Fundamentals Of Calculus

Fundamentals Of Calculus

giancarlo black

Published by giancarlo black, 2024.

While every precaution has been taken in the preparation of this book, the publisher assumes no responsibility for errors or omissions, or for damages resulting from the use of the information contained herein.

FUNDAMENTALS OF CALCULUS

First edition. October 19, 2024.

Copyright © 2024 giancarlo black.

Written by giancarlo black.

Fundamentals of   Calculus

Giancarlo Black

Giancarlo Black

FUNDAMENTALS OF CALCULUS

CONTENTS

Preface ix

About the Authors xiii

1  Linear Equations and Functions 1

1.1  Solving Linear Equations, 2

1.2  Linear Equations and their Graphs, 7

1.3  Factoring and the Quadratic Formula, 16

1.4  Functions and their Graphs, 25

1.5  Laws of Exponents, 34

1.6  Slopes and Relative Change, 37

2  The Derivative 43

2.1  Slopes of Curves, 44

2.2  Limits, 46

2.3  Derivatives, 52

2.4  Differentiability and Continuity, 59

2.5  Basic Rules of Differentiation, 63

2.6  Continued Differentiation, 66

2.7  Introduction to Finite Differences, 70

3  Using The Derivative 76

3.1  Describing Graphs, 77

3.2  First and Second Derivatives, 83

3.3  Curve Sketching, 92

3.4  Applications of Maxima and Minima, 95

3.5  Marginal Analysis, 103

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Giancarlo Black

vi CONTENTS

4  Exponential and Logarithmic Functions 109

4.1  Exponential Functions, 109

4.2  Logarithmic Functions, 113

4.3  Derivatives of Exponential Functions, 119

4.4  Derivatives of Natural Logarithms, 121

4.5  Models of Exponential Growth and Decay, 123

4.6  Applications to Finance, 129

5  Techniques of Differentiation 138

5.1  Product and Quotient Rules, 139

5.2  The Chain Rule and General Power Rule, 144

5.3  Implicit Differentiation and Related Rates, 147

5.4  Finite Differences and Antidifferences, 153

6  Integral Calculus 166

6.1  Indefinite Integrals, 168

6.2  Riemann Sums, 174

6.3  Integral Calculus – The Fundamental Theorem, 178

6.4  Area Between Intersecting Curves, 184

7  Techniques of Integration 192

7.1  Integration by Substitution, 193

7.2  Integration by Parts, 196

7.3  Evaluation of Definite Integrals, 199

7.4  Partial Fractions, 201

7.5  Approximating Sums, 205

7.6  Improper Integrals, 210

8  Functions of Several Variables 214

8.1  Functions of Several Variables, 215

8.2  Partial Derivatives, 217

8.3  Second-Order Partial Derivatives – Maxima and Minima, 223

8.4  Method of Least Squares, 228

8.5  Lagrange Multipliers, 231

8.6  Double Integrals, 235

9  Series and Summations 240

9.1  Power Series, 241

9.2  Maclaurin and Taylor Polynomials, 245

9.3  Taylor and Maclaurin Series, 250

9.4  Convergence and Divergence of Series, 256

9.5  Arithmetic and Geometric Sums, 263

Giancarlo Black

CONTENTS vii

10  Applications to Probability 269

10.1  Discrete and Continuous Random Variables, 270

10.2  Mean and Variance; Expected Value, 278

10.3  Normal Probability Density Function, 283

Answers to Odd Numbered Exercises 295

Index 349

Giancarlo Black

x PREFACE

Some students may skip Chapter 1, Linear Equations and Functions while others find it a useful review.

In Chapter 2, The Derivative finite differences are introduced naturally in forming derivatives and is a topic in its own right. Usually absent from applied calculus texts, finite calculus emphasizes understanding calculus as the mathematics of change (not simply rote techniques) and is an aid to popular spreadsheet modeling.

In Chapter 3, Using the Derivative students’ newly acquired knowledge of a derivative appears in everyday contexts including marginal economic analysis. Early on, it shows students an application of calculus.

