Calculus study solutions guide 8e ron larson (1)

Calculus required: slope of tangent line at is rate of change, and equals about 0.16.. a The limit does not exist at b The function is not defined at c The limit exists at but it is no

C H A P T E R P Preparation for Calculus

Section P.1 Graphs and Models 2

Section P.2 Linear Models and Rates of Change 7

Section P.3 Functions and Their Graphs 14

Section P.4 Fitting Models to Data 18

Review Exercises 19

Problem Solving 23

P A R T I

Preparation for Calculus

Solutions to Odd-Numbered Exercises

y

x

4 6 8

x

3 2 1

1

, 2

y  3x  2

0, 1)

y

2

) , ( 1 0 (

x

1 2 3 4 5

− 2

6

− 4 − 3 − 1 1 2 3 4 (−2, 0) (0, 0)

x

3 2 1

59. yx 2x 4x 6(other answers possible) 61 Some possible equations:

y= −x2 + 3x− 1

y=x3 − 2x2 +x− 1 (2, 1) (0, 1) −

This problem can also be solved by using a graphing utility and finding the intersection of the graphs of C and R.

77 (a) Using a graphing utility, you obtain

400

100

85 Distance to the origin Distance to

Note: This is the equation of a circle!

81 False; x-axis symmetry means that if 1, 2is on the graph, then 1, 2is also on the graph

83 True; the x-intercepts are

b±b2 4ac

Section P.2 Linear Models and Rates of Change

− 1

− 2

3 4 5 6

− 1

(2, 1) (2, 5)

y  3x  10

y 7  3x 1

19 Given a line L, you can use any two distinct points to calculate its slope Since a line is straight, the ratio of the change in

y-values to the change in x-values will always be the same See Section P.2 Exercise 93 for a proof.

(b) The slopes of the line segments are

The population increased most rapidly from 1991 to 1992

1 2 3 4

− 1

2 3 4

− 2

6 7 9

− 2 − 1

− 3

( )1 7 ,

1 2

y  2x  3

2x  y  3  0

The lines are perpendicular because their slopes 1 and are negative reciprocals of each other.

You must use a square setting in order for perpendicular lines to appear perpendicular

y5

3x 5x  3y  0

You can use the graphing utility to determine that the points of intersection are and Analytically,

m12 1 01 1

73 Equations of perpendicular bisectors:

Letting in either equation gives the point of

100, 212

0, 32

79 (a)

(c) Both jobs pay $17 per hour if 6 units are produced

For someone who can produce more than 6 units per

hour, the second offer would pay more For a worker

who produces less than 6 units per hour, the first offer

81 (a) Two points are and The slope is

89 If then is the horizontal line The distance to is

(Note that A and B cannot both be zero.)

to is:

The point of intersection of these two lines is:

(1)(2)(By adding equations (1) and (2))

(3)(4)(By adding equations (3) and (4))

91 For simplicity, let the vertices of the rhombus be

slopes of the diagonals are then

and Since the sides of the Rhombus are equal,

(0, 0)

y

x

93 Consider the figure below in which the four points are

collinear Since the triangles are similar, the result

Section P.3 Functions and Their Graphs

2

2 4

2 2 4

2 3

− 1

1 2

gt 2 sin t

29.

y is not a function of x Some vertical lines intersect

the graph twice

x  y2 0 ⇒ y±x 31 y is a function of x Vertical lines intersect the graph

d

47 (a) The graph is shifted

3 units to the left

1 unit to the right

2 1

3

x y

Horizontal shift 2 units to theright

6 4 3

3 2

1 1

− 1

− 2 2 4

(c) If Ht Tt 1,then the overall temperature would be reduced 1 degree

63 (a) If f is even, then 3 is on the graph.

