**Single Variable Calculus**

view a copy of this license, visit http://creativecommons.org/licenses/by-nc-sa/3.0/or send a letter to Creative Commons, 543 Howard Street, 5th Floor, San Francisco, California, 94105, USA. If you distribute this work or a derivative, include the history of the document.

This text was initially written by David Guichard. The single variable material in chapters 1–9 is a mod-ification and expansion of notes written by Neal Koblitz at the University of Washington, who generously gave permission to use, modify, and distribute his work. New material has been added, and old material has been modified, so some portions now bear little resemblance to the original.

The book includes some exercises and examples from*Elementary Calculus: An Approach Using *
*Infinitesi-mals*, by H. Jerome Keisler, available athttp://www.math.wisc.edu/~keisler/calc.htmlunder a Creative
Commons license. In addition, the chapter on diﬀerential equations (in the multivariable version) and the
section on numerical integration are largely derived from the corresponding portions of Keisler’s book.
Albert Schueller, Barry Balof, and Mike Wills have contributed additional material.

This copy of the text was compiled from source at 8:19 on 8/26/2019.

*For Kathleen,*

*without whose encouragement*
*this book would not have*

**Contents**

**1**

**Analytic Geometry**

**13**

**1.1** Lines . . . 14

**1.2** Distance Between Two Points; Circles . . . 19

**1.3** Functions . . . 20

**1.4** Shifts and Dilations . . . 25

**2**

**Instantaneous Rate of Change: The Derivative**

**29**

**2.1**The slope of a function . . . 29

**2.2** An example . . . 34

**2.3** Limits . . . 36

**2.4** The Derivative Function . . . 46

**2.5** Adjectives For Functions . . . 51

**3**

**Rules for Finding Derivatives**

**55**

**3.1** The Power Rule . . . 55

**3.2** Linearity of the Derivative . . . 58

**3.3** The Product Rule . . . 60

**3.4** The Quotient Rule . . . 62

**3.5** The Chain Rule . . . 65

**4**

**Transcendental Functions**

**71**

**4.1**Trigonometric Functions . . . 71

**4.2** The Derivative of sin*x* . . . 74

**4.3** A hard limit . . . 75

**4.4** The Derivative of sin*x*, continued . . . 78

**4.5** Derivatives of the Trigonometric Functions . . . 79

**4.6** Exponential and Logarithmic functions . . . 80

**4.7** Derivatives of the exponential and logarithmic functions . . . 82

**4.8** Implicit Diﬀerentiation . . . 87

**4.9** Inverse Trigonometric Functions . . . 92

**4.10** Limits revisited . . . 95

**4.11** Hyperbolic Functions . . . 100

**5**

**Curve Sketching**

**105**

**5.1**Maxima and Minima . . . 105

**5.2** The first derivative test . . . 109

**5.3** The second derivative test . . . 110

**5.4** Concavity and inflection points . . . 111

**Contents** **7**

**6**

**Applications of the Derivative**

**117**

**6.1** Optimization . . . 117

**6.2** Related Rates . . . 129

**6.3** Newton’s Method . . . 137

**6.4** Linear Approximations . . . 141

**6.5** The Mean Value Theorem . . . 143

**7**

**Integration**

**147**

**7.1**Two examples . . . 147

**7.2** The Fundamental Theorem of Calculus . . . 151

**7.3** Some Properties of Integrals . . . 158

**8**

**Techniques of Integration**

**163**

**8.1**Substitution . . . 164

**8.2** Powers of sine and cosine . . . 169

**8.3** Trigonometric Substitutions . . . 171

**8.4** Integration by Parts . . . 174

**8.5** Rational Functions . . . 178

**8.6** Numerical Integration . . . 182

**9**

**Applications of Integration**

**189**

**9.1** Area between curves . . . 189

**9.2** Distance, Velocity, Acceleration . . . 194

**9.3** Volume . . . 197

**9.4** Average value of a function . . . 204

**9.5** Work . . . 207

**9.6** Center of Mass . . . 211

**9.7** Kinetic energy; improper integrals . . . 217

**9.8** Probability . . . 221

**9.9** Arc Length . . . 230

**9.10** Surface Area . . . 232

**10**

**Polar Coordinates, Parametric Equations**

**239**

**10.1**Polar Coordinates . . . 239

**10.2** Slopes in polar coordinates . . . 243

**10.3** Areas in polar coordinates . . . 245

**10.4** Parametric Equations . . . 248

**Contents** **9**

**11**

**Sequences and Series**

**255**

**11.1** Sequences . . . 256

**11.2** Series . . . 262

**11.3** The Integral Test . . . 266

**11.4** Alternating Series . . . 271

**11.5** Comparison Tests . . . 273

**11.6** Absolute Convergence . . . 276

**11.7** The Ratio and Root Tests . . . 277

**11.8** Power Series . . . 280

**11.9** Calculus with Power Series . . . 283

**11.10** Taylor Series . . . 285

**11.11** Taylor’s Theorem . . . 288

**11.12** Additional exercises . . . 294

**A**

**Selected Answers**

**297**

**B**

**Useful Formulas**

**313**

**Introduction**

The emphasis in this course is on problems—doing calculations and story problems. To master problem solving one needs a tremendous amount of practice doing problems. The more problems you do the better you will be at doing them, as patterns will start to emerge in both the problems and in successful approaches to them. You will learn fastest and best if you devote some time to doing problems every day.

Typically the most diﬃcult problems are story problems, since they require some eﬀort before you can begin calculating. Here are some pointers for doing story problems:

**1.** Carefully read each problem twice before writing anything.

**2.** Assign letters to quantities that are described only in words; draw a diagram if
appropriate.

**3.** Decide which letters are constants and which are variables. A letter stands for a
constant if its value remains the same throughout the problem.

**4.** Using mathematical notation, write down what you know and then write down
what you want to find.

**5.** Decide what category of problem it is (this might be obvious if the problem comes
at the end of a particular chapter, but will not necessarily be so obvious if it comes
on an exam covering several chapters).

**6.** Double check each step as you go along; don’t wait until the end to check your
work.

**7.** Use common sense; if an answer is out of the range of practical possibilities, then
check your work to see where you went wrong.

**Suggestions for Using This Text**

**1.** Read the example problems carefully, filling in any steps that are left out (ask
someone for help if you can’t follow the solution to a worked example).

**2.** Later use the worked examples to study by covering the solutions, and seeing if
you can solve the problems on your own.

**3.** Most exercises have answers in AppendixA; the availability of an answer is marked
by “*⇒*” at the end of the exercise. In the pdf version of the full text, clicking
on the arrow will take you to the answer. The answers should be used only as
a final check on your work, not as a crutch. Keep in mind that sometimes an
answer could be expressed in various ways that are algebraically equivalent, so
don’t assume that your answer is wrong just because it doesn’t have exactly the
same form as the answer in the back.

**1**

**Analytic Geometry**

Much of the mathematics in this chapter will be review for you. However, the examples will be oriented toward applications and so will take some thought.

