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DEMYSTIFIED

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RHONDA HUETTENMUELLER

McGRAW-HILL

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DOI: 10.1036/0071451579

v

CHAPTER 7 Implicit Differentiation and

This book was written to help you solve problems and understand concepts

covered in a business calculus course To make the material easy to absorb, only

one idea is covered in each section Examples and solutions are given in detail so

that you will not be distracted by missing algebra and/or calculus steps Topics

that students find difficult are written with extra care

Each section contains an explanation of a concept along with worked out

examples At the end of each section is a set of practice problems to help you

master the computations, and solutions are given in detail Each chapter ends with

a chapter test so that you can see how well you have learned the material, and

there is a final exam at the end of the book

If you have recently taken an algebra course, you can probably skip the algebra

review at the beginning of the book The material in Chapters 1 and 2 lay the

foundation for the concept of the derivative, which is introduced in Chapter 3 The

formulas in Chapter 4 are used throughout the book and should be memorized

Calculus techniques and other formulas are covered in Chapters 6, 7, 8, 9, and 11

Calculus can solve many business problems, such as finding the price (or quantity)

that maximizes revenue, finding the dimensions that minimize the cost to construct

a box, and finding how fast the profit is changing at different production levels

These applications and others can be found in Chapters 5, 7, 10, 11, and 12

Integral calculus and its applications are introduced in the last three chapters

I hope you find this book easy to use and that you come to appreciate the beauty

of this powerful subject

Rhonda Huettenmueller

vii

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DEMYSTIFIED

Algebra Review

Success in calculus requires a solid algebra background Although most of the

algebra steps (as well as calculus steps) are provided in the book, it is worth

reviewing algebra basics In this chapter, we will briefly review how to factor,

simplify fractions, solve equations, find equations of lines, and more

Factoring

One of the most important properties in mathematics is the distributive property:

a(b +c) = ab+ac This property allows us to either add b and c before multiplying

by a or to multiply b and c by a before adding For example, 2(4 + 5) could be

computed either as 2× 9 or as 8 + 10 Factoring is working with the distributive

property in reverse To factor an expression means to write the sum or difference

as a product

1

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Factor the expression The common factor is in bold print

• 6x + 9xy

Each of 6x and 9xy is divisible by 3x: 6x = 2 · 3x and 9xy = 3 · 3x · y.

When we divide 6x by 3x, we are left with 2 When we divide 9xy by 3x,

we are left with 3y.

6x + 9xy = 2 · 3x + 3 · 3x · y = 3x(2 + 3y)

• x2y + 4xy2+ 5x = x · xy + 4 · x · y2+ 5 · x = x(xy + 4y2+ 5)

• 8xh−2xyh+7yh2+h = 8x·h−2xy·h+7yh·h+1·h = h(8x−2xy+7yh+1)

Using the distributive property on such quantities as (x + 2)(y − 3) requires

several steps In this book, we will use the FOIL method The letters in “FOIL”help us to keep track of which quantities are multiplied and which are added.The “F” stands for “first times first.” We multiply the first two quantities In

(x + 2)(y − 3), this means we multiply x and y “O” stands for “outer times outer.” We multiply the outside quantities: x and−3 “I” stands for “inner times

inner.” We multiply the inside quantities: 2 and y “L” stands for “last times last.”

We multiply the last quantities: 2 and−3

a , b, and c stand for fixed numbers.

Many quadratic expressions can be factored with little trouble We will begin

with expressions of the form x2+bx +c The first step is to write (x )(x )

so that when we use the FOIL method, the first term is x · x = x2 Next, we will

choose two numbers whose product is c For example, if we factor x2+ 6x + 5,

we would try 5 and 1: (x 5)(x 1) Finally, we will decide if we need to use

two plus signs, two minus signs, or one of each The second sign in x2+ 6x + 5

tells us whether or not the signs are the same If the second sign is plus, then both

signs are the same The second sign is plus, so both signs in (x 5)(x 1) are

the same If the signs are the same, then they will be the first sign In x2+ 6x + 5 the first sign is plus, so we need to plus signs in (x 5)(x 1): (x + 5)(x + 1).