Chapter 4, Exponential and Logarithmic Functions delineates their principles. This chapter appears earlier than in other texts for two reasons. One, it allows for more complex derivatives to be discussed in Chapter 5 to include exponentials and logarithms. Two, it allows for the discussion on finite differences in Chapter 5 to be a lead in for integration in Chapter 6, Integral Calculus.

Chapter 5, Techniques of Differentiation treats the derivatives of products and quo- tients, maxima, and minima. Finite differences (or finite calculus) appear again in a brief section of this chapter. The anti-differences introduced in Chapter 5 anticipate the basics of integration in Chapter 6.

Chapters 7 and 8, Integration Techniques and Functions of Several Variables respec- tively, typify most texts. An exception is our inclusion of partial fractions.

Chapter 9, Series and Summations includes important insight to applications. Chapter 10, Applications to Probability links calculus and probability.

The table below suggests sample topic choices for a basic calculus course:

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SUPPLEMENTS

A modestly priced Student Solutions Manual contains complete solutions.

1  Linear Equations and Functions

1.1  Solving Linear Equations 2

Example 1.1.1 Solving a Linear Equation 3

Example 1.1.2 Solving for y 4

Example 1.1.3 Simple Interest 4

Example 1.1.4 Investment 4

Example 1.1.5 Gasoline Prices 5

1.2  Linear Equations and Their Graphs 7

Example 1.2.1 Ordered Pair Solutions 8

Example 1.2.2 Intercepts and Graph of a Line 8

Example 1.2.3 Intercepts of a Demand Function 9

Example 1.2.4 Slope-Intercept Form 11

Example 1.2.5 Point-Slope Form 12

Example 1.2.6 Temperature Conversion 12

Example 1.2.7 Salvage Value 13

Example 1.2.8 Parallel or Perpendicular Lines 13

1.3  Factoring and the Quadratic Formula 16

Example 1.3.1 Finding the GCF 16

Example 1.3.2 Sum and Difference of Squares 17

Example 1.3.3 Sum and Difference of Cubes 18

Example 1.3.4 Factoring Trinomials 19

Example 1.3.5 Factoring Trinomials (revisited) 20

Example 1.3.6 Factoring by Grouping 21

The Quadratic Formula 21

Example 1.3.7 The Quadratic Formula 22

Example 1.3.8 Zeros of Quadratics 23

Example 1.3.9 A Quadratic Supply Function 24

1.4  Functions and their Graphs 25

Example 1.4.1 Interval Notation 26

Functions 27

Example 1.4.2 Finding Domains 27

Example 1.4.3 Function Values 27

Example 1.4.4 Function Notation and Piecewise Intervals 28

Example 1.4.5 Determining Functions 29

Graphs of Functions 29

Example 1.4.6 Graph of a Parabola 29

Example 1.4.7 A Piecewise (segmented) Graph 30

The Algebra of Functions 31

Example 1.4.8 Algebra of Functions 32

Example 1.4.9 Composite Functions 32

1.5  Laws of Exponents 34

Fundamentals of Calculus, First Edition. Carla C. Morris and Robert M. Stark.

© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc. Companion Website: http://www.wiley.com/go/morris/calculus

1

Giancarlo Black

Example 1.5.1 Using Exponent Laws 35

Example 1.5.2 Using Exponent Laws (revisited) 35

Example 1.5.3 Using Fractional Exponents 36

1.6  Slopes and Relative Change 37

Example 1.6.1 Another Difference Quotient 38

Example 1.6.2 Difference Quotients 38

Historical Notes — René Descartes 41

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1.1  SOLVING LINEAR EQUATIONS

Mathematical descriptions, often as algebraic expressions, usually consist of alphanumeric characters and special symbols.

↑ The name algebra has fascinating origins in early Arabic language (Historical Notes).

For example, physicists describe the distance, s, that an object falls under gravity in a time, t, by s = (1∕2)gt². Here, the letters s and t represent variables since their values may change while, g, the acceleration of gravity, is considered as constant. While any letters can represent variables, typically, later letters of the alphabet are customary. Use of x and y is generic. Sometimes, it is convenient to use a letter that is descriptive of a variable, as t for time.

