Calculus early transcendentals 2nd edition briggs test bank

Choose the one alternative that best completes the statement or answers the question... A The limit of fx as x→a from the left exists, the limit of fx as x→a from the right exists, andth

MULTIPLE CHOICE Choose the one alternative that best completes the statement or answers the question Find the average velocity of the function over the given interval.

1) y = x2 + 8x, [5, 8]

638

Use the table to find the instantaneous velocity of y at the specified value of x.

0.48

1.08

1.92

34.32

5.88

11)

t 1.9 1.99 1.999 2.001 2.01 2.1s(t) 5.043 5.364 5.396 5.404 5.436 5.763 ; instantaneous velocity is 5.40C)

t 1.9 1.99 1.999 2.001 2.01 2.1s(t) 5.043 5.364 5.396 5.404 5.436 5.763 ; instantaneous velocity is ∞D)

s(t) 16.810 17.880 17.988 18.012 18.120 19.210 ; instantaneous velocity is 18.0

14)

t -0.1 -0.01 -0.001 0.001 0.01 0.1s(t) -2.9910 -2.9999 -3.0000 -3.0000 -2.9999 -2.9910 ; instantaneous velocity is -3.0C)

t -0.1 -0.01 -0.001 0.001 0.01 0.1s(t) -1.4970 -1.4999 -1.5000 -1.5000 -1.4999 -1.4970 ; instantaneous velocity is -15.0D)

t -0.1 -0.01 -0.001 0.001 0.01 0.1s(t) -1.4970 -1.4999 -1.5000 -1.5000 -1.4999 -1.4970 ; instantaneous velocity is ∞

x→0f(x) does not exist.

21)

24) What conditions, when present, are sufficient to conclude that a function f(x) has a limit as x

approaches some value of a?

A) The limit of f(x) as x→a from the left exists, the limit of f(x) as x→a from the right exists, andthese two limits are the same

B) Either the limit of f(x) as x→a from the left exists or the limit of f(x) as x→a from the right

exists

C) The limit of f(x) as x→a from the left exists, the limit of f(x) as x→a from the right exists, and

at least one of these limits is the same as f(a)

D) f(a) exists, the limit of f(x) as x→a from the left exists, and the limit of f(x) as x→a from the

right exists

24)

Use the graph to evaluate the limit.

25) lim

x→-1f(x)

x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6

y

1

-1

x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-1 -2 -3 -4

x

y 4 3 2 1

-1 -2 -3 -4

26)

27) lim

x→0f(x)

x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6

y 6 5 4 3 2 1 -1 -2 -3 -4 -5 -6

x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6

y 6 5 4 3 2 1 -1 -2 -3 -4 -5 -6

10

8 6 4 2

10

8 6 4 2

-2

-4

28)

-1 -2 -3 -4

x

y 4 3 2 1

-1 -2 -3 -4

-1 -2 -3 -4

x

y 4 3 2 1

-1 -2 -3 -4

30)

-1 -2 -3 -4

x

y 4 3 2 1

-1 -2 -3 -4

-1 -2 -3 -4

x

y 4 3 2 1

-1 -2 -3 -4

32)

-1 -2 -3 -4

x

y 4 3 2 1

-1 -2 -3 -4

Use the table of values of f to estimate the limit.

35) Let f(x) = x2 + 8x - 2, find lim

x 1.9 1.99 1.999 2.001 2.01 2.1f(x) 5.043 5.364 5.396 5.404 5.436 5.763 ; limit = 5.40C)

f(x) 16.692 17.592 17.689 17.710 17.808 18.789 ; limit = 17.70D)

f(x) 1.19245 1.19925 1.19993 1.20007 1.20075 1.20745 ; limit = 1.20C)

f(x) 3.97484 3.99750 3.99975 4.00025 4.00250 4.02485 ; limit = 4.0D)

f(x) 5.07736 5.09775 5.09978 5.10022 5.10225 5.12236 ; limit = 5.10

36)

37) Let f(x) = x2 - 5, find lim

f(x) 0.7802 0.7780 0.7778 0.7778 0.7775 0.7753 ; limit = 0.7778C)

f(x) 3.2222 3.0202 3.0020 2.9980 2.9802 2.8182 ; limit = 3D)

2 - 2 cos(x) < 1 hold for all values of x close

to zero What, if anything, does this tell you about x sin(x)

2 - 2 cos(x) ? Explain.

