Book preview

Series

CHAPTER 1

Fundamentals

1.1 Number Systems

The real number system can be broken down into several parts and each of these parts have certain operations which can be performed on them. First, let us define the components of the real number system.

The natural numbers, denoted N, are 1, 2, 3, 4, .... The integers, denoted Z, are ...-3, -2, -2, 0, 1, 2, 3, .... The rational numbers, denoted Q, are all numbers of the form p/q where p and q are integers and q≠0. A real number x is a non-terminating decimal (with a sign + or -).

Six basic algebraic properties of rational numbers:

The closure property: If x and y are rational numbers, then x+v and x·y are also rational numbers.

Additive and multiplicative identity elements: If x is a rational number, then x+0=x and x·1=x.

Associative property: If x, y and z are rational numbers, then x+(y+z)=(x+y)+z, x·(y·z)=(x·y)·z.

Additive and multiplicative inverses : For each rational number, x, there exists a rational number, denoted - x such that x+(-x)=0; if x≠0, there exists a rational number x-1 such that x·x-1=1.

Commutative property: If x and y are rational numbers, then x+y=y+x, x·y=y·x.

Distributive property: If x, y and z are rational numbers, then

x·(y+z) = (x·y) + (x·z)

If q and p are rational numbers and p-q is negative, then q is greater than p, (q>p) or p is less than q, (p

Problem Solving, Examples:

Find all real numbers satisfying the inequality |3x + 2| > 5.

Writing the given equation into the form which can be dealt with algebraically, we note that the given inequality expressed with an absolute number is equal to the following inequalities:

3x + 2 > 5 or 3x + 2 < -5.

Considering, now, the first inequality, we have:

3x + 2 > 5,

or

x > 1.

Therefore, the interval (1, + ∞) is a solution.

From the second inequality, we have:

3x + 2 < -5,

or

is a solution.

or, expressed in another way, all x

Show that the set A = {x ∈ R|0 < x < 1 } is uncountable. Conclude that R is uncountable.

A set S is countable if it is in a one to one correspondence with a subset of the natural numbers

N = {1, 2, 3, 4, ...}.

S is uncountable if it is not countable. Since

A is at least countably infinite. Suppose A is countably infinite. Then we could list the members of A (represented as infinite decimals) as follows:

a1 = .a11 a12 a13 a14 ...

a2 = .a21 a22 a23 a24 ...

a3 = .a31 a32 a33 a34 ...

where

Let

b = .b1 b2 b3 b4 ...

where

bi =5 if aii ≠ 5

and

Hence b, which differs from each ai in the ith decimal place, is not in the list. Since b ∈ A, A is not countably infinite, therefore it is uncountable. If

f: A → R

is defined as the one-to-one map onto the real numbers by

then it is seen that R is of the same uncountable order as A. In order to check that f is onto, let c be any real number. If c < 0, let x be chosen so that

i.e., 2x - 1 = cx or x(2 - c

By construction f(x) = c. If c ≥ 0, let x be such that

or

2x + cx = 1 + c

or

Again f(x) = c.

This shows that f is onto. To show f is one-one note that for

On the other hand if

), then

(respectively

and this is positive, i.e., f(y) > f(x).

This means that if

x ≠ y, f(x) ≠ f(y)

showing f is one-one.

1.1.1 Properties of Rational Numbers

Trichotomy property: If p and q are rational numbers, then one and only one of the relations q=p, q>p or q

Transitive property: If p, q and r are rational numbers, and if p

If p, q and r are rational numbers and p

If p, q and r are rational numbers and if p

Problem Solving Examples:

Show that the set Q of rational numbers x such that 0

A set A is countably infinite if it is in a one to one correspondence with the natural numbers (i.e., {1, 2, 3, ...}). Construct a table of Q in the following manner:

The numerators in successive rows of this table are 1, 2, 3 .... The denominators in each row are increasing but so that each fraction is proper (i.e., in Q) and in lowest terms (i.e., appears only once in the table). Now match the above table of Q with this table of the natural numbers:

Consequently the set Q is in a one to one correspondence with the natural numbers.

Remark: Let An = {p/n : p an integer} for n ≥ 1, i.e., An is the set of all rational numbers with denominator n. Then and by construction each An is countable. A countable union of countable sets is countable. Hence Q is countable. This shows that Q is countably infinite.

1.2 Inequalities

To solve a linear inequality draw a number line, dashed for x<-b/a and solid for x>-b/a.

ax+b>0 or x>-b/a , where a >0

To solve (ax+b)(cx+d) > 0 graphically, where a>0 and c>0:

The solution lies in the interval where both lines are dashed and both lines are solid.

is the solution to the above inequality.

Problem Solving Examples:

Solve the inequality |3 - 2x |< 1.

|3 - 2x |< 1 can be represented as -1 < 3 - 2x < 1 By subtracting 3 from all the terms, we have

and remembering to reverse 2 the signs of the inequalities, we have

2 > x > 1

or all values of x in the open interval (1, 2).

Graph the following two inequalities and show where the two graphs coincide:

The first inequality consists of an infinite strip between the lines x = 2 and x = 3. Note that the points on the line x = 2 are included, but the points on the x = 3 are not.

For the second inequality, two cases must be considered, depending on whether (y - 2) is positive or negative. This may be expressed

Adding 2 to each term to simplify, gives:

The result is actually a square. Therefore, all points inside this square satisfy both inequalities. However, in addition, the points on the boundary of the square along the line x = 2 (left boundary), except for the corners, also satisfy both inequalities.