Earlier letters of the alphabet are customary for fixed values or constants. However, exceptions are common. The equal sign, a special symbol, is used to form an equation. An equation equates algebraic expressions. Numerical values for variables that preserve equality are called solutions to the equations.

For example, 5x + 1 = 11 is an equation in a single variable, x. It is a condi- tional equation since it is only true when x = 2. Equations that hold for all values of the variable are called identities. For example, (x + 1)2 = x² + 2x + 1 is an iden- tity. By solving an equation, values of the variables that satisfy the equation are determined.

An equation in which only the first powers of variables appear is a linear equation. Every linear equation in a single variable can be solved using some or all of these properties:

Substitution – Substituting one expression for an equivalent one does not alter the orig- inal equation. For example, 2(x − 3) + 3(x − 1) = 21 is equivalent to

2x − 6 + 3x − 3 = 21 or 5x − 9 = 21.

Addition  –  Adding (or subtracting) a quantity to each side of an equation leaves it unchanged. For example, 5x − 9 = 21 is equivalent to 5x − 9 + 9 = 21 + 9 or 5x = 30.

Multiplication – Multiplying (or dividing) each side of an equation by a non-zero quan- tity leaves it unchanged. For example, 5x = 30 is equivalent to (5x)(1∕5) = (30)(1∕5) or x = 6.

↓ Here are examples of linear equations: 5x − 3 = 11, y = 3x + 5, 3x + 5y + 6z = 4. They are linear in one, two, or three variables, respectively. It is the unit exponent on the vari- ables that identifies them as linear.

↓ By solving an equation we generally intend the numerical values of its variables.

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Solve (3x∕2) − 8 = (2∕3)(x − 2).

Solution:

To remove fractions, multiply both sides of the equation by 6, the least common denominator of 2 and 3. (Step 1 above)

The revised equation becomes

9x − 48 = 4(x − 2).

Next, remove grouping symbols (Step 2). That leaves

9x − 48 = 4x − 8.

Now, subtract 4x and add 48 to both sides (Step 3). Now,

9x − 4x − 48 + 48 = 4x − 4x − 8 + 48 or 5x = 40.

Finally, divide both sides by the coefficient 5 (Step 4). One obtains x = 8. The result, x = 8, is checked by substitution in the original equation (Step 5):

3(8)∕2 − 8 = (2∕3)(8 − 2)

4 = 4 checks!

The solution x = 8 is correct!

Giancarlo Black

Equations often have more than one variable. To solve linear equations in several vari- ables simply bring a variable of interest to one side. Proceed as for a single variable regard- ing the other variables as constants for the moment.

↓ If y is the variable of interest in 3x + 5y + 6z = 2 , it can be written as y = (2 − 3x − 6z)∕5 regarding x and z as constants for now.

Solve for y: 5x + 4y = 20.

Solution:

Move terms with y to one side of the equation and any remaining terms to the opposite side. Here, 4y = 20 − 5x. Next, divide both sides by 4 to yield y = 5 − (5∕4)x.

Interest equals Principal times Rate times Time expresses the well-known Simple Interest Formula, I = PRT. Solve for the time, T.

Solution:

Grouping, I = (PR)T so PR becomes a coefficient of T. Dividing by PR gives T = I∕PR.

Mathematics is often called the language of science or the universal language. To study phenomena or situations of interest, mathematical expressions and equations are used to create mathematical models. Extracting information from the mathematical model provides solutions and insights. Mathematical modeling ideas appear throughout the text. These suggestions may aid your modeling skills.

Ms. Brown invests $5000 at 6% annual interest. Model her resulting capital for one year.

Solution:

Here the principal (original investment) is $5000. The interest rate is 0.06 (expressed as a decimal) and the time is 1 year.

Using the simple interest formula, I = PRT, Ms. Brown’s interest is I = ($5000)(0.06)(1) = $300.

After one year a model for her capital is P + PRT = $5000 + $300 = $5300.