42)

MULTIPLE CHOICE Choose the one alternative that best completes the statement or answers the question.

43) Write the formal notation for the principle "the limit of a quotient is the quotient of the limits" andinclude a statement of any restrictions on the principle

A) If lim

x→a g(x) = M and limx→a f(x) = L, then limx→a

g(x)f(x) =

limx→a g(x)limx→a f(x)

C) lim

x→a

g(x)f(x) = g(a)f(a), provided that f(a) ≠ 0.

D) If lim

x→a g(x) = M and limx→a f(x) = L, then limx→a

g(x)f(x) =

limx→a g(x)limx→a f(x)

B) The sum or the difference of two functions is the sum of two limits

C) The limit of a sum or a difference is the sum or the difference of the functions

D) The sum or the difference of two functions is continuous

44)

45) The statement "the limit of a constant times a function is the constant times the limit" follows from

a combination of two fundamental limit principles What are they?

A) The limit of a function is a constant times a limit, and the limit of a constant is the constant

B) The limit of a product is the product of the limits, and a constant is continuous

C) The limit of a product is the product of the limits, and the limit of a quotient is the quotient ofthe limits

D) The limit of a constant is the constant, and the limit of a product is the product of the limits

Give an appropriate answer.

5

58)

65) lim

x→0

1 + x - 1x

78) lim

x→-1

x2 - 6x - 7x2 - 2x - 3

78)

79) lim

h → 0

(x + h)3 - x3h

Provide an appropriate response.

81) It can be shown that the inequalities -x ≤ x cos 1

x ≤ x hold for all values of x ≥ 0

x 1.9 1.99 1.999 2.001 2.01 2.1f(x) 5.043 5.364 5.396 5.404 5.436 5.763 ; limit = ∞C)

f(x) 16.692 17.592 17.689 17.710 17.808 18.789 ; limit = 17.70D)

f(x) 16.810 17.880 17.988 18.012 18.120 19.210 ; limit = 18.0

84)

For the function f whose graph is given, determine the limit.

91) Find lim

x→4-f(x) and limx→4+f(x).

x -2 -1 1 2 3 4 5 6 7 8 9 10 11

y 8 6 4 2

-2 -4 -6 -8

x -2 -1 1 2 3 4 5 6 7 8 9 10 11

y 8 6 4 2

-2 -4 -6 -8

y 5 4 3 2 1

-1 -2 -3 -4 -5

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

y 5 4 3 2 1

-1 -2 -3 -4 -5

92)

93) Find lim

x→3f(x).

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

y 5 4 3 2 1

-1 -2 -3 -4 -5

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

y 5 4 3 2 1

-1 -2 -3 -4 -5

93)

94) Find lim

x→-4f(x).

x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6

y 6 5 4 3 2 1 -1 -2 -3 -4 -5 -6

x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6

y 6 5 4 3 2 1 -1 -2 -3 -4 -5 -6

Find the limit.

105)

Find all vertical asymptotes of the given function.