In the (*x, y*) coordinate system we normally write the*x*-axis horizontally, with positive
numbers to the right of the origin, and the *y*-axis vertically, with positive numbers above
the origin. That is, unless stated otherwise, we take “rightward” to be the positive *x*
-direction and “upward” to be the positive*y*-direction. In a purely mathematical situation,
we normally choose the same scale for the*x*- and*y*-axes. For example, the line joining the
origin to the point (*a, a*) makes an angle of 45*◦* with the *x*-axis (and also with the *y*-axis).
In applications, often letters other than*x*and*y*are used, and often diﬀerent scales are
chosen in the horizontal and vertical directions. For example, suppose you drop something
from a window, and you want to study how its height above the ground changes from
second to second. It is natural to let the letter *t* denote the time (the number of seconds
since the object was released) and to let the letter *h* denote the height. For each *t* (say,
at one-second intervals) you have a corresponding height *h*. This information can be
tabulated, and then plotted on the (*t, h*) coordinate plane, as shown in figure 1.0.1.

We use the word “quadrant” for each of the four regions into which the plane is divided by the axes: the first quadrant is where points have both coordinates positive, or the “northeast” portion of the plot, and the second, third, and fourth quadrants are counted oﬀ counterclockwise, so the second quadrant is the northwest, the third is the southwest, and the fourth is the southeast.

Suppose we have two points *A* and*B* in the (*x, y*)-plane. We often want to know the
change in *x*-coordinate (also called the “horizontal distance”) in going from *A* to *B*. This

seconds 0 1 2 3 4 meters 80 75.1 60.4 35.9 1.6

20 40 60 80

0 1 2 3 4

*t*
*h*

..._{...}
..._{...}

..._{...}
...

..._{...}
..._{...}

..._{...}
..._{...}

..._{...}
..._{...}

..._{...}
...

*•* _{•}

*•*

*•*

*•*

**Figure 1.0.1** A data plot, height versus time.

is often written ∆*x*, where the meaning of ∆ (a capital delta in the Greek alphabet) is
“change in”. (Thus, ∆*x* can be read as “change in*x*” although it usually is read as “delta

*x*”. The point is that ∆*x*denotes a single number, and should not be interpreted as “delta
times *x*”.) For example, if *A* = (2*,*1) and *B* = (3*,*3), ∆*x* = 3 *−*2 = 1. Similarly, the
“change in *y*” is written ∆*y*. In our example, ∆*y* = 3*−*1 = 2, the diﬀerence between the

*y*-coordinates of the two points. It is the vertical distance you have to move in going from

*A* to *B*. The general formulas for the change in *x* and the change in *y* between a point
(*x*1*, y*1) and a point (*x*2*, y*2) are:

∆*x*=*x*2*−x*1*,* ∆*y*=*y*2*−y*1*.*

Note that either or both of these might be negative.

### 1.1 Lines

If we have two points *A*(*x*1*, y*1) and *B*(*x*2*, y*2), then we can draw one and only one line

through both points. By the *slope*of this line we mean the ratio of ∆*y* to ∆*x*. The slope
is often denoted *m*: *m*= ∆*y/*∆*x*= (*y*2*−y*1)*/*(*x*2*−x*1). For example, the line joining the

points (1*,−*2) and (3*,*5) has slope (5 + 2)*/*(3*−*1) = 7*/*2.

**1.1** **Lines** **15**

information (15%, 28%, or 33%) using mathematical terminology, and graph the tax on
the*y*-axis against the taxable income on the *x*-axis.

The percentages, when converted to decimal values 0.15, 0.28, and 0.33, are the *slopes*

of the straight lines which form the graph of the tax for the corresponding tax brackets.
The tax graph is what’s called a *polygonal line*, i.e., it’s made up of several straight line
segments of diﬀerent slopes. The first line starts at the point (0,0) and heads upward
with slope 0.15 (i.e., it goes upward 15 for every increase of 100 in the *x*-direction), until
it reaches the point above *x* = 26050. Then the graph “bends upward,” i.e., the slope
changes to 0.28. As the horizontal coordinate goes from*x* = 26050 to*x* = 67200, the line
goes upward 28 for each 100 in the*x*-direction. At *x*= 67200 the line turns upward again
and continues with slope 0.33. See figure1.1.1.

10000 20000 30000

50000 100000 134930

...

... ...

... ...

... ...

... ...

... ...

... ...

... ...

... ...

... ...

... ...

*•*

*•*

*•*

**Figure 1.1.1** Tax vs. income.

The most familiar form of the equation of a straight line is: *y*=*mx*+*b*. Here*m*is the
slope of the line: if you increase*x* by 1, the equation tells you that you have to increase *y*

by *m*. If you increase *x* by ∆*x*, then *y* increases by ∆*y* = *m*∆*x*. The number *b* is called
the **y-intercept**, because it is where the line crosses the *y*-axis. If you know two points
on a line, the formula*m*= (*y*2*−y*1)*/*(*x*2*−x*1) gives you the slope. Once you know a point

and the slope, then the*y*-intercept can be found by substituting the coordinates of either
point in the equation: *y*1 = *mx*1 +*b*, i.e., *b* = *y*1*−mx*1. Alternatively, one can use the

“point-slope” form of the equation of a straight line: start with (*y−y*1)*/*(*x−x*1) =*m*and

then multiply to get (*y−y*1) =*m*(*x−x*1), the point-slope form. Of course, this may be

further manipulated to get*y* =*mx−mx*1+*y*1, which is essentially the “*mx*+*b*” form.

It is possible to find the equation of a line between two points directly from the relation
(*y−y*1)*/*(*x−x*1) = (*y*2*−y*1)*/*(*x*2*−x*1), which says “the slope measured between the point

and any other point (*x, y*) on the line.” For example, if we want to find the equation of
the line joining our earlier points *A*(2*,*1) and*B*(3*,*3), we can use this formula:

*y−*1

*x−*2 =
3*−*1

3*−*2 = 2*,* so that *y−*1 = 2(*x−*2)*,* i.e., *y*= 2*x−*3*.*

Of course, this is really just the point-slope formula, except that we are not computing *m*

in a separate step.

The slope*m* of a line in the form*y*=*mx*+*b* tells us the direction in which the line is
pointing. If *m* is positive, the line goes into the 1st quadrant as you go from left to right.
If *m* is large and positive, it has a steep incline, while if *m* is small and positive, then the
line has a small angle of inclination. If *m* is negative, the line goes into the 4th quadrant
as you go from left to right. If *m*is a large negative number (large in absolute value), then
the line points steeply downward; while if *m*is negative but near zero, then it points only
a little downward. These four possibilities are illustrated in figure 1.1.2.

... ...

... ...

... ...

... ...

... ...

... ..

*−*4

*−*2
0
2
4

*−*4 *−*2 0 2 4

... ...

*−*4

*−*2
0
2
4

*−*4 *−*2 0 2 4

..._{...}
..._{...}

..._{...}
..._{...}

..._{...}
..._{...}

..._{...}
..._{...}

...
*−*4

*−*2
0
2
4

*−*4 *−*2 0 2 4

..._{...}

*−*4

*−*2
0
2
4

*−*4 *−*2 0 2 4

**Figure 1.1.2** Lines with slopes 3, 0*.*1,*−*4, and *−*0*.*1.

If *m*= 0, then the line is horizontal: its equation is simply *y*=*b*.

There is one type of line that cannot be written in the form *y* = *mx* +*b*, namely,
vertical lines. A vertical line has an equation of the form *x* =*a*. Sometimes one says that
a vertical line has an “infinite” slope.

Sometimes it is useful to find the *x*-intercept of a line*y* =*mx*+*b*. This is the *x*-value
when *y*= 0. Setting *mx*+*b* equal to 0 and solving for *x* gives: *x* = *−b/m*. For example,
the line *y* = 2*x−*3 through the points*A*(2*,*1) and*B*(3*,*3) has *x*-intercept 3*/*2.