We will use the FOIL method on (x + 5)(x + 1) to see if our factorization is

correct

(x + 5)(x + 1) = x · x + x · 1 + 5 · x + 5 · 1 = x2+ 6x + 5

EXAMPLE

• Factor x2− 2x − 15.

We have several choices for (x )(x ) Beginning with the factors

of 15, we need to choose between 1 and 15 or 3 and 5 That is, we

either want (x 1)(x 15) or (x 3)(x 5) Because the second sign in

x2− 2x − 15 is a minus sign, the signs in the factors are different We have

four possibilities

(x −1)(x +15) (x +1)(x −15) (x −3)(x +5) (x +3)(x −5) The last possibility is correct: (x + 3)(x − 5) = x2 − 5x + 3x − 15 =

x2− 2x − 15.

If both signs in the factors are the same, b is the sum of the factors of c If the signs in the factors are different, the difference of the factors of c is b In the first

example, the sum of 1 and 5 is 6 In the second example, the difference of 5 and

other has a minus sign Now we have (3x + )(x− ) and (3x− )(x+ ).

We have two pairs of factors of 8 to try: 1 and 8 and 2 and 4 There areeight possibilities

( 3x + 1)(x − 8) ( 3x − 1)(x + 8) ( 3x − 2)(x + 4) ( 3x + 2)(x − 4)

( 3x + 8)(x − 1) ( 3x − 8)(x + 1) ( 3x − 4)(x + 2) ( 3x + 4)(x − 2) The correct factorization is (3x − 2)(x + 4) = 3x2 + 12x − 2x − 8 = 3x2+ 10x − 8.

• 9x2− 4

We factor quadratic expressions of the form (ax)2 − c2 with the formula

A2− B2 = (A − B)(A + B) In this example, 9x2 is (3x)2 and 4 is 22

the middle terms always cancel

( 3x − 2)(3x + 2) = 9x2+ 6x − 6x − 4 = 9x2− 4

Some quadratic expressions do not factor easily For example, x2+ x + 1

cannot be factored using the techniques we have learned so far

A fraction is reduced to its lowest terms, or simplified, when the numerator anddenominator have no common factors The fraction2x6 is not reduced to its lowest

terms because the numerator, 2x, and denominator, 6, are each divisible by 2.

We simplify fractions by factoring the numerator and denominator, using theircommon factors, and canceling

the same as x nor as 1 + x We can rewrite the expression as the sum of two

fractions and reduce one of them

4h2

h = h · 4h

h· 1 = 4h3

of two separate fractions Remember that the fractiona b is another way of writing

We will begin by writing x +h1 −1

We will replace x+h1 −1

x with −h x(x+h).

5x −5x−5h x(x +h) h

=

−5h x(x +h)

Exponents and Roots

In order to use two important formulas in calculus, we need exponent and rootproperties to rewrite expressions as quantities raised to a power Properties 5–7below are the most important

The union symbol, “∪,” is used to describe two or more regions For example,

( −∞, 3) ∪ (5, ∞) describes all numbers smaller than 3 or all numbers larger

than 5 (see Figure R.4)

<

Fig R.4.

The Greek letter sigma, “
,” is used in Chapter 14 to describe a sum There is usually a subscript and a superscript on
The subscript tells us where the sum

begins, and the superscript tells us where the sum ends

The absolute value of a number is its distance from 0 The absolute value of−5

is 5 because it is 5 units away from 0 The absolute value of a positive number isthe number itself The absolute value of a quantity is denoted with absolute valuebars, “| |.” The absolute value of −5 is denoted “| − 5|.” This notation is used onoccasion beginning in Chapter 13

EXAMPLES

• 4x + 10 = 0 4x + 10 = 0

−10 − 10 Move the non-x term to the right side.

4x = −10

x = −10

4 Divide both sides by 4, the coefficient of x.

x = −52

A quadratic equation is an equation that can be put in the form ax2+bx+c = 0.