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

2

-2

-4

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

y 4

2

-2

-4

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

y 4

2

-2

-4

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

y 4

2

-2

-4

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

2

-2

-4

117)

118) f(x) = x

x2 + x + 4

A)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

2

-2

-4

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

y 4

2

-2

-4

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

y 4

2

-2

-4

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

y 4

2

-2

-4

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

2

-2

-4

118)

119) f(x) = x2 - 3

x3

A)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

2

-2

-4

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

y 4

2

-2

-4

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

y 4

2

-2

-4

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

y 4

2

-2

-4

x -10 -8 -6 -4 -2 2 4 6 8 10

y 4

2

-2

-4

119)

120) f(x) = 1

x + 1

A)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

B)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

C)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

D)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

120)

121) f(x) = x - 1

x + 1

A)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

B)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

C)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

D)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

121)

122) f(x) = 1

(x + 2)2

A)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

B)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

C)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

D)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

122)

123) f(x) = 2x2

4 - x2A)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

B)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

C)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

D)

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

x -10 -8 -6 -4 -2 2 4 6 8 10

y 10 8 6 4 2

-2 -4 -6 -8 -10

127) lim

x→∞

x2 + 7x + 3x3 + 6x2 + 4

1-4

135)

136) lim

x→∞

-3x-1 - 2x-3-2x-2 + x-5

107

x y

153)

154) f(0) = 5, f(1) = -5, f(-1) = -5, lim

x→±∞f(x) = 0.

x y

159) Find the horizontal asymptote, if any, of the given function.

f(x) = 2x3 - 3x - 9

9x3 - 5x + 3A) y = 3

Provide an appropriate response.

169) Is f continuous at f(1)?

f(x) =

-x2 + 1,4x,-5,-4x + 81,

d 6 5 4 3 2 1 -1 -2 -3 -4 -5 -6 (1, -5)

t -6 -5 -4 -3 -2 -1 1 2 3 4 5 6

d 6 5 4 3 2 1 -1 -2 -3 -4 -5 -6 (1, -5)

d 6 5 4 3 2 1 -1 -2 -3 -4 -5 -6

(1, -2)

t -6 -5 -4 -3 -2 -1 1 2 3 4 5 6

d 6 5 4 3 2 1 -1 -2 -3 -4 -5 -6 (1, -2)

d 10 8 6 4 2

-2 -4 -6 -8 -10

(2, 0)

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

d 10 8 6 4 2

-2 -4 -6 -8 -10 (2, 0)

171)

172) Is f continuous at x = 4?

f(x) =

x3,-2x,3,0,

d 10 8 6 4 2

-2 -4 -6 -8 -10

(2, 0)

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

d 10 8 6 4 2

-2 -4 -6 -8 -10 (2, 0)

172)

173) Is the function given by f(x) = x + 2

x2 - 3x + 2 continuous at x = 1? Why or why not?

A) Yes, lim

x→ 1f(x) = f(1)B) No, f(1) does not exist and lim

x→ 1 f(x) does not exist

175) Is the function given by f(x) = x2 - 3, for x < 0

-4, for x ≥ 0 continuous at x = -3? Why or why not?

continuous at x = 2? Why or why not?

C) discontinuous only when x = -7 D) discontinuous only when x = 7

177)

(x + 5)2 + 10

A) discontinuous only when x = 35 B) discontinuous only when x = -5

178)

179) y = x + 3

x2 - 5x + 4

A) discontinuous only when x = -4 or x = 1 B) discontinuous only when x = 1

C) discontinuous only when x = -1 or x = 4 D) discontinuous only when x = 1 or x = 4

179)

180) y = 1

x2 - 9

A) discontinuous only when x = -3 B) discontinuous only when x = -9 or x = 9

C) discontinuous only when x = 9 D) discontinuous only when x = -3 or x = 3

180)

181) y = 1

x + 4 - x2

5A) discontinuous only when x = -4 B) discontinuous only when x = -9

181)

182) y = sin (4θ)

2θ

C) discontinuous only when θ = π

A) continuous on the interval - 67, ∞ B) continuous on the interval -∞, - 67

C) continuous on the interval 6

7, ∞ D) continuous on the interval -

C) continuous on the interval -∞, 2

2

3, ∞

185)

186) y = x2 - 7

A) continuous on the interval [ 7, ∞)

B) continuous on the interval [- 7, 7]

C) continuous on the intervals (-∞, - 7] and [ 7, ∞)

187)

188) lim

x→∞

5x - 1x3

d 10 8 6 4 2

-2 -4 -6 -8 -10

(2, 0)

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

d 10 8 6 4 2

-2 -4 -6 -8 -10 (2, 0)

x > -3

201)

Solve the problem.

206) Select the correct statement for the definition of the limit: lim

x→x0f(x) = Lmeans that

A) if given a number ε > 0, there exists a number δ > 0, such that for all x,

207) Identify the incorrect statements about limits

I The number L is the limit of f(x) as x approaches x0 if f(x) gets closer to L as x approaches x0

II The number L is the limit of f(x) as x approaches x0 if, for any ε > 0, there corresponds a δ > 0

such that f(x) - L < ε whenever 0 < x - x0 < δ

III The number L is the limit of f(x) as x approaches x0 if, given any ε > 0, there exists a value of x

for which f(x) - L < ε

207)

Use the graph to find a δ > 0 such that for all x, 0 < x - x0 < δ ⇒ f(x) - L < ε.

208)

x y

y

0

y = x + 34.2

y

0

y = 4x - 35.2

5

4.8

 2 1.95 2.05

NOT TO SCALE

f(x) = 4x - 3x0 = 2

L = 5

ε = 0.2

209)

x y

x y

L = 2

ε = 14

214)

215)

x y

L = 2

ε = 14

215)

x y

y

0

y = 2x23

y

0

y = x2 - 14

-2 -4 -6 -8

x

y 8 6 4 2

-2 -4 -6 -8

153) Answers may vary One possible answer:

x

y 8 6 4 2

-2 -4 -6 -8

x

y 8 6 4 2

-2 -4 -6 -8

Answer Key

Testname: UNTITLED2

154) Answers may vary One possible answer:

x -12 -10 -8 -6 -4 -2 2 4 6 8 10 12

y 12 10 8 6 4 2 -2 -4 -6 -8 -10 -12

x -12 -10 -8 -6 -4 -2 2 4 6 8 10 12

y 12 10 8 6 4 2 -2 -4 -6 -8 -10 -12

155) Answers may vary One possible answer:

x

y 2

-2

x

y 2

198) Let f(x) = 8x4 + 4x3 - 7x - 5 and let y0 = 0 f(-1) = 6 and f(0) = -5 Since f is continuous on [-1, 0] and since y0 = 0 isbetween f(-1) and f(0), by the Intermediate Value Theorem, there exists a c in the interval (-1, 0) with the propertythat f(c) = 0 Such a c is a solution to the equation 8x4 + 4x3 - 7x - 5 = 0.

199) Let f(x) = x(x - 6)2 and let y0 = 6 f(5) = 5 and f(7) = 7 Since f is continuous on [5, 7] and since y0 = 6 is between f(5)and f(7), by the Intermediate Value Theorem, there exists a c in the interval (5, 7) with the property that f(c) = 6 Such a

c is a solution to the equation x(x - 6)2 = 6

200) Let f(x) = sin x

x and let y0 = 16 f π

2 ≈ 0.6366 and f(π) = 0 Since f is continuous on π

2, π and since y0 = 16 is between fπ

2 and f(π), by the Intermediate Value Theorem, there exists a c in the interval

π

2, π , with the property that f(c) =

1

6.Such a c is a solution to the equation 6 sin x = x

224) Let ε > 0 be given Choose δ = ε Then 0 < x - 7 < δ implies that

x - 7 - 14 < ε225) Let ε > 0 be given Choose δ = ε/2 Then 0 < x - 9 < δ implies that

x - 9 - 21 < ε226) Let ε > 0 be given Choose δ = min{7/2, 49ε/2} Then 0 < x - 7 < δ implies that

1

7 ∙

49ε

2 = εThus, 0 < x - 7 < δ implies that 1

x - 1

7 < ε