Section P.4 Fitting Models to Data

5 (a), (b)

Yes The cancer mortality increases linearly with

increased exposure to the carcinogenic substance

(c) If x  3, then y  136.

x y

3 6 9 12 15 50

9 (a) Let per capita energy usage (in millions of Btu)

per capita gross national product (in thousands)

(b)

(c) Denmark, Japan, and Canada

(d) Deleting the data for the three countries above,

(e) The slope represents the average increase per year

in the number of people (in millions) in HMOs

0 70

y3 0.4297t2 0.5994t  32.9745

Review Exercises for Chapter P

y  1.81x3 14.58x2 16.39x  10 17 (a) Yes, y is a function of t At each time t, there is one

and only one displacement y.

(b) The amplitude is approximately

The period is approximately

(d)

0 4

1 2

x

y

6 5

2 5

y2

5x6 5

5 10

y  7  6x  x2

− 1

− 2

3 4

2 2

2 3

41 —CONTINUED—

(c)

4 2 1

2 1 1 2 3

x

y

0 2

c c

2

c

fx cx3, c 2, 0, 2

43 (a) Odd powers:

The graphs of f, g, and h all rise to the right and fall to

the left As the degree increases, the graph rises and

falls more steeply All three graphs pass through the

2

fx x, gx x3, hx x5 Even powers:

The graphs of f, g, and h all rise to the left and to the

right As the degree increases, the graph rises moresteeply All three graphs pass through the points

4

fx x2, gx x4, hx x6

will look like hx x6,but rise even more steeply

area is attained when the rectangle is a square In this

47 (a) 3 (cubic), negative leading coefficient

(b) 4 (quartic), positive leading coefficient

(c) 2 (quadratic), negative leading coefficient

(d) 5, positive leading coefficient

49 (a) Yes, y is a function of t At each time t, there is one

and only one displacement y.

(b) The amplitude is approximately

.The period is approximately 1.1

(c) One model is (d)

Problem Solving for Chapter P

1 (a)

Slope of tangent line is Hence,

0< x < 100

Ax xy  x100 x

2  x22 50x

x  2y  100 ⇒ y1002 x 7 The length of the trip in the water is and the

the total time is

T42 x2143 x2 hours

13 x2

22 x2,

9 (a) Slope of tangent line is less than 5

− 2

− 2

− 1

1 2

2

d1d2 1

C H A P T E R 1 Limits and Their Properties

Section 1.1 A Preview of Calculus 27 Section 1.2 Finding Limits Graphically and Numerically 27 Section 1.3 Evaluating Limits Analytically 31 Section 1.4 Continuity and One-Sided Limits 37 Section 1.5 Infinite Limits 42 Review Exercises 47

Problem Solving 49

Limits and Their Properties

Solutions to Odd-Numbered Exercises

1 Precalculus: 20 ftsec15 seconds 300 feet 3 Calculus required: slope of tangent line at is rate of

change, and equals about 0.16

(b) The graphs of are approximations to the tangent line to at

(c) The slope is approximately 2 For a better approximation make the list numbers smaller:

Actual limit is 1 

16

lim

15. tan x does not exist since the function increases and

decreases without bound as x approaches 2

lim

between 1and 1 as x approaches 0.

Using the first series of equivalent inequalities, you obtain

whereas for values of x to the right of 5, x 5x 5equals 1

x 0

utility does not show the hole at 9, 6.

45 (i) The values of f approach different

numbers as x approaches c from

2 3 4 5

x

y

(iii) The values of f oscillate

between two fixed numbers as

y

Section 1.3 Evaluating Limits Analytically

53 Answers will vary.

Therefore,L1 L2 < 2.Since  > 0is arbitrary, it follows that L1 L2

21 lim

x→ 4x 324  32 1 23 (a)

(b)(c) lim

t→ 0 sin 3t 3t

fx x cos x

fx x sin 1x

95 We say that two functions f and g agree at all but one

interval except for x  c, where c is in the interval.

fx gx 97 An indeterminant form is obtained when evaluating a limitusing direct substitution produces a meaningless fractional

expression such as That is,

for which lim

for any  > 0,then any value of  > 0will work

fx b fx b,b  b 0 <   > 0  > 0 fx b <  x  c < 

109 If then the property is true because both sides are equal to 0 If let be given Since

we have

orwhich implies that limbfx bL.

does not exist.