Recently East Coast regular grade gasoline was priced about $3.50 per gallon. West Coast prices were about $0.50/gallon higher.

a)  What was the average regular grade gasoline price on the East Coast for 10 gallons?

b)  What was the average regular grade gasoline price on the West Coast for 15 gallons?

Solution:

On average, a model for the East Coast cost of ten gallons was (10)(3.50) = $35.00.

On average, a model for the West Coast of fifteen gallons was (15)($4.00) = $60.00.

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EXERCISES 1.1

In Exercises 1 – 6 identify equations as an identity, a conditional equation, or a contradiction.

Giancarlo Black

1. 3x + 1 = 4x − 5

2. 2(x + 1) = x + x + 2

3. 5(x + 1) + 2(x − 1) = 7x + 6

4. 4x + 3(x + 2) = x + 6

5. 4(x + 3) = 2(2x + 5)

6. 3x + 7 = 2x + 4

In Exercises 7 – 27 solve the equations.

7. 5x − 3 = 17

8. 3x + 2 = 2x + 7

9. 2x = 4x − 10

10.  x∕3 = 10

11. 4x − 5 = 6x − 7

12.  5x + (1∕3) = 7

13. 0.6x = 30

14.  (3x∕5) − 1 = 2 − (1∕5)(x − 5)

15.  2∕3 = (4∕5)x − (1∕3)

16. 4(x − 3) = 2(x − 1)

17. 5(x − 4) = 2x + 3(x − 7)

18. 3x + 5(x − 2) = 2(x + 7)

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19. 3s − 4 = 2s + 6

20. 5(z − 3) + 3(z + 1) = 12

21. 7t + 2 = 4t + 11

22.  (1∕3)x + (1∕2)x = 5

23. 4(x + 1) + 2(x − 3) = 7(x − 1)

24.  1∕3 = (3∕5)x − (1∕2)

25. x + 8  = 2

2x − 5

26. 3x − 1 = x − 3

7

27. 8 − {4[x − (3x − 4) − x] + 4}

= 3(x + 2)

In Exercises 28 – 35 solve for the indicated variable.

Solve: 5x − 2y + 18 = 0 for y.

Solve: 6x − 3y = 9 for x.

Solve: y = mx + b for x.

Solve: 3x + 5y = 15 for y.

Solve: A = P + PRT for P.

Solve: V = LWH for W.

Solve: C = 2𝜋r for r.

Solve: Z = x − 𝜇 for x.

𝜎

Exercises 36 – 45 feature mathematical models.

The sum of three consecutive positive integers is 81. Determine the largest integer.

Sally purchased a used car for $1300 and paid $300 down. If she plans to pay the balance in five equal monthly installments, what is the monthly payment?

A suit, marked down 20%, sold for $120. What was the original price?

If the marginal propensity to consume is m = 0.75 and consumption, C, is $11 when disposable income is $2, develop the consumption function.

A new addition to a fire station costs $100,000. The annual maintenance cost increases by $2500 with each fire engine housed. If $115,000 has been allocated for the addition and maintenance next year, how many additional fire engines can be housed?

Lightning is seen before thunder is heard as the speed of light is much greater than the speed of sound. The flash’s distance from an observer can be calculated from the time between the flash and the sound of thunder.

The distance, d (in miles), from the storm can be modeled as d = 4.5t where time,

t, is in seconds.

If thunder is heard two seconds after lightning is seen, how far is the storm?

If a storm is 18 miles distant, how long before thunder is heard?

A worker has forty hours to produce two types of items, A and B. Each unit of A takes three hours to produce and each item of B takes two hours. The worker made eight items of B and with the remaining time produced items of A. How many of item A were produced?

An employee’s Social Security Payroll Tax was 6.2% for the first $87,000 of earn- ings and was matched by the employer. Develop a linear model for an employee’s portion of the Social Security Tax.

An employee works 37.5 hours at a $10 hourly wage. If Federal tax deductions are 6.2% for Social Security, 1.45% for Medicare Part A, and 15% for Federal taxes, what is the take-home pay?