Most quadratic equations in this book can be solved by factoring We will use the

quadratic formula on others No matter which method we use, we need to have

a zero on one side of the equation Once this is done, we will try to factor the

quadratic expression ax2 + bx + c If it factors easily, we will set each factor

equal to zero and will solve for x If it does not factor easily, we will use the

quadratic formula Most of these equations have two solutions

The quadratic formula can solve any quadratic equation If ax2+bx+c = 0,

1−√334When an equation is in the form “fraction = fraction,” we will cross-multiply

to solve for x That is, we will multiply the numerator of each fraction by

the denominator of the other fraction

4x2 = 5 is a quadratic equation We could use the quadratic formula for

4x2− 5 = 0, but we can solve it more quickly by dividing both sides ofthe equation by 4 and then taking the square root of each side

x2 = 54

x = ±

5

4 = −

5

4,

54

= −3 +

√33

−3 −√332

= −7 +

√41

−7 −√4146

2x

3 = 5

7x 2x · 7x = 3 · 5 14x2 = 15

x2 = 1514

x = ±

15

14 = −

15

14,

1514

The Equation of a Line

Throughout much of the book, we find equations of lines Although there are

several forms for the equation of a line, we will use the form y = mx + b We will be given an x-value, a y-value, and m (Later, we will use a formula to find

m ) Having values for x, y, and m, gives us enough information to find b.

Functions and Their Graphs

The definition of a function is a relation between two sets A and B such that every

element in set A is paired with exactly one element in set B Functions in this

book are equations, usually with the variables x and y At times, we will use the

name of the function such as f (x), C(x), R(x), etc., instead of y Here are some

examples of functions

• y = 2x − 1 • f (x) = x2+ x − 2 • R(x) =√x

To evaluate a function at a number or expression means to substitute the number

or expression for x This gives us the y-value, also called the functional value, for a particular x-value The notation “f (6)” means that 6 has been substituted

in the equation for x.

• f (x) = 100; 6, 28

f (x) = 100 is a linear function whose slope is 0 No matter what value x has, the functional value (the y-value) is always 100.

f ( 6)= 100 f ( 28)= 100

In Chapter 3, we evaluate functions at algebraic expressions Again, we will

substitute the given quantity for x.

f (l − 4) =√l− 4 + 9 =√l+ 5 f ( 3l)=√3l+ 95.

4 For example, (3, 1) is on the graph because 3+ 1 = 4

The graph of a line having a zero slope is a horizontal line The graph in

Figure R.9 is the graph of y = 5 (or y = 0x + 5).

-3 -2 -1 1 2 3 4 5

-3-2-1

12345

246810

y = 5

< >

Fig R.9.

We can look at the graph of a function to find functional values The

y-coordinate of the point (the second number) is the functional value for the

x -coordinate of the point (the first number) For example, the point (2, 5) is

on the graph of f (x) = x2+ 1 because f (2) = 22+ 1 = 5

The point ( −1, −2) is on the graph, so f (−1) = −2 The point (0, −3) is

on the graph, so f (0) = −3 The point (2, 1) is on the graph, so f (2) = 1.

• The graph in Figure R.11 is the graph of a function f (x) Find f (4), f (−3), and f ( −4).

The point (4, 1) is on the graph, so f (4) = 1 The point (−3, 1) is on the graph, so f (−3) = 1 There is a hole in the curve at x = −4, so the curve does not give us the functional value at x = −4 The dot at (−4, −2) indicates that the function is defined there for x = −4 The dot is the point

( −4, −2), so f (−4) = −2.

PRACTICE

1 The graph of f (x)=√x is given in Figure R.12 Find f (4) and f (9).

2 The graph of f (x) is given in Figure R.13 Find f (2), f (4), and f (1).

-5 -4 -3 -2 -1 1 2 3 4 5

-4-3-2-1

123456

12345

-5 -4 -3 -2 -1 1 2 3 4 5

-5-4-3-2-1

1234

The Slope of a Line

and the Average Rate of Change

Calculus is the study of the rate of change We use the slope of a line to describe

the rate of change of a function Instead of thinking of the slope of a line as a

simple “rise over run,” we need to think of it as a number that measures how one

variable changes compared to a change in the other variable The numerator of

the slope describes the change in y, and the denominator describes the change

in x For example, a slope of 35 says that as x increases by 5, y increases by 3.