**EXAMPLE 1.1.2** Suppose that you are driving to Seattle at constant speed, and notice
that after you have been traveling for 1 hour (i.e., *t* = 1), you pass a sign saying it is 110
miles to Seattle, and after driving another half-hour you pass a sign saying it is 85 miles
to Seattle. Using the horizontal axis for the time *t* and the vertical axis for the distance *y*

from Seattle, graph and find the equation *y*=*mt*+*b*for your distance from Seattle. Find
the slope, *y*-intercept, and *t*-intercept, and describe the practical meaning of each.

**1.1** **Lines** **17**

110)*/*(1*.*5*−*1) =*−*50. The meaning of the slope is that you are traveling at 50 mph; *m* is
negative because you are traveling*toward* Seattle, i.e., your distance *y* is *decreasing*. The
word “velocity” is often used for*m*=*−*50, when we want to indicate direction, while the
word “speed” refers to the magnitude (absolute value) of velocity, which is 50 mph. To
find the equation of the line, we use the point-slope formula:

*y−*110

*t−*1 =*−*50*,* so that *y* =*−*50(*t−*1) + 110 =*−*50*t*+ 160*.*

The meaning of the*y*-intercept 160 is that when*t*= 0 (when you started the trip) you were
160 miles from Seattle. To find the*t*-intercept, set 0 =*−*50*t*+160, so that*t*= 160*/*50 = 3*.*2.
The meaning of the*t*-intercept is the duration of your trip, from the start until you arrive
in Seattle. After traveling 3 hours and 12 minutes, your distance *y* from Seattle will be 0.

**Exercises 1.1.**

**Exercises 1.1.**

**1.** Find the equation of the line through (1*,*1) and (*−*5*,−*3) in the form *y*=*mx*+*b*. *⇒*

**2.** Find the equation of the line through (*−*1*,*2) with slope*−*2 in the form *y*=*mx*+*b*. *⇒*

**3.** Find the equation of the line through (*−*1*,*1) and (5*,−*3) in the form *y*=*mx*+*b*. *⇒*

**4.** Change the equation *y* *−*2*x* = 2 to the form *y* = *mx*+*b*, graph the line, and find the

*y*-intercept and *x*-intercept. *⇒*

**5.** Change the equation*x*+*y*= 6 to the form*y* =*mx*+*b*, graph the line, and find the*y*-intercept
and *x*-intercept. *⇒*

**6.** Change the equation *x* = 2*y* *−*1 to the form *y* = *mx*+*b*, graph the line, and find the

*y*-intercept and *x*-intercept. *⇒*

**7.** Change the equation 3 = 2*y* to the form*y*=*mx*+*b*, graph the line, and find the*y*-intercept
and *x*-intercept. *⇒*

**8.** Change the equation 2*x*+ 3*y*+ 6 = 0 to the form *y* =*mx*+*b*, graph the line, and find the

*y*-intercept and *x*-intercept. *⇒*

**9.** Determine whether the lines 3*x*+ 6*y*= 7 and 2*x*+ 4*y*= 5 are parallel. *⇒*

**10.** Suppose a triangle in the *x, y*–plane has vertices (*−*1*,*0), (1*,*0) and (0*,*2). Find the equations
of the three lines that lie along the sides of the triangle in *y*=*mx*+*b*form. *⇒*

**11.** Suppose that you are driving to Seattle at constant speed. After you have been traveling
for an hour you pass a sign saying it is 130 miles to Seattle, and after driving another 20
minutes you pass a sign saying it is 105 miles to Seattle. Using the horizontal axis for the
time *t* and the vertical axis for the distance *y* from your starting point, graph and find the
equation *y* =*mt*+*b* for your distance from your starting point. How long does the trip to
Seattle take? *⇒*

**13.** A car rental firm has the following charges for a certain type of car: $25 per day with 100
free miles included, $0.15 per mile for more than 100 miles. Suppose you want to rent a
car for one day, and you know you’ll use it for more than 100 miles. What is the equation
relating the cost *y* to the number of miles *x* that you drive the car? *⇒*

**14.** A photocopy store advertises the following prices: 5/c per copy for the first 20 copies, 4/c per
copy for the 21st through 100th copy, and 3/c per copy after the 100th copy. Let *x* be the
number of copies, and let *y* be the total cost of photocopying. (a) Graph the cost as *x* goes
from 0 to 200 copies. (b) Find the equation in the form *y* =*mx*+*b* that tells you the cost
of making *x* copies when*x* is more than 100. *⇒*

**15.** In the Kingdom of Xyg the tax system works as follows. Someone who earns less than 100
gold coins per month pays no tax. Someone who earns between 100 and 1000 gold coins
pays tax equal to 10% of the amount over 100 gold coins that he or she earns. Someone
who earns over 1000 gold coins must hand over to the King all of the money earned over
1000 in addition to the tax on the first 1000. (a) Draw a graph of the tax paid *y* versus the
money earned *x*, and give formulas for *y* in terms of *x* in each of the regions 0 *≤x* *≤* 100,
100*≤x≤*1000, and *x≥*1000. (b) Suppose that the King of Xyg decides to use the second
of these line segments (for 100 *≤x≤* 1000) for *x* *≤*100 as well. Explain in practical terms
what the King is doing, and what the meaning is of the *y*-intercept. *⇒*

**16.** The tax for a single taxpayer is described in the figure 1.1.3. Use this information to graph
tax versus taxable income (i.e.,*x*is the amount on Form 1040, line 37, and*y*is the amount on
Form 1040, line 38). Find the slope and*y*-intercept of each line that makes up the polygonal
graph, up to *x*= 97620. *⇒*

**1990 Tax Rate Schedules**

**Schedule X**

—Use if your filing status is
**Single**

If the amount Enter on of the on Form 1040 But not Form 1040 amount line 37 is over: over: line 38 over:

$0 $19,450 **15%** **$0**

19,450 47,050 **$2,917.50+28%** **19,450**

47,050 97,620 **$10,645.50+33%** **47,050**

Use**Worksheet**

97,620 ... below to figure your tax

**Schedule Z**

—Use if your filing status is
**Head of household**

If the amount Enter on of the on Form 1040 But not Form 1040 amount line 37 is over: over: line 38 over:

$0 $26,050 **15%** **$0**

26,050 67,200 **$3,907.50+28%** **26,050**

67,200 134,930 **$15,429.50+33%** **67,200**

Use**Worksheet**

134,930 ... below to figure your tax

**Figure 1.1.3** Tax Schedule.

**17.** Market research tells you that if you set the price of an item at $1.50, you will be able to sell
5000 items; and for every 10 cents you lower the price below $1.50 you will be able to sell
another 1000 items. Let *x* be the number of items you can sell, and let*P* be the price of an
item. (a) Express*P* linearly in terms of*x*, in other words, express*P* in the form*P* =*mx*+*b*.
(b) Express *x* linearly in terms of *P*. *⇒*

**1.2** **Distance Between Two Points; Circles** **19**

will be linear. Let*x*be the exam score, and let*y*be the corresponding grade. Find a formula
of the form *y*=*mx*+*b* which applies to scores*x* between 40 and 90. *⇒*

### 1.2 Distance Between Two Points; Circles

Given two points (*x*1*, y*1) and (*x*2*, y*2), recall that their horizontal distance from one another

is ∆*x*=*x*2*−x*1and their vertical distance from one another is ∆*y* =*y*2*−y*1. (Actually, the

word “distance” normally denotes “positive distance”. ∆*x* and ∆*y* are *signed* distances,
but this is clear from context.) The actual (positive) distance from one point to the other
is the length of the hypotenuse of a right triangle with legs *|*∆*x|* and *|*∆*y|*, as shown in
figure1.2.1. The Pythagorean theorem then says that the distance between the two points
is the square root of the sum of the squares of the horizontal and vertical sides:

distance =√(∆*x*)2_{+ (∆}_{y}_{)}2 _{=}√_{(}_{x}

2*−x*1)2+ (*y*2*−y*1)2*.*

For example, the distance between points*A*(2*,*1) and*B*(3*,*3) is√(3*−*2)2_{+ (3}*−*_{1)}2 _{=}*√*_{5.}

... ...

... ...

... ...

... ...

... ..

(*x*1*, y*1)

(*x*2*, y*2)

∆*x*

∆*y*

**Figure 1.2.1** Distance between two points, ∆*x*and ∆*y* positive.

As a special case of the distance formula, suppose we want to know the distance of a
point (*x, y*) to the origin. According to the distance formula, this is√(*x−*0)2_{+ (}_{y}−_{0)}2 _{=}

√

*x*2_{+}* _{y}*2

_{.}

A point (*x, y*) is at a distance *r* from the origin if and only if √*x*2_{+}* _{y}*2

_{=}

_{r}_{, or, if we}

square both sides: *x*2 _{+}* _{y}*2

_{=}

*2*

_{r}_{. This is the equation of the circle of radius}

_{r}_{centered at}

the origin. The special case *r*= 1 is called the unit circle; its equation is *x*2+*y*2 = 1.
Similarly, if *C*(*h, k*) is any fixed point, then a point (*x, y*) is at a distance *r* from the
point*C* if and only if √(*x−h*)2_{+ (}_{y}−_{k}_{)}2 _{=}_{r}_{, i.e., if and only if}

(*x−h*)2+ (*y−k*)2 =*r*2*.*

This is the equation of the circle of radius *r* centered at the point (*h, k*). For example, the
circle of radius 5 centered at the point (0*,−*6) has equation (*x−*0)2+ (*y− −*6)2 = 25, or

**EXAMPLE 1.2.1** Graph the circle *x*2_{−}_{2}_{x}_{+}* _{y}*2

_{+ 4}

_{y}_{−}_{11 = 0. With a little thought}

we convert this to (*x−*1)2+ (*y*+ 2)2*−*16 = 0 or (*x−*1)2+ (*y*+ 2)2 = 16. Now we see
that this is the circle with radius 4 and center (1*,−*2), which is easy to graph.

**Exercises 1.2.**

**Exercises 1.2.**

**1.** Find the equation of the circle of radius 3 centered at:

**a)** (0*,*0) **d)** (0*,*3)

**b)** (5*,*6) **e)** (0*,−*3)

**c)** (*−*5*,−*6) **f )** (3*,*0)

*⇒*

**2.** For each pair of points *A*(*x*1*, y*1) and *B*(*x*2*, y*2) find (i) ∆*x* and ∆*y* in going from *A* to *B*,
(ii) the slope of the line joining *A* and *B*, (iii) the equation of the line joining *A* and *B* in
the form *y* =*mx*+*b*, (iv) the distance from *A* to*B*, and (v) an equation of the circle with
center at *A* that goes through*B*.

**a)** *A*(2*,*0),*B*(4*,*3) **d)** *A*(*−*2*,*3), *B*(4*,*3)

**b)** *A*(1*,−*1), *B*(0*,*2) **e)** *A*(*−*3*,−*2), *B*(0*,*0)

**c)** *A*(0*,*0),*B*(*−*2*,−*2) **f )** *A*(0*.*01*,−*0*.*01),*B*(*−*0*.*01*,*0*.*05)

*⇒*

**3.** Graph the circle *x*2+*y*2+ 10*y* = 0.

**4.** Graph the circle *x*2*−*10*x*+*y*2= 24.

**5.** Graph the circle *x*2*−*6*x*+*y*2*−*8*y*= 0.

**6.** Find the standard equation of the circle passing through (*−*2*,*1) and tangent to the line
3*x−*2*y*= 6 at the point (4*,*3). Sketch. (Hint: The line through the center of the circle and
the point of tangency is perpendicular to the tangent line.) *⇒*

### 1.3 Functions

A **function** *y* = *f*(*x*) is a rule for determining *y* when we’re given a value of *x*. For
example, the rule *y* = *f*(*x*) = 2*x* + 1 is a function. Any line *y* = *mx*+ *b* is called a

**linear**function. The graph of a function looks like a curve above (or below) the *x*-axis,
where for any value of *x* the rule *y* = *f*(*x*) tells us how far to go above (or below) the

*x*-axis to reach the curve.

Functions can be defined in various ways: by an algebraic formula or several algebraic formulas, by a graph, or by an experimentally determined table of values. (In the latter case, the table gives a bunch of points in the plane, which we might then interpolate with a smooth curve, if that makes sense.)

**1.3** **Functions** **21**

is also true that if*x* is any number not 1 there is no *y* which corresponds to*x*, but that is
not a problem—only multiple *y* values is a problem.

In addition to lines, another familiar example of a function is the parabola*y* =*f*(*x*) =

*x*2. We can draw the graph of this function by taking various values of *x* (say, at regular
intervals) and plotting the points (*x, f*(*x*)) = (*x, x*2). Then connect the points with a
smooth curve. (See figure 1.3.1.)

The two examples *y* = *f*(*x*) = 2*x*+ 1 and *y* = *f*(*x*) = *x*2 are both functions which
can be evaluated at *any* value of *x* from negative infinity to positive infinity. For many
functions, however, it only makes sense to take *x* in some interval or outside of some
“forbidden” region. The interval of*x*-values at which we’re allowed to evaluate the function
is called the**domain** of the function.

..._{...}
..._{...}

..._{...}
..._{...}

..._{...}

..._{...}...
...

... ...

... ...

... ...

....

*y*=*f*(*x*) =*x*2

... ...

... ...

...

*y* =*f*(*x*) =*√x*

..._{...}
..._{...}

..._{...}
..._{...}

..._{...}
...

..._{...}
..._{...}

..._{...}
..._{...}

..._{...}
...

*y*=*f*(*x*) = 1*/x*

**Figure 1.3.1** Some graphs.

For example, the square-root function*y* =*f*(*x*) =*√x* is the rule which says, given an

*x*-value, take the nonnegative number whose square is *x*. This rule only makes sense if *x*

is positive or zero. We say that the domain of this function is *x* *≥* 0, or more formally

*{x* *∈*R *|* *x* *≥* 0*}*. Alternately, we can use interval notation, and write that the domain is
[0*,∞*). (In interval notation, square brackets mean that the endpoint is included, and a
parenthesis means that the endpoint is not included.) The fact that the domain of*y*=*√x*

is [0*,∞*) means that in the graph of this function ((see figure 1.3.1) we have points (*x, y*)
only above *x*-values on the right side of the *x*-axis.

Another example of a function whose domain is not the entire *x*-axis is: *y* = *f*(*x*) =
1*/x*, the reciprocal function. We cannot substitute *x* = 0 in this formula. The function
makes sense, however, for any nonzero *x*, so we take the domain to be: *{x* *∈*R *|* *x* *̸*= 0*}*.
The graph of this function does not have any point (*x, y*) with *x* = 0. As *x* gets close to
0 from either side, the graph goes oﬀ toward infinity. We call the vertical line *x* = 0 an

**asymptote**.

of a negative number. We will encounter some other ways in which functions might be undefined later.

Another reason why the domain of a function might be restricted is that in a given
situation the*x*-values outside of some range might have no practical meaning. For example,
if *y* is the area of a square of side *x*, then we can write *y* = *f*(*x*) = *x*2. In a purely
mathematical context the domain of the function *y* = *x*2 is all of R. But in the
story-problem context of finding areas of squares, we restrict the domain to positive values of *x*,
because a square with negative or zero side makes no sense.

In a problem in pure mathematics, we usually take the domain to be all values of *x*

at which the formulas can be evaluated. But in a story problem there might be further
restrictions on the domain because only certain values of*x*are of interest or make practical
sense.

In a story problem, often letters diﬀerent from *x* and *y* are used. For example, the
volume*V* of a sphere is a function of the radius*r*, given by the formula *V* =*f*(*r*) = 4_{/}

3*πr*3.

Also, letters diﬀerent from *f* may be used. For example, if *y* is the velocity of something
at time *t*, we may write *y* =*v*(*t*) with the letter *v* (instead of *f*) standing for the velocity
function (and *t* playing the role of *x*).

The letter playing the role of *x* is called the **independent variable**, and the letter
playing the role of *y* is called the **dependent variable** (because its value “depends on”
the value of the independent variable). In story problems, when one has to translate from
English into mathematics, a crucial step is to determine what letters stand for variables.
If only words and no letters are given, then we have to decide which letters to use. Some
letters are traditional. For example, almost always, *t* stands for time.

**EXAMPLE 1.3.1** An open-top box is made from an*a×b*rectangular piece of cardboard
by cutting out a square of side *x* from each of the four corners, and then folding the sides
up and sealing them with duct tape. Find a formula for the volume *V* of the box as a
function of *x*, and find the domain of this function.

The box we get will have height*x* and rectangular base of dimensions*a−*2*x*by*b−*2*x*.
Thus,

*V* =*f*(*x*) =*x*(*a−*2*x*)(*b−*2*x*)*.*

Here *a* and*b* are constants, and*V* is the variable that depends on *x*, i.e.,*V* is playing the
role of *y*.

This formula makes mathematical sense for any*x*, but in the story problem the domain
is much less. In the first place, *x* must be positive. In the second place, it must be less
than half the length of either of the sides of the cardboard. Thus, the domain is

*{x* *∈*R*|*0*< x <* 1

**1.3** **Functions** **23**

In interval notation we write: the domain is the interval (0*,*min(*a, b*)*/*2). (You might think
about whether we could allow 0 or min(*a, b*)*/*2 to be in the domain. They make a certain
physical sense, though we normally would not call the result a box. If we were to allow
these values, what would the corresponding volumes be? Does that make sense?)

**EXAMPLE 1.3.2** **Circle of radius** *r* **centered at the origin** The equation for
this circle is usually given in the form *x*2 +*y*2 = *r*2. To write the equation in the form

*y* = *f*(*x*) we solve for *y*, obtaining *y* = *±√r*2*− _{x}*2

_{. But}

_{this is not a function}_{, because}

when we substitute a value in (*−r, r*) for*x* there are two corresponding values of*y*. To get
a function, we must choose one of the two signs in front of the square root. If we choose
the positive sign, for example, we get the upper semicircle *y* = *f*(*x*) = *√r*2*− _{x}*2

_{(see}

figure1.3.2). The domain of this function is the interval [*−r, r*], i.e., *x*must be between*−r*

and *r* (including the endpoints). If *x* is outside of that interval, then *r*2 * _{−}_{x}*2

_{is negative,}

and we cannot take the square root. In terms of the graph, this just means that there are
no points on the curve whose *x*-coordinate is greater than *r* or less than *−r*.

*−r* *r*

... ... ... ... ... .... .... .... ... ... ... ... ... ... .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .. .. .. .. ... ... ... ... ... ... ... ... ... ... ...

**Figure 1.3.2** Upper semicircle*y*=*√r*2*− _{x}*2

**EXAMPLE 1.3.3** Find the domain of

*y*=*f*(*x*) = *√* 1
4*x−x*2*.*

To answer this question, we must rule out the*x*-values that make 4*x−x*2 negative (because
we cannot take the square root of a negative number) and also the *x*-values that make
4*x−x*2 zero (because if 4*x−x*2 = 0, then when we take the square root we get 0, and
we cannot divide by 0). In other words, the domain consists of all *x* for which 4*x−x*2 is
strictly positive. We give two diﬀerent methods to find out when 4*x−x*2 *>*0.

or else *x <* 0 and 4*−x <* 0. The latter alternative is impossible, since if *x* is negative,
then 4*−x* is greater than 4, and so cannot be negative. As for the first alternative, the
condition 4*−x >*0 can be rewritten (adding*x* to both sides) as 4 *> x*, so we need: *x >*0
and 4*> x* (this is sometimes combined in the form 4*> x >*0, or, equivalently, 0*< x <*4).
In interval notation, this says that the domain is the interval (0*,*4).

**Second method.** Write 4*x* *−* *x*2 as *−*(*x*2 *−* 4*x*), and then complete the square,
obtaining *−*

(

(*x−*2)2*−*4

)

= 4*−*(*x−*2)2. For this to be positive we need (*x−*2)2 *<*4,
which means that *x−*2 must be less than 2 and greater than *−*2: *−*2*< x−*2*<*2. Adding
2 to everything gives 0 *< x <*4. Both of these methods are equally correct; you may use
either in a problem of this type.

A function does not always have to be given by a single formula, as we have already
seen (in the income tax problem, for example). Suppose that *y* = *v*(*t*) is the velocity
function for a car which starts out from rest (zero velocity) at time *t* = 0; then increases
its speed steadily to 20 m/sec, taking 10 seconds to do this; then travels at constant speed
20 m/sec for 15 seconds; and finally applies the brakes to decrease speed steadily to 0,
taking 5 seconds to do this. The formula for *y*=*v*(*t*) is diﬀerent in each of the three time
intervals: first *y* = 2*x*, then *y* = 20, then *y* = *−*4*x*+ 120. The graph of this function is
shown in figure 1.3.3.

10 25 30

0 10 20

... ...

... ...

... ...

... ...

... ...

... ...

...

...
..._{...}

..._{...}
..._{...}

..._{...}
..._{...}

..._{...}
..._{...}

..._{...}
..._{...}

..._{...}
*t*
*v*

**Figure 1.3.3** A velocity function.

**1.4** **Shifts and Dilations** **25**

**Exercises 1.3.**

**Exercises 1.3.**

Find the domain of each of the following functions:

**1.** *y* =*f*(*x*) =*√*2*x−*3 *⇒*

**2.** *y* =*f*(*x*) = 1*/*(*x*+ 1) *⇒*

**3.** *y* =*f*(*x*) = 1*/*(*x*2*−*1)*⇒*

**4.** *y* =*f*(*x*) =√*−*1*/x* *⇒*

**5.** *y* =*f*(*x*) = *√*3_{x}*⇒*

**6.** *y* =*f*(*x*) = *√*4
*x* *⇒*

**7.** *y* =*f*(*x*) =√*r*2*−*_{(}_{x}−_{h}_{)}2 _{, where}_{r}_{is a positive constant.} *⇒*

**8.** *y* =*f*(*x*) =√1*−*(1*/x*) *⇒*

**9.** *y* =*f*(*x*) = 1*/*√1*−*(3*x*)2 *⇒*

**10.** *y* =*f*(*x*) =*√x*+ 1*/*(*x−*1)*⇒*

**11.** *y* =*f*(*x*) = 1*/*(*√x−*1)*⇒*

**12.** Find the domain of *h*(*x*) =

{

(*x*2*−*9)*/*(*x−*3) *x̸*= 3

6 if*x*= 3. *⇒*

**13.** Suppose *f*(*x*) = 3*x−*9 and *g*(*x*) = *√x*. What is the domain of the composition (*g◦f*)(*x*)?
(Recall that **composition** is defined as (*g* *◦* *f*)(*x*) = *g*(*f*(*x*)).) What is the domain of
(*f◦g*)(*x*)? *⇒*

**14.** A farmer wants to build a fence along a river. He has 500 feet of fencing and wants to enclose
a rectangular pen on three sides (with the river providing the fourth side). If *x* is the length
of the side perpendicular to the river, determine the area of the pen as a function of*x*. What
is the domain of this function? *⇒*

**15.** A can in the shape of a cylinder is to be made with a total of 100 square centimeters of
material in the side, top, and bottom; the manufacturer wants the can to hold the maximum
possible volume. Write the volume as a function of the radius *r* of the can; find the domain
of the function. *⇒*

**16.** A can in the shape of a cylinder is to be made to hold a volume of one liter (1000 cubic
centimeters). The manufacturer wants to use the least possible material for the can. Write
the surface area of the can (total of the top, bottom, and side) as a function of the radius *r*

of the can; find the domain of the function. *⇒*

### 1.4 Shifts and Dilations

Many functions in applications are built up from simple functions by inserting constants in various places. It is important to understand the eﬀect such constants have on the appearance of the graph.

**Horizontal shifts.** *If we replacex* *by* *x−C* *everywhere it occurs in the formula forf*(*x*)*,*

*then the graph shifts overC* *to the right.* (If *C* is negative, then this means that the graph

shifts over*|C|*to the left.) For example, the graph of*y*= (*x−*2)2 _{is the}* _{x}*2

_{-parabola shifted}

parabola shifted over to the left so as to have its vertex at *−*1 on the *x*-axis. Note well:
when replacing *x* by *x−C* we must pay attention to meaning, not merely appearance.
Starting with *y* =*x*2 and literally replacing *x* by*x−*2 gives *y*=*x−*22. This is *y*=*x−*4,
a line with slope 1, not a shifted parabola.

**Vertical shifts.** *If we replace* *y* *by* *y−D, then the graph moves up* *D* *units.* (If *D* is
negative, then this means that the graph moves down *|D|* units.) If the formula is written
in the form *y*=*f*(*x*) and if*y*is replaced by *y−D* to get*y−D*=*f*(*x*), we can equivalently
move *D* to the other side of the equation and write *y* = *f*(*x*) +*D*. Thus, this principle
can be stated: *to get the graph of* *y* =*f*(*x*) +*D, take the graph of* *y* = *f*(*x*) *and move it*

*D* *units up.* For example, the function *y* = *x*2*−*4*x* = (*x−*2)2*−*4 can be obtained from

*y* = (*x−*2)2 (see the last paragraph) by moving the graph 4 units down. The result is the

*x*2-parabola shifted 2 units to the right and 4 units down so as to have its vertex at the
point (2*,−*4).

**Warning.** Do not confuse*f*(*x*) +*D* and*f*(*x*+*D*). For example, if*f*(*x*) is the function*x*2,
then *f*(*x*) + 2 is the function *x*2+ 2, while *f*(*x*+ 2) is the function (*x*+ 2)2 =*x*2+ 4*x*+ 4.

**EXAMPLE 1.4.1** **Circles** An important example of the above two principles starts
with the circle *x*2_{+}* _{y}*2

_{=}

*2*

_{r}_{. This is the circle of radius}

_{r}_{centered at the origin. (As we}

saw, this is not a single function *y* = *f*(*x*), but rather two functions *y* = *±√r*2*− _{x}*2

_{put}

together; in any case, the two shifting principles apply to equations like this one that are
not in the form *y* =*f*(*x*).) If we replace*x* by *x−C* and replace *y* by *y−D*—getting the
equation (*x−C*)2+ (*y−D*)2 = *r*2—the eﬀect on the circle is to move it *C* to the right
and *D* up, thereby obtaining the circle of radius *r* centered at the point (*C, D*). This tells
us how to write the equation of any circle, not necessarily centered at the origin.

We will later want to use two more principles concerning the eﬀects of constants on the appearance of the graph of a function.

**Horizontal dilation.** *If* *x* *is replaced by* *x/A* *in a formula and* *A >*1*, then the eﬀect on*

*the graph is to expand it by a factor of* *A* *in the* *x-direction (away from the* *y-axis).* If *A*

is between 0 and 1 then the eﬀect on the graph is to contract by a factor of 1*/A* (towards
the *y*-axis). We use the word “dilate” to mean expand or contract.

For example, replacing*x* by*x/*0*.*5 =*x/*(1*/*2) = 2*x*has the eﬀect of contracting toward
the *y*-axis by a factor of 2. If *A* is negative, we dilate by a factor of *|A|* and then flip
about the *y*-axis. Thus, replacing *x*by *−x* has the eﬀect of taking the mirror image of the
graph with respect to the *y*-axis. For example, the function *y* = *√−x*, which has domain

**1.4** **Shifts and Dilations** **27**

**Vertical dilation.** *If* *y* *is replaced by* *y/B* *in a formula and* *B >* 0*, then the eﬀect on*

*the graph is to dilate it by a factor of* *B* *in the vertical direction.* As before, this is an

expansion or contraction depending on whether*B* is larger or smaller than one. Note that
if we have a function*y*=*f*(*x*), replacing*y* by*y/B* is equivalent to multiplying the function
on the right by*B*: *y*=*Bf*(*x*). The eﬀect on the graph is to expand the picture away from
the*x*-axis by a factor of*B* if *B >*1, to contract it toward the *x*-axis by a factor of 1*/B* if
0*< B <*1, and to dilate by *|B|* and then flip about the *x*-axis if *B* is negative.

**EXAMPLE 1.4.2** **Ellipses** A basic example of the two expansion principles is given
by an **ellipse of semimajor axis** *a* **and semiminor axis** *b*. We get such an ellipse by
starting with the unit circle—the circle of radius 1 centered at the origin, the equation
of which is *x*2 +*y*2 = 1—and dilating by a factor of *a* horizontally and by a factor of *b*

vertically. To get the equation of the resulting ellipse, which crosses the *x*-axis at *±a* and
crosses the *y*-axis at *±b*, we replace *x* by *x/a* and *y* by *y/b* in the equation for the unit
circle. This gives

(_{x}*a*

)2

+

(_{y}*b*

)2

= 1 or *x*

2

*a*2 +

*y*2
*b*2 = 1*.*

Finally, if we want to analyze a function that involves both shifts and dilations, it
is usually simplest to work with the dilations first, and then the shifts. For instance, if
we want to dilate a function by a factor of *A* in the *x*-direction and then shift *C* to the
right, we do this by replacing *x* first by *x/A* and then by (*x−C*) in the formula. As an
example, suppose that, after dilating our unit circle by*a*in the *x*-direction and by *b*in the

*y*-direction to get the ellipse in the last paragraph, we then wanted to shift it a distance

*h* to the right and a distance *k* upward, so as to be centered at the point (*h, k*). The new
ellipse would have equation

(
*x−h*

*a*
)2

+

(
*y−k*

*b*
)2

= 1*.*

Note well that this is diﬀerent than first doing shifts by *h* and *k* and then dilations by *a*

and*b*: _{(}

*x*
*a* *−h*

)2

+

(_{y}*b* *−k*

)2

= 1*.*

1 2 3
*−*1
1
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0 1 2 3 4

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**Figure 1.4.1** Ellipses: (*x−*_{2}1)2+(*y−*_{3}1)2 = 1 on the left, (*x*_{2} *−*1)2+(*y*_{3}*−*1)2 = 1 on the
right.