The body surface area (BSA) and weight (Wt) in infants and children weighing between 3 kg and 30 kg has been modeled by the linear relationship

BSA = 1321 + 0.3433Wt (where BSA is in square centimeters and weight in grams)

Determine the BSA for a child weighing 20 kg.

A child’s BSA is 10,320 cm². Estimate its weight in kilograms.

Current, J.D.,A Linear Equation for Estimating the Body Surface Area in Infants and Children., The Internet Journal of Anesthesiology 1998:Vol2N2.

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1.2  LINEAR EQUATIONS AND THEIR GRAPHS

Mathematical models express features of interest. In the managerial, social, and natural sciences and engineering, linear equations often relate quantities of interest. Therefore, a thorough understanding of linear equations is important.

The standard form of a linear equation is ax + by = c where a, b, and c are real valued constants. It is characterized by the first power of the exponents.

Giancarlo Black

Do the points (3, 5) and (1, 7) satisfy the linear equation 2x + y = 9?

Solution:

A point satisfies an equation if equality is preserved. The point (3, 5) yields: 2(3) + 5 ≠ 9. Therefore, the ordered pair (3, 5) is not a solution to the equation 2x + y = 9.

For (1, 7), the substitution yields 2(1) + 7 = 9. Therefore, (1, 7) is a point on the line 2x + y = 9.

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↓ An ordered (coordinate) pair, (x, y) describes a (graphical) point in the x, y plane. By convention, the x value always appears first.

A graph is a pictorial representation of a function. It consists of points that satisfy the function. Cartesian Coordinates are used to represent the relative positions of points in a plane or in space. In a plane, a point P is specified by the coordinates or ordered pair (x, y) representing its distance from two perpendicular intersecting straight lines, called the x-axis and the y-axis, respectively (figure).

x-axis

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Cartesian coordinates are so named to honor the mathematician René Descartes

(Historical Notes).

The graph of a linear equation is a line. It is uniquely determined by two distinct points. Any additional points can be checked as the points must be collinear (i.e., lie on the same line). The coordinate axes may be differently scaled. To determine the x-intercept of a line (its intersection with the x-axis), set y = 0 and solve for x. Likewise, for the y-intercept set x = 0 and solve for y.

↓ For the linear equation 2x + y = 9, set y = 0 for the x-intercept (x = 4.5) and x = 0 for the y-intercept (y = 9) . As noted, intercepts are intersections of the line with the respective axes.

Locate the x and y-intercepts of the line 2x + 3y = 6 and graph its equation.

Solution:

When x = 0, 3y = 6 so the y-intercept is y = 2. When y = 0, 2x = 6 so the x-intercept is x = 3. The two intercepts, (3, 0) and (0, 2), as two points, uniquely determine the line. As

a check, arbitrarily choose a value for x, say x = −1. Then, 2(−1) + 3y = 6 or 3y = 8 so y = 8∕3. Therefore (−1,   8∕3) is another point on the line. Check that these three points lie on the same line.

y

x

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↑ Besides the algebraic representation of linear equations used here, many applications use elegant matrix representations. So 2x + 3y = 6 (algebraic) can also be expressed as

2  3 x 6 in matrix format.

y

Price and quantity often arise in economic models. For instance, the demand D(p) for an item is related to its unit price, p, by the equation D(p) = 240 − 3p. In graphs of economic models price appears on the (vertical) y-axis and quantity on the (horizontal) x-axis.

Given D(p) = 240 − 3p.

Determine demand when price is 10.

What is demand when the goods are free?

At what price will consumers no longer purchase the goods?

For what values of price is D(p) meaningful?

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Solution:

Substituting p = 10 yields a demand of 210 units.

When the goods are free p = 0 and D(p) = 240. Note that this is an intercept.

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Giancarlo Black

Here, D(p) = 0 and p = 80 is the price that is too high and results in no demand for the goods. Note, this is an intercept.

Since price is at least zero, and the same for demand, therefore, 0 ≤ p ≤ 80 and 0 ≤ D(p) ≤ 240.

When either a or b in ax + by = c is zero, the standard equation reduces to a single value for the remaining variable. If y = 0, ax = c so x = c∕a; a vertical line. If x = 0, by = c so y = c∕b; a horizontal line.