1.3 Absolute Value

Definition: The absolute value of a real number x is defined as

For real numbers a and b:

lal = |-a|

|ab| = |a|·|b|

- |a| ≤a≤|a|

ab ≤ |a||b|

|a+b|² = (a+b)²

|a+b| ≤ |a| + |b| (Triangle Inequality)

|a-b| ≥ ||a| - |b||

For positive values of b

|a| < b if and only if -b

|a| > b if and only if a > b ora<-b

|a| = b if and only if a = b or a = -b

Problem Solving Examples:

Solve for x when |5 - 3x |= -2.

This problem has no solution, since the absolute value can never be negative and we need not proceed further.

Solve for x when |3x + 2 |= 5.

First we write expressions which replace the absolute symbols in forms of equations that can be manipulated algebraically. Thus this equation will be satisfied if either

3x+2=5 or 3x + 2 = -5.

Considering each equation separately, we find

Accordingly the given equation has two solutions.

1.4 Set Notation

A set is a collection of objects called elements. Let A and B be sets.

A : indicates that x is an element of A

B : indicates that x is not an element of B

A is a subset of B , (A ⊂ B), means that A is contained in another set B and each element of A is also an element of B.

A is equal to B (A=B) , if and only if A ⊂ B and B ⊂ A.

A U B : The union of A and B ; the set consists of all elements of A and B .

A ∩ B : the intersection of A and B ; the set consists of elements, common to both A and B .

A ∩ B = ø: It is the set that has no elements common to both A and B; thus it is an empty set. In this case A and B are said to be disjoint.

These notations may be used to describe intervals of numbers such as:

The open interval (a,b) = {x : a < x < b }

The closed interval [a,b] = {x : a ≤ x ≤ b}

The half-open intervals [a,b) = {x : a ≤ x < b}

and (a, b] = {x : a < x ≤ b}

1.5 Summation Notation

ai, that is,

c = nc for every real number c, and

For any positive integer n and the sets of numbers {a1, a2, a3, ...,an} and {b1, b2, b3 ,..., bn},

Problem Solving Examples:

Find the numerical value of the following:

A(r) means

is the Greek letter sigma and is a shorthand way to denote the sum. It avoids having to write the sum A(0) + A(1) + A(2) + ... + A(n).

For a) successively replace j by 1,...,7 and add up the terms 7

For b) successively replace j by 1,2,3,...,21 and add up the terms.

Establish the convergence or divergence of the series:

< n for n > 1. Therefore the given series is divergent.

CHAPTER 2

Functions

2:1 Functions

Definition: function is a correspondence between two sets, the domain and the range, such that for each value in the domain there corresponds exactly one value in the range.

A function has three distinct features:

the set x which is the domain,

the set y which is the co-domain or range,

X. We write y = f(x) to denote the functional value y at x.

Consider Figure 2.1. The machine f transforms the domain X, element by element, into the co-domain Y.

Fig. 2.1

Problem Solving Exgmples:

Determine the domain of the function:

is defined only if the denominator is not equal to zero.

The denominator is equal to zero when (x - 2) (x + 1) = 0, or, equivalently, when (x - 2) = 0 or (x + 1) = 0. (x - 2) = 0 when x = 2, and (x + 1) = 0 when x = -1.

Therefore, the domain of the function is all values of x where x ≠ -1 and x ≠ 2.

Determine the domain of the function:

Since we are restricted to the real number system, there are two cases to be considered:

Case 1. When the denominator equals 0, the fraction is undefined. This occurs when x = 2.

the square root gives an imaginary number. Therefore,

only if x< 0, and x >2.

Combining the conditions in cases 1 and 2, the domain of x is the interval 0 ≤ x < 2.

2.2 Combination of Functions

Let f and g represent functions, then

the sum (f+g)(x) = f(x) + g(x),

the difference (f-g)(x) = f(x) - g(x),

the product (fg)(x) = f(x)g(x),

g(x) ≠ 0,

the composite function (go f) (x) = g(f(x)) where f(x) must be in the domain of g.

A polynomial function of degree n is denoted as

where an is the leading coefficient and not equal to zero and akxk is the kth term of the polynomial.

Problem Solving Examples:

Find the general expression for the nth derivative of:

First derivative:

f’(x) = -1(3x + 2)-2 (3) = -3(3x + 2)-2.

Second derivative:

f"(x) = 6(3x + 2)-3 (3) = 18(3x + 2)-3.

Third derivative:

f" ’(x) = -54(3x + 2)-4 (3) = 162(3x + 2)-4.

Fourth derivative:

f (x) = 648(3x + 2)-5 (3) = 1944(3x + 2)-5.

To express the nth derivative, a pattern that the sequence of derivatives follows must be found.

1) First, it is noted that the derivatives are alternately negative and positive. The odd order of differentiation results in a negative value, and the even order of differentiation, in a positive value. This property can be expressed by (-1)n for the nth derivative.

2) The coefficients of (3x + 2) (ignoring the sign) are 3, 18, 162, 1944,...

First derivative:

n = 1: 3 = 3¹ · = 3n · n!

Second derivative:

n = 2: 18 = 3² (2!) = 3n · n!

Third derivative:

n = 3: 162 = 3³ (3!) = 3n · n!

Fourth derivative:

n = 4: 1944 = 3⁴ (4!) = 3n · n!

3) The power of (3x + 2) is considered.

Combining these results, the nth derivative is:

We first rewrite the given function as

Substituting, we have: