# Whitman Calculus

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### Single Variable Calculus

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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 fromElementary 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.

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For Kathleen,

without whose encouragement this book would not have

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### 13

1.1 Lines . . . 14

1.2 Distance Between Two Points; Circles . . . 19

1.3 Functions . . . 20

1.4 Shifts and Dilations . . . 25

### 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

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### 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

### 71

4.1 Trigonometric Functions . . . 71

4.2 The Derivative of sinx . . . 74

4.3 A hard limit . . . 75

4.4 The Derivative of sinx, 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

### 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

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Contents 7

### 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

### 147

7.1 Two examples . . . 147

7.2 The Fundamental Theorem of Calculus . . . 151

7.3 Some Properties of Integrals . . . 158

### 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

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### 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

### 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

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Contents 9

### 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

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### 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.

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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.

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## 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 thex-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 positivey-direction. In a purely mathematical situation, we normally choose the same scale for thex- andy-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 thanxandyare 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 andB 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

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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

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

...... ...

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

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

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

[image:14.595.143.432.71.303.2]

...... ...

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 inx” although it usually is read as “delta

x”. The point is that ∆xdenotes 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 = 31 = 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 (x1, y1) and a point (x2, y2) are:

x=x2−x1,y=y2−y1.

Note that either or both of these might be negative.

### 1.1 Lines

If we have two points A(x1, y1) and B(x2, y2), then we can draw one and only one line

through both points. By the slopeof this line we mean the ratio of ∆y to ∆x. The slope is often denoted m: m= ∆y/x= (y2−y1)/(x2−x1). For example, the line joining the

points (1,−2) and (3,5) has slope (5 + 2)/(31) = 7/2.

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1.1 Lines 15

information (15%, 28%, or 33%) using mathematical terminology, and graph the tax on they-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 fromx = 26050 tox = 67200, the line goes upward 28 for each 100 in thex-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

...

... ...

... ...

... ...

... ...

... ...

... ...

... ...

... ...

... ...

... ...

[image:15.595.99.544.276.486.2]

Figure 1.1.1 Tax vs. income.

The most familiar form of the equation of a straight line is: y=mx+b. Heremis the slope of the line: if you increasex 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 = mx. 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 formulam= (y2−y1)/(x2−x1) gives you the slope. Once you know a point

and the slope, then they-intercept can be found by substituting the coordinates of either point in the equation: y1 = mx1 +b, i.e., b = y1−mx1. Alternatively, one can use the

“point-slope” form of the equation of a straight line: start with (y−y1)/(x−x1) =mand

then multiply to get (y−y1) =m(x−x1), the point-slope form. Of course, this may be

further manipulated to gety =mx−mx1+y1, 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−y1)/(x−x1) = (y2−y1)/(x2−x1), which says “the slope measured between the point

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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) andB(3,3), we can use this formula:

y−1

x−2 = 31

32 = 2, so that y−1 = 2(x−2), i.e., y= 2x−3.

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

in a separate step.

The slopem of a line in the formy=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 mis a large negative number (large in absolute value), then the line points steeply downward; while if mis 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 liney =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 = 2x−3 through the pointsA(2,1) andB(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+bfor your distance from Seattle. Find the slope, y-intercept, and t-intercept, and describe the practical meaning of each.

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1.1 Lines 17

110)/(1.51) =50. The meaning of the slope is that you are traveling at 50 mph; m is negative because you are travelingtoward Seattle, i.e., your distance y is decreasing. The word “velocity” is often used form=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 =50t+ 160.

The meaning of they-intercept 160 is that whent= 0 (when you started the trip) you were 160 miles from Seattle. To find thet-intercept, set 0 =50t+160, so thatt= 160/50 = 3.2. The meaning of thet-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.

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 slope2 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 2x = 2 to the form y = mx+b, graph the line, and find the

y-intercept and x-intercept.

5. Change the equationx+y= 6 to the formy =mx+b, graph the line, and find they-intercept and x-intercept.

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

y-intercept and x-intercept.

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

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

y-intercept and x-intercept.

9. Determine whether the lines 3x+ 6y= 7 and 2x+ 4y= 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+bform.

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?

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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 whenx 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.,xis the amount on Form 1040, line 37, andyis the amount on Form 1040, line 38). Find the slope andy-intercept of each line that makes up the polygonal graph, up to x= 97620.

### 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

UseWorksheet

97,620 ... below to figure your tax

### Schedule Z

—Use if your filing status is

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

UseWorksheet

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 letP be the price of an item. (a) ExpressP linearly in terms ofx, in other words, expressP in the formP =mx+b. (b) Express x linearly in terms of P.

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1.2 Distance Between Two Points; Circles 19

will be linear. Letxbe the exam score, and letybe the corresponding grade. Find a formula of the form y=mx+b which applies to scoresx between 40 and 90.

### 1.2 Distance Between Two Points; Circles

Given two points (x1, y1) and (x2, y2), recall that their horizontal distance from one another

is ∆x=x2−x1and their vertical distance from one another is ∆y =y2−y1. (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−x1)2+ (y2−y1)2.

For example, the distance between pointsA(2,1) andB(3,3) is√(32)2+ (31)2 =5.

... ...

... ...

... ...

... ...

... ..

(x1, y1)

(x2, y2)

x

y

Figure 1.2.1 Distance between two points, ∆xand ∆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+ (y0)2 =

x2+y2.

A point (x, y) is at a distance r from the origin if and only if √x2+y2 =r, or, if we

square both sides: x2 +y2 = r2. 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 x2+y2 = 1. Similarly, if C(h, k) is any fixed point, then a point (x, y) is at a distance r from the pointC if and only if √(x−h)2+ (yk)2 =r, i.e., if and only if

(x−h)2+ (y−k)2 =r2.

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

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EXAMPLE 1.2.1 Graph the circle x22x+y2+ 4y11 = 0. With a little thought

we convert this to (x−1)2+ (y+ 2)216 = 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.

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(x1, y1) and B(x2, y2) 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 toB, and (v) an equation of the circle with center at A that goes throughB.

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 x2+y2+ 10y = 0.

4. Graph the circle x210x+y2= 24.

5. Graph the circle x26x+y28y= 0.

6. Find the standard equation of the circle passing through (2,1) and tangent to the line 3x−2y= 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) = 2x + 1 is a function. Any line y = mx+ b is called a

linearfunction. 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.)

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1.3 Functions 21

is also true that ifx is any number not 1 there is no y which corresponds tox, 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 parabolay =f(x) =

x2. 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, x2). Then connect the points with a smooth curve. (See figure 1.3.1.)

The two examples y = f(x) = 2x+ 1 and y = f(x) = x2 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 ofx-values at which we’re allowed to evaluate the function is called thedomain of the function.

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

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

......

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

... ...

... ...

... ...

....

y=f(x) =x2

... ...

... ...

...

y =f(x) =√x

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

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

...... ...

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

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

...... ...

y=f(x) = 1/x

Figure 1.3.1 Some graphs.

For example, the square-root functiony =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 ofy=√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.

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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 thex-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) = x2. In a purely mathematical context the domain of the function y = x2 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 ofxare of interest or make practical sense.

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

3πr3.

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 ana×brectangular 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 heightx and rectangular base of dimensionsa−2xbyb−2x. Thus,

V =f(x) =x(a−2x)(b−2x).

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

This formula makes mathematical sense for anyx, 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

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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 x2 +y2 = r2. To write the equation in the form

y = f(x) we solve for y, obtaining y = ±√r2x2. But this is not a function, because

when we substitute a value in (−r, r) forx there are two corresponding values ofy. 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) = √r2x2 (see

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

and r (including the endpoints). If x is outside of that interval, then r2 x2 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 semicircley=√r2x2

EXAMPLE 1.3.3 Find the domain of

y=f(x) = 1 4x−x2.

To answer this question, we must rule out thex-values that make 4x−x2 negative (because we cannot take the square root of a negative number) and also the x-values that make 4x−x2 zero (because if 4x−x2 = 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 4x−x2 is strictly positive. We give two diﬀerent methods to find out when 4x−x2 >0.

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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 (addingx 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 4x x2 as (x2 4x), and then complete the square, obtaining

(

(x−2)24

)

= 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 = 2x, then y = 20, then y = 4x+ 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.

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1.4 Shifts and Dilations 25

### Exercises 1.3.

Find the domain of each of the following functions:

1. y =f(x) =2x−3

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

3. y =f(x) = 1/(x21)

4. y =f(x) =√1/x

5. y =f(x) = 3x

6. y =f(x) = 4 x

7. y =f(x) =√r2(xh)2 , wherer is a positive constant.

8. y =f(x) =√1(1/x)

9. y =f(x) = 1/√1(3x)2

10. y =f(x) =√x+ 1/(x−1)

11. y =f(x) = 1/(√x−1)

12. Find the domain of h(x) =

{

(x29)/(x−3) = 3

6 ifx= 3.

13. Suppose f(x) = 3x−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 ofx. 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 ofy= (x−2)2 is thex2-parabola shifted

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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 =x2 and literally replacing x byx−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 ifyis replaced by y−D to gety−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 = x24x = (x−2)24 can be obtained from

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

x2-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 confusef(x) +D andf(x+D). For example, iff(x) is the functionx2, then f(x) + 2 is the function x2+ 2, while f(x+ 2) is the function (x+ 2)2 =x2+ 4x+ 4.

EXAMPLE 1.4.1 Circles An important example of the above two principles starts with the circle x2+y2 = r2. 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 = ±√r2x2 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 replacex by x−C and replace y by y−D—getting the equation (x−C)2+ (y−D)2 = r2—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, replacingx byx/0.5 =x/(1/2) = 2xhas 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 xby −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

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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 whetherB is larger or smaller than one. Note that if we have a functiony=f(x), replacingy byy/B is equivalent to multiplying the function on the right byB: y=Bf(x). The eﬀect on the graph is to expand the picture away from thex-axis by a factor ofB 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 x2 +y2 = 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

a2 +

y2 b2 = 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 byain the x-direction and by bin 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

andb: (

x a −h

)2

+

(y b −k

)2

= 1.

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[image:28.595.69.520.67.343.2]

1 2 3 1 1 2 3 4 2 1 ... ... ... ... ... ... .... .... .... ... ... ... ... ... ... ... .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .. .. .. ... ... ... ... ... ... ... ... ... ... ... ...... ...... ...... ...... ...... ...... ...... ... ......... ... ... ... ... ... ... ... ... ... ... ...

0 1 2 3 4

0 1 2 3 4 5 6 ... ... ... ... ... ... .... .... .... ... ... ... ... ... ... ... .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .. .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ...... ...... ...... ...... ...... ...... ... ......... ... ... ... ... ... ... ... ... ... ... ...

Figure 1.4.1 Ellipses: (x−21)2+(y−31)2 = 1 on the left, (x2 1)2+(y31)2 = 1 on the right.

### Exercises 1.4.

Starting with the graph of y = √x, the graph of y = 1/x, and the graph of y = √1−x2 (the upper unit semicircle), sketch the graph of each of the following functions:

1. f(x) =√x−2 2. f(x) =11/(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) =√10025(x−1)2+ 2 The graph off(x) is shown below. Sketch the graphs of the following functions.

13. y =f(x−1)

1 2 3

1 0 1 2 ... ... ... ... ... ... ... ... ... ... ... ... ...... ...... ...... ...... ... ... ... ... ... ... ...... ......

14. y = 1 +f(x+ 2)

15. y = 1 + 2f(x)

16. y = 2f(3x)

17. y = 2f(3(x−2)) + 1

18. y = (1/2)f(3x−3)

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## 2

### 2.1 The slope of a function

Suppose thaty is a function ofx, sayy =f(x). It is often necessary to know how sensitive the value ofy is to small changes in x.

EXAMPLE 2.1.1 Take, for example, y = f(x) =√625−x2 (the upper semicircle of

radius 25 centered at the origin). Whenx= 7, we find that y=62549 = 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 liney=mx+b, the slopem= ∆y/xmeasures the change in

yper unit change in x. This can be interpreted as a measure of “sensitivity”; for example, if y = 100x+ 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−x2 when x

changes from 7 to 7.1. Here ∆x = 7.17 = 0.1 is the change in x, and

y=f(x+ ∆x)−f(x) =f(7.1)−f(7)

=√6257.12 62572 23.970624 =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.

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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)224

x .

For example, if x changes only from 7 to 7.01, then the diﬀerence quotient (slope of the chord) is approximately equal to (23.99708124)/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 ∆xused 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.

... ... ... ... ... ... ... .... .... .... ... ... ... ... ... ... ... ... ... .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 10 15 20 25

[image:30.595.202.370.404.596.2]

5 10 15 20 25 ...... ...... ...... ...... ...... ...... ...... ...... ... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...

Figure 2.1.1 Chords approximating the tangent line. (AP)

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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)224

x =

625(7 + ∆x)224

x

625(7 + ∆x)2+ 24

625(7 + ∆x)2+ 24

= 625(7 + ∆x)

2242

x(√625(7 + ∆x)2+ 24)

= 494914∆x−x

2

x(√625(7 + ∆x)2+ 24)

= ∆x(14x)

x(√625(7 + ∆x)2+ 24)

= √ 14x

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 to14 while the denominator gets very close to

62572+ 24 = 48. Is the fraction therefore very close to14/48 =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 to7/24, and so the slope of the tangent line is exactly 7/24.

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example, the “24” in the calculation came from √62572, so we’ll need to fix that too.

625(x+ ∆x)2−√625x2

x =

=

625(x+ ∆x)2−√625x2

x

625(x+ ∆x)2 +625x2

625(x+ ∆x)2 +625x2

= 625(x+ ∆x)

2625 +x2

x(√625(x+ ∆x)2+625x2)

= 625−x

22xxx2625 +x2

x(√625(x+ ∆x)2+625x2)

= ∆x(2x−x)

x(√625(x+ ∆x)2+625x2)

= √ 2x−x

625(x+ ∆x)2+625x2

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

2x

625−x2+625x2 =

2x

2625−x2 =

−x

625−x2.

Replacing xby 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−x2, that tells us the slope of the

tangent line for any value of x. This slope, in turn, tells us how sensitive the value ofy 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−x2, we have derived, by means of some slightly nasty

algebra, a new function, −x/√625−x2, 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 ythen the derivative is often written f′ or y′ and pronounced “f prime” or “y prime”, so in this case we might writef′(x) =−x/√625−x2.

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 is7/24.”

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

f(x+ ∆x)−f(x)

x , (2.1.1)

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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−x2) has slope625x2/x, so the slope of the

tangent line is −x/√625−x2, 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 off, 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+bis f′(x) =m; see exercise 6.

### Exercises 2.1.

1. Draw the graph of the function y=f(x) =√169−x2 betweenx= 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−x2 in the text for each of the followingx: (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 betweenx= 1/2 and x= 4. Find the slope of the chord between (a)x= 3 andx= 3.1, (b)x= 3 andx= 3.01, (c)x= 3 andx= 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 whenf(x) =

x2(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) =x3 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=A3+ 3A2B+ 3AB2+B3.) Determine what happens as ∆x

approaches 0, and in your graph of y = x3 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))/xwhenf(x) =

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

Figure

+7

References

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