↓ Remember: horizontal lines have zero slopes while vertical lines have infinite slopes.

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It is often useful to express equations of lines in different (and equivalent) algebraic formats. The slope, m, of a line can be described in several ways; the rise divided by the run, or "the change in y, denoted by Δy, divided by the change in x, Δx. From left to right, positive sloped lines rise (∕) while negative sloped lines fall" (∖ ).

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The equation of a line can be expressed in different, but equivalent, ways. The slope-intercept form of a line is y = mx + b, where m is the slope and b its y-intercept. A horizontal line has zero slope. A vertical line has an infinite (undefined) slope, as there is no change in x for any value of y.

y y

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x x

Positive slope Negative slope

y y

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x x

Undefined slope Zero slope

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A linear equation in standard form, ax + by = c, is written in slope-intercept form by solving for y.

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Write 2x + 3y = 6 in slope-intercept form and identify the slope and y-intercept.

Solution:

Solving, y = (−2∕3)x + 2. By inspection, the slope is −2 ∕3 (line falls) and the y-intercept is (0, 2); in agreement with the previous Example.

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Linear equations are also written in point-slope form: y − y1 = m(x − x1). Here, (x1, y1)

is a given point on the line and m is the slope.

Giancarlo Black

Determine the equation of a line in point-slope form passing through (2, 4) and (5, 13).

Solution:

First, the slope m =

5 − 2 = 3 = 3. Now, using (2, 4) in the point-slope form we have

y − 4 = 3(x − 2). [Using the point (5, 13) yields the equivalent y − 13 = 3(x − 5)]. For the slope-intercept form, solving for y yields y = 3x − 2. The standard form is 3x − y = 2.

↓ Various representations of linear equations are equivalent but can seem confusing. Usage depends on the manner in which information is provided. If the slope and y-intercept are known, use the slope-intercept form. If coordinates of a point through which the line

passes is known, use the point-slope form. Using a bit of algebraic manipulation, you can simply remember to use y = mx + b.

Water boils at 212 ∘F(100 ∘C) and freezes at 32 ∘F (0 ∘C). What linear equation relates Celsius and Fahrenheit temperatures?

Solution:

Denote Celsius temperatures by C and Fahrenheit temperatures by F. The ordered

pairs are (100, 212) and (0, 32). Using the slope-intercept form of a line, m = 32 − 212

−180 9

0 − 100

=  −100  =  5 . The second ordered pair, (0, 32) is its y-intercept. Therefore, F = 9∕5C + 32 is the widely used relation to enable conversion of Celsius to Fahrenheit temperatures. An

Exercise seeks the Fahrenheit to Celsius relation.

Equipment value at time t is V(t) = −10,000t + 80,000 and its useful life expectancy is 6 years. Develop a model for the original value, salvage value, and annual depreciation.

Solution:

The original value, at t = 0, is $80,000. The salvage value, at t = 6, the end of useful life, is

$20,000 = (−10,000(6) + 80,000). The slope, which is the annual depreciation, is $10,000.

Lines having the same slope are parallel. Two lines are perpendicular if their slopes are negative reciprocals.

Are these pairs of lines parallel, perpendicular, or neither?

a)  3x − y = 1 and y = (1 ∕3)x − 4

b)  y = 2x + 3  and y = (−1 ∕2)x + 5

c) y = 7x + 1 and y = 7x + 3

Giancarlo Black

Solution:

The two slopes are required. By inspection, the slope of the second line is 1/3. The line 3x − y = 1 in slope-intercept form is y = 3x − 1 so its slope is 3. Since the slopes are neither equal nor negative reciprocals the lines intersect.

The slopes are 2 and (−1 ∕2). Since these are negative reciprocals the two lines are perpendicular.

The lines have the same slope (and different intercepts) so they are parallel.