**Exercises 1.4.**

**Exercises 1.4.**

Starting with the graph of *y* = *√x*, the graph of *y* = 1*/x*, and the graph of *y* = √1*−x*2 _{(the}
upper unit semicircle), sketch the graph of each of the following functions:

**1.** *f*(*x*) =*√x−*2 **2.** *f*(*x*) =*−*1*−*1*/*(*x*+ 2)

**3.** *f*(*x*) = 4 +*√x*+ 2 **4.** *y* =*f*(*x*) =*x/*(1*−x*)

**5.** *y* =*f*(*x*) =*−√−x* **6.** *f*(*x*) = 2 +√1*−*(*x−*1)2

**7.** *f*(*x*) =*−*4 +√*−*(*x−*2) **8.** *f*(*x*) = 2√1*−*(*x/*3)2

**9.** *f*(*x*) = 1*/*(*x*+ 1) **10.** *f*(*x*) = 4 + 2√1*−*(*x−*5)2_{/}_{9}

**11.** *f*(*x*) = 1 + 1*/*(*x−*1) **12.** *f*(*x*) =√100*−*25(*x−*1)2_{+ 2}
The graph of*f*(*x*) is shown below. Sketch the graphs of the following functions.

**13.** *y* =*f*(*x−*1)

1 2 3

*−*1
0
1
2
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**14.** *y* = 1 +*f*(*x*+ 2)

**15.** *y* = 1 + 2*f*(*x*)

**16.** *y* = 2*f*(3*x*)

**17.** *y* = 2*f*(3(*x−*2)) + 1

**18.** *y* = (1*/*2)*f*(3*x−*3)

**2**

**Instantaneous Rate of Change:**

**The Derivative**

### 2.1 The slope of a function

Suppose that*y* is a function of*x*, say*y* =*f*(*x*). It is often necessary to know how sensitive
the value of*y* is to small changes in *x*.

**EXAMPLE 2.1.1** Take, for example, *y* = *f*(*x*) =√625*−x*2 _{(the upper semicircle of}

radius 25 centered at the origin). When*x*= 7, we find that *y*=*√*625*−*49 = 24. Suppose
we want to know how much *y* changes when *x* increases a little, say to 7.1 or 7.01.

In the case of a straight line*y*=*mx*+*b*, the slope*m*= ∆*y/*∆*x*measures the change in

*y*per unit change in *x*. This can be interpreted as a measure of “sensitivity”; for example,
if *y* = 100*x*+ 5, a small change in *x* corresponds to a change one hundred times as large
in *y*, so *y* is quite sensitive to changes in *x*.

Let us look at the same ratio ∆*y/*∆*x* for our function *y* =*f*(*x*) =√625*−x*2 _{when} _{x}

changes from 7 to 7*.*1. Here ∆*x* = 7*.*1*−*7 = 0*.*1 is the change in *x*, and

∆*y*=*f*(*x*+ ∆*x*)*−f*(*x*) =*f*(7*.*1)*−f*(7)

=√625*−*7*.*12 *−*√_{625}*−*_{7}2 *≈*_{23}_{.}_{9706}*−*_{24 =}*−*_{0}_{.}_{0294}_{.}

Thus, ∆*y/*∆*x* *≈ −*0*.*0294*/*0*.*1 = *−*0*.*294. This means that *y* changes by less than one
third the change in *x*, so apparently *y* is not very sensitive to changes in *x* at *x* = 7.
We say “apparently” here because we don’t really know what happens between 7 and 7*.*1.
Perhaps *y* changes dramatically as *x* runs through the values from 7 to 7*.*1, but at 7*.*1 *y*

just happens to be close to its value at 7. This is not in fact the case for this particular function, but we don’t yet know why.

One way to interpret the above calculation is by reference to a line. We have computed
the slope of the line through (7*,*24) and (7*.*1*,*23*.*9706), called a **chord** of the circle. In
general, if we draw the chord from the point (7*,*24) to a nearby point on the semicircle
(7 + ∆*x, f*(7 + ∆*x*)), the slope of this chord is the so-called **diﬀerence quotient**

slope of chord = *f*(7 + ∆*x*)*−f*(7)
∆*x* =

√

625*−*(7 + ∆*x*)2*−*_{24}

∆*x* *.*

For example, if *x* changes only from 7 to 7.01, then the diﬀerence quotient (slope of the
chord) is approximately equal to (23*.*997081*−*24)*/*0*.*01 = *−*0*.*2919. This is slightly less
steep than the chord from (7*,*24) to (7*.*1*,*23*.*9706).

As the second value 7 + ∆*x* moves in towards 7, the chord joining (7*, f*(7)) to (7 +
∆*x, f*(7 + ∆*x*)) shifts slightly. As indicated in figure2.1.1, as ∆*x* gets smaller and smaller,
the chord joining (7*,*24) to (7 + ∆*x, f*(7 + ∆*x*)) gets closer and closer to the **tangent line**

to the circle at the point (7*,*24). (Recall that the tangent line is the line that just grazes
the circle at that point, i.e., it doesn’t meet the circle at any second point.) Thus, as ∆*x*

gets smaller and smaller, the slope ∆*y/*∆*x* of the chord gets closer and closer to the slope
of the tangent line. This is actually quite diﬃcult to see when ∆*x* is small, because of the
scale of the graph. The values of ∆*x*used for the figure are 1, 5, 10 and 15, not really very
small values. The tangent line is the one that is uppermost at the right hand endpoint.

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[image:30.595.202.370.404.596.2]5 10 15 20 25
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**Figure 2.1.1** Chords approximating the tangent line. (AP)

**2.1** **The slope of a function** **31**

Instead of looking at more particular values of ∆*x*, let’s see what happens if we do
some algebra with the diﬀerence quotient using just ∆*x*. The slope of a chord from (7*,*24)
to a nearby point is given by

√

625*−*(7 + ∆*x*)2*−*_{24}

∆*x* =

√

625*−*(7 + ∆*x*)2*−*_{24}

∆*x*

√

625*−*(7 + ∆*x*)2_{+ 24}

√

625*−*(7 + ∆*x*)2_{+ 24}

= 625*−*(7 + ∆*x*)

2_{−}_{24}2

∆*x*(√625*−*(7 + ∆*x*)2_{+ 24)}

= 49*−*49*−*14∆*x−*∆*x*

2

∆*x*(√625*−*(7 + ∆*x*)2_{+ 24)}

= ∆*x*(*−*14*−*∆*x*)

∆*x*(√625*−*(7 + ∆*x*)2_{+ 24)}

= √ *−*14*−*∆*x*

625*−*(7 + ∆*x*)2_{+ 24}

Now, can we tell by looking at this last formula what happens when ∆*x* gets very close to
zero? The numerator clearly gets very close to*−*14 while the denominator gets very close to

√

625*−*72_{+ 24 = 48. Is the fraction therefore very close to}*−*_{14}_{/}_{48 =}*−*_{7}_{/}_{24}*∼*_{=}*−*_{0}_{.}_{29167?}

It certainly seems reasonable, and in fact it is true: as ∆*x* gets closer and closer to zero,
the diﬀerence quotient does in fact get closer and closer to*−*7*/*24, and so the slope of the
tangent line is exactly *−*7*/*24.

example, the “24” in the calculation came from √625*−*72_{, so we’ll need to fix that too.}

√

625*−*(*x*+ ∆*x*)2*−√*_{625}*− _{x}*2

∆*x* =

=

√

625*−*(*x*+ ∆*x*)2*−√*_{625}*− _{x}*2

∆*x*

√

625*−*(*x*+ ∆*x*)2 _{+}*√*_{625}*− _{x}*2

√

625*−*(*x*+ ∆*x*)2 _{+}*√*_{625}*− _{x}*2

= 625*−*(*x*+ ∆*x*)

2_{−}_{625 +}* _{x}*2

∆*x*(√625*−*(*x*+ ∆*x*)2_{+}*√*_{625}*− _{x}*2

_{)}

= 625*−x*

2_{−}_{2}_{x}_{∆}_{x}_{−}_{∆}* _{x}*2

_{−}_{625 +}

*2*

_{x}∆*x*(√625*−*(*x*+ ∆*x*)2_{+}*√*_{625}*− _{x}*2

_{)}

= ∆*x*(*−*2*x−*∆*x*)

∆*x*(√625*−*(*x*+ ∆*x*)2_{+}*√*_{625}*− _{x}*2

_{)}

= √ *−*2*x−*∆*x*

625*−*(*x*+ ∆*x*)2_{+}*√*_{625}*− _{x}*2

Now what happens when ∆*x* is very close to zero? Again it seems apparent that the
quotient will be very close to

*−*2*x*
*√*

625*−x*2_{+}*√*_{625}*− _{x}*2 =

*−*2*x*

2*√*625*−x*2 =

*−x*
*√*

625*−x*2*.*

Replacing *x*by 7 gives *−*7*/*24, as before, and now we can easily do the computation for 12
or any other value of *x* between *−*25 and 25.

So now we have a single, simple formula,*−x/*√625*−x*2_{, that tells us the slope of the}

tangent line for any value of *x*. This slope, in turn, tells us how sensitive the value of*y* is
to changes in the value of *x*.

What do we call such a formula? That is, a formula with one variable, so that
substi-tuting an “input” value for the variable produces a new “output” value? This is a function.
Starting with one function, √625*−x*2_{, we have derived, by means of some slightly nasty}

algebra, a new function, *−x/*√625*−x*2_{, that gives us important information about the}

original function. This new function in fact is called the **derivative** of the original
func-tion. If the original is referred to as *f* or *y*then the derivative is often written *f′* or *y′* and
pronounced “f prime” or “y prime”, so in this case we might write*f′*(*x*) =*−x/*√625*−x*2_{.}

At a particular point, say *x*= 7, we say that *f′*(7) =*−*7*/*24 or “*f* prime of 7 is *−*7*/*24” or
“the derivative of *f* at 7 is*−*7*/*24.”

To summarize, we compute the derivative of *f*(*x*) by forming the diﬀerence quotient

*f*(*x*+ ∆*x*)*−f*(*x*)

∆*x* *,* (2*.*1*.*1)

**2.1** **The slope of a function** **33**

We should note that in the particular case of a circle, there’s a simple way to find the
derivative. Since the tangent to a circle at a point is perpendicular to the radius drawn
to the point of contact, its slope is the negative reciprocal of the slope of the radius. The
radius joining (0*,*0) to (7*,*24) has slope 24/7. Hence, the tangent line has slope *−*7*/*24. In
general, a radius to the point (*x,*√625*−x*2_{) has slope} √_{625}*− _{x}*2

_{/x}_{, so the slope of the}

tangent line is *−x/*√625*−x*2_{, as before. It is} _{NOT}_{always true that a tangent line is}

perpendicular to a line from the origin—don’t use this shortcut in any other circumstance.
As above, and as you might expect, for diﬀerent values of *x* we generally get diﬀerent
values of the derivative *f′*(*x*). Could it be that the derivative always has the same value?
This would mean that the slope of*f*, or the slope of its tangent line, is the same everywhere.
One curve that always has the same slope is a line; it seems odd to talk about the tangent
line to a line, but if it makes sense at all the tangent line must be the line itself. It is not
hard to see that the derivative of *f*(*x*) =*mx*+*b*is *f′*(*x*) =*m*; see exercise 6.

**Exercises 2.1.**

**Exercises 2.1.**

**1.** Draw the graph of the function *y*=*f*(*x*) =√169*−x*2 _{between}_{x}_{= 0 and} _{x}_{= 13. Find the}
slope ∆*y/*∆*x* of the chord between the points of the circle lying over (a)*x*= 12 and *x*= 13,
(b) *x* = 12 and *x* = 12*.*1, (c) *x* = 12 and *x* = 12*.*01, (d) *x* = 12 and *x* = 12*.*001. Now use
the geometry of tangent lines on a circle to find (e) the exact value of the derivative *f′*(12).
Your answers to (a)–(d) should be getting closer and closer to your answer to (e). *⇒*

**2.** Use geometry to find the derivative *f′*(*x*) of the function *f*(*x*) =√625*−x*2 _{in the text for}
each of the following*x*: (a) 20, (b) 24, (c)*−*7, (d)*−*15. Draw a graph of the upper semicircle,
and draw the tangent line at each of these four points. *⇒*

**3.** Draw the graph of the function *y*=*f*(*x*) = 1*/x* between*x*= 1*/*2 and *x*= 4. Find the slope
of the chord between (a)*x*= 3 and*x*= 3*.*1, (b)*x*= 3 and*x*= 3*.*01, (c)*x*= 3 and*x*= 3*.*001.
Now use algebra to find a simple formula for the slope of the chord between (3*, f*(3)) and
(3 + ∆*x, f*(3 + ∆*x*)). Determine what happens when ∆*x* approaches 0. In your graph of

*y* = 1*/x*, draw the straight line through the point (3*,*1*/*3) whose slope is this limiting value
of the diﬀerence quotient as ∆*x* approaches 0. *⇒*

**4.** Find an algebraic expression for the diﬀerence quotient

(

*f*(1 + ∆*x*)*−f*(1)

)

*/*∆*x* when*f*(*x*) =

*x*2*−*(1*/x*). Simplify the expression as much as possible. Then determine what happens as
∆*x* approaches 0. That value is *f′*(1). *⇒*

**5.** Draw the graph of *y* =*f*(*x*) =*x*3 between *x* = 0 and *x* = 1*.*5. Find the slope of the chord
between (a) *x* = 1 and *x* = 1*.*1, (b) *x* = 1 and *x* = 1*.*001, (c) *x* = 1 and *x* = 1*.*00001.
Then use algebra to find a simple formula for the slope of the chord between 1 and 1 + ∆*x*.
(Use the expansion (*A*+*B*)3=*A*3+ 3*A*2*B*+ 3*AB*2+*B*3.) Determine what happens as ∆*x*

approaches 0, and in your graph of *y* = *x*3 draw the straight line through the point (1*,*1)
whose slope is equal to the value you just found. *⇒*

**6.** Find an algebraic expression for the diﬀerence quotient (*f*(*x*+ ∆*x*)*−f*(*x*))*/*∆*x*when*f*(*x*) =

*mx*+*b*. Simplify the expression as much as possible. Then determine what happens as ∆*x*