No matter how “close to” 0 x is, there are still an infinite number of rational and irrational numbers so that does notexist

When x is “close to” 0, both parts of the function are “close to” 0.

Section 1.4 Continuity and One-Sided Limits

The function is NOT continuous

35. fx 3x  cos x is continuous for all real x. 37. is not continuous at Since

does not exist, and has a removable discontinuity atsince

51. has nonremovable discontinuities at integer

The graph appears to be continuous on the interval

Since is not defined, we know that f has

a discontinuity at This discontinuity is removable

so it does not show up on the graph

1 and 2

fc 0

f2 4

f133 16

1, 2

fx 1

16x4 x3 3

77. is continuous on

one value of c between 0 and 1 Using a graphing utility,

one value c between 0 and 1 Using a graphing utility, we

The Intermediate Value Theorem applies

The Intermediate Value Theorem applies

fx x3 x2 x  2 87 (a) The limit does not exist at

(b) The function is not defined at (c) The limit exists at but it is not equal to thevalue of the function at

(d) The limit does not exist at x  c.

However, for non-integer values of x, the functions

2 4 6 8 10 12 10

20 30 40 50

N

Time (in months)

93.

Discontinuous at every positive even integer.The

company replenishes its inventory every two months

Nt 25 2t 2

95 Let be the volume of a sphere of radius r.

Since 4.19 < 275 < 523.6, the Intermediate Value Theorem implies that there is at least one value r between 1 and 5 such

that (In Vr 275 fact,r 4.0341.)

97 Let c be any real number Then does not exist since there are both rational and

irrational numbers arbitrarily close to c Therefore, f is not continuous at c.

at least one of these limits (if they exist) does not equal

the corresponding function at x  c.

b b

b b y

Section 1.5 Infinite Limits

Therefore,x 1is a vertical asymptote.

fx tan 2x  cos 2x sin 2x 19.

Therefore,t 0is a vertical asymptote

Therefore,x 1is a vertical asymptote

a nonzero integer There is no vertical asymptote at since

lim

t→ 0

t sin t 1

t 0

t  n,

stsin t t

53 A limit in which increases or decreases without

bound as x approaches c is called an infinite limit is

not a number Rather, the symbol

says how the limit fails to exist

0.32920.1585

fx

0.00010.001

0.010.1

0.20.5

0.04110.1585

fx

0.00010.001

0.010.1

0.20.5

0.03330.0823

0.1585

fx

0.00010.001

0.010.1

0.20.5

0.16670.1666

0.16630.1646

0.1585

fx

0.00010.001

0.010.1

0.20.5

67 (a) Because the circumference of the motor is

half that of the saw arbor, the saw makes

revolutions per minute

The angle subtended in each circle is

Thus, the length of the belt around the pulleys is

lim →  2   L 60

0 450

This is not possible Thus,lim does not exist

Review Exercises for Chapter 1

1 Calculus required Using a graphing utility, you can estimate the length to be 8.3.

Or, the length is slightly longer than the distance between the two points, 8.25

1 cos x

x 10 0

Nonremovable discontinuity at each integer k

Continuous on k, k 1for all integers k

Therefore by the Intermediate Value

Theorem, there is at least one value c in such

5 (a)

(b) Slope of tangent line is

Tangent line(c)

(d) The area under the graph of u, and above the x-axis, is 1.