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EXERCISES 1.2

Determine the x and y-intercepts for the following:

a) 5x − 3y = 15

b) y = 4x − 5

c) 2x + 3y = 24

d) 9x − y = 18

x = 4

y = −2

Determine slopes and y-intercepts for the following:

a)  y = (2∕3)x + 8

b) 3x + 4y = 12

c) 2x − 3y − 6 = 0

d) 6y = 4x + 3

e) 5x = 2y + 10

f) y = 7

Determine the slopes of lines defined by the points:

a) (3, 6) and (−1, 4)

b) (1, 6) and (2, 11)

c) (6, 3) and (12, 7)

Determine the equation for the line

with slope 4 that passes through (1, 7).

passing through (2, 7) and (5, 13).

d) (2, 3) and (2, 7)

e) (2, 6) and (5, 6)

f) (5/3, 2/3) and (10/3, 1)

with undefined slope passing through (2, 5/2).

with x-intercept 6 and y-intercept −2.

with slope 5 and passing through (0, −7).

passing through (4, 9) and (7, 18).

Giancarlo Black

Plot graphs of:

y = 2x − 5

x = 4

Plot graphs of: a) 2x − 3y = 6

b) y = −3

c)  y = (−2∕3)x + 2

d) y = 4x − 7

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c) 3x + 5y = 15

d) 2x + 7y = 14

Are the pairs of lines parallel, perpendicular, or neither? a)  y = (5∕3)x + 2 and 5x − 3y = 10

b)  6x + 2y = 4 and y = (1∕3)x + 1

c) 2x − 3y = 6 and 4x − 6y = 15

y = 5x − 4 and 3x − y = 4

y = 5 and x = 3

Determine equation for the line

through (2, 3) and parallel to y = 5x − 1.

through (1, 4) and perpendicular to 2x + 3y = 6.

through (5, 7) and perpendicular to x = 6.

through (4, 1) and parallel to x = 1.

through (2, 3) and parallel to 2y = 5x + 4.

When does a linear equation lack an x-intercept? Have more than one x-intercept? Lack a y-intercept? Have more than one y-intercept?

Model the conversion of Fahrenheit to Celsius temperatures. Hint: Use the freezing and boiling temperatures for water.

A new machine cost $75,000 and has a salvage value of $21,000 after nine years. Model its straight line depreciation.

A new car cost $28,000 and has a trade-in value of $3000 after 5 years. Use straight line depreciation.

The distance a car travels depends on the quantity of gasoline available. A car requires seven gallons to travel 245 miles and 12 gallons to travel 420 miles. What linear rela- tionship expresses distance (miles) as a function of gasoline usage (gallons)?

Office equipment is purchased for $50,000 and after ten years has a salvage value of

$5000. Model its depreciation with a linear equation.

Monthly rent on a building is $1100 (a fixed cost). Each unit of the firm’s product costs

$5 (a variable cost). Form a linear model for the total monthly cost to produce x items.

A skateboard sells for $24. Determine the revenue function for selling x skateboards.

A car rental company charges $50 per day for a medium sized car and 28 cents per mile driven.

Model the cost for renting a medium sized car for a single day.

How many miles can be driven that day for $92?

At the ocean’s surface water and air pressures are equal (15 lb ∕in²). Below the surface water pressure increases by 4.43 lb ∕in² for every 10 feet of depth.

Express water pressure as a function of ocean depth.

At what depth is the water pressure 80 lb ∕in²?

A man’s suit sells for $84. The cost to the store is $70. A woman’s dress sells for $48 and costs the store $40. If the store’s markup policy is linear, and is reflected in the price of these items, model the relationship between retail price, R, and store cost, C.

Demand for a product is linearly related to its price. If the product is priced at $1.50 each, 40 items can be sold. Priced at $6, only 22 items are sold. Let x be the price and y the number of items sold. Model a linear relationship between price and items sold.

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1.3  FACTORING AND THE QUADRATIC FORMULA

Factoring is a most useful mathematical skill. A first step in factoring expressions is to seek the greatest common factor (GCF) among terms of an algebraic expression. For example, for 5x + 10, the GCF is 5. So, to factor this expression one writes 5(x + 2).

Often one is unsure whether an expression is factorable. One aid is to note that a linear (first