2

a

1

y

C H A P T E R 2 Differentiation

Section 2.1 The Derivative and the Tangent Line Problem 53

Section 2.2 Basic Differentiation Rules and Rates of Change 60

Section 2.3 The Product and Quotient Rules and

Higher-Order Derivatives 67 Section 2.4 The Chain Rule 73 Section 2.5 Implicit Differentiation 79 Section 2.6 Related Rates 85

Problem Solving 98

Differentiation

Solutions to Odd-Numbered Exercises

6 5 4 3 2 1 1

y

x

1 ) 1

f )

1 )x)

) ,

f )

) 1

25 (a)

At the slope of the tangent line is

The equation of the tangent line is

27 (a)

The equation of the tangent line is

At the slope of the tangent line is

The equation of the tangent line is

(1, 1)

31 (a)

At the slope of the tangent line is

The equation of the tangent line is

33 From Exercise 27 we know that Since the

slope of the given line is 3, we have

fx 3x2 35 Using the limit definition of derivative,

Since the slope of the given line is , we have

Therefore, at the point the tangent line is parallel to

The equation of this line is

5, 2

− 2 − 1

− 3

y

(b) If and fc 3 fis even, then fc fc 3

fc fc 3

f

fc 3

47 Let be a point of tangency on the graph of f By the limit definition for the derivative, The slope of the

Therefore, the points of tangency are and and the corresponding slopes are 2 and The equations of the tangentlines are

4 3 5

7 6

1

(1, 3) (3, 3) (2, 5)

(f) No, it is not possible All you can say is that g is decreasing (falling) at x 2

1 32

1 4

27 32

(Sharp turn in the graph)

x 3

fx

77. is differentiable on the interval

(At x 1 the tangent line is vertical)

The derivative from the left is

The derivative from the right is

The one-sided limits are not equal Therefore, f is not

The derivative from the left is

The derivative from the right is

These one-sided limits are equal Therefore, f is differentiable at x 1 f1 0

85 Note that f is continuous at

The derivative from the left is

The derivative from the right is

The one-sided limits are equal Therefore, f is differentiable at x 2 f2 4

The function d is not differentiable at This corresponds to the line

which passes through the point 3, 1

1 2 3 4 1

y

mx  y  4  0

3, 1

Section 2.2 Basic Differentiation Rules and Rates of Change

89 False the slope is lim

x→ 0

f2 x f2

x .

93.

Using the alternative form of the derivative we have

Since this limit does not exist (it oscillates between and 1), the function is not differentiable at

continu-ous at Using the alternative form of the derivative again we have

91 False If the derivative from the left of a point does not equal the derivative from the right of a point, then the derivative does

x 0

fxx,

y13 2

cannot equal zero.

Therefore, there are no horizontal tangents

67 (a) The slope appears to be steepest between A and B. (c)

(b) The average rate of change between A and B is

greater than the instantaneous rate of change at B.

x

f C A B E D

2

x

f f

y

73 Let and be the points of tangency on and respectively The derivatives of

these functions are

) 2

, )

) 0 1

) ,

1) 1,

0.77 3.64

f1 1

79 (a) One possible secant is between and :

(b)

is an approximation of the tangent line

(c) As you move further away from the accuracy of the approximation T gets worse.

Instantaneous rate of change is the constant 2

Average rate of change:

(These are the same because f is a line of slope 2.)

Instantaneous rate of change:

Average rate of change:

f2 f1

2 1 122 11

12

2, 12 ⇒ f214

1, 1 ⇒ f1 1

fxx12

fx 1x, 1, 2

Time (in minutes)

t

2 4 6 8 10 2

4 6 8 10

(10, 6) (8, 4) (6, 4)

T B R

105 (a) is the rate of change of the amount of gasoline sold when the price is $1.47 per gallon

(b) f1.47 is usually negative As prices go up, sales go down

f1.47

Tangent lines through

Tangent lines:

9x  y  0 9x  4y  27  0

y  9x y 9

4x27 4

f must be continuous at to be differentiable at

For f to be differentiable at the left derivative must equal the right derivative

Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives

Function Rewrite Derivative Simplify

2sec x tan x sec2 x3

2sec xtan x  sec x

 4x cos x  2 sin x  x2 sin x

y   2x cos x  2 sin x  x2sin x 2x cos x

Section 2.4 The Chain Rule

does not exist since the left and right derivatives

are not equal

Review Exercises for Chapter 1

1 Calculus required Using a graphing utility, you can estimate the length to be 8.3.

Or,... 53

Differentiation

Solutions to Odd-Numbered Exercises

6 1

y