Calculus 2nd edition briggs test bank

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

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.

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)

s(t) 16.692 17.592 17.689 17.710 17.808 18.789 ; instantaneous velocity is 17.70C)

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)

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 5.40

14)

s(t) -4.9900 -4.9999 -5.0000 -5.0000 -4.9999 -4.9900 ; instantaneous velocity is-5.0

C)

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 ∞D)

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-3.0

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) 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

B) 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)

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, andthese two limits are the same

D) 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

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 = ∞C)

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

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 = 5.40

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

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) -1.0690 -1.0067 -1.0007 -0.9993 -0.9934 -0.9355 ; limit = -1C)

f(x) 0.2304 0.2416 0.2427 0.2430 0.2441 0.2549 ; limit = 0.2429D)

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).

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 limit of a sum or a difference is the sum or the difference of the functions

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

D) The limit of a sum or a difference is the sum or the difference of the limits

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 product is the product of the limits, and a constant is continuous

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

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

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.

58)

65) lim

x→0

1 + x - 1x

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

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

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 = 5.40D)

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 = ∞

84)

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

91) Find lim

x→5-f(x) and limx→5+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→-3f(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.

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 + 2

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

Divide numerator and denominator by the highest power of x in the denominator to find the limit.

x y

156)

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

x→±∞ f(x) = 0.

x y

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)

175)

176) Is f continuous at f(0)?

f(x) =

-x2 + 1,2x,-5,-2x + 4 1,

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 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)

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)

178)

179) Is the function given by f(x) = x + 1

x2 - 8x + 12 continuous at x = 2? Why or why not?

A) Yes, lim

x→ 2f(x) = f(2)B) No, f(2) does not exist and lim

x→ 2 f(x) does not exist

179)

180) Is the function given by f(x) = 10x + 9 continuous at x = - 109 ? Why or why not?

A) No, lim

x→ - 910

x→- 910

f(x) = f - 9

10

180)

181) Is the function given by f(x) = x2 - 4, for x < 0

-2, for x ≥ 0 continuous at x = -2? Why or why not?

continuous at x = 3? Why or why not?

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

183)

(x + 4)2 + 8

C) discontinuous only when x = -4 D) discontinuous only when x = -32

184)

185) y = x + 3

x2 - 7x + 12

A) discontinuous only when x = 3 B) discontinuous only when x = 3 or x = 4

C) discontinuous only when x = -3 or x = 4 D) discontinuous only when x = -4 or x = 3

185)

186) y = 2

x2 - 16

A) discontinuous only when x = -4

B) discontinuous only when x = 16

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

186)

187) y = 2

x + 1 -

x22

C) discontinuous only when x = -3 D) discontinuous only when x = -2 or x = -1

187)

188) y = sin (3θ)

2θ

C) discontinuous only when θ = π

188)

189) y = 2 cos θ

θ + 9A) discontinuous only when θ = π

C) discontinuous only when θ = 9 D) discontinuous only when θ = -9

189)

190) y = 2x + 2

A) continuous on the interval 1, ∞ B) continuous on the interval - 1, ∞

C) continuous on the interval -∞, - 1 D) continuous on the interval - 1, ∞

190)

191) y = 49x - 9

A) continuous on the interval 1, ∞ B) continuous on the interval -∞, 1

C) continuous on the interval 1, ∞ D) continuous on the interval - 1, ∞

191)

192) y = x2 - 10

A) continuous on the intervals (-∞, - 10] and [ 10, ∞)

B) continuous on the interval [- 10, 10]

C) continuous on the interval [ 10, ∞)

193)

194) lim

x→∞

11x - 1x3

194)

Provide an appropriate response.

195) Is f continuous on (-2, 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)

SHORT ANSWER Write the word or phrase that best completes each statement or answers the question Provide an appropriate response.

203) Use the Intermediate Value Theorem to prove that 2x3 - 7x2 - 9x + 4 = 0 has a solution

Solve the problem.

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

x→x0f(x) = L means that

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

213) 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

y

0

y = 2x + 35.2

L = 5

ε = 0.2

214)

x y

y

0

y = 4x - 35.2

5

4.8

 2 1.95 2.05

NOT TO SCALE

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

NOT TO SCALE

216)

0

y = 32x + 12.7

L = 2.5

ε = 0.2

218)

L = 2 3

ε = 14

220)

x y

y

0

y = x - 21.25

L = 1

ε = 14

221)

222)

x y

y

0

y = x25

L = 4

ε = 1

222)

x y

y

0

y = x2 - 37

SHORT ANSWER Write the word or phrase that best completes each statement or answers the question.

Prove the limit statement

-2 -4 -6 -8

x

y 8 6 4 2

-2 -4 -6 -8

156) 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

157) 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

158) Answers may vary One possible answer:

x

y 2

-2

x

y 2

204) Let f(x) = 3x4 - 2x3 + 7x - 5 and let y0 = 0 f(-2) = 45 and f(-1) = -7 Since f is continuous on [-2, -1] and since y0 = 0 isbetween f(-2) and f(-1), by the Intermediate Value Theorem, there exists a c in the interval (-2, -1) with the propertythat f(c) = 0 Such a c is a solution to the equation 3x4 - 2x3 + 7x - 5 = 0.

205) Let f(x) = x(x - 7)2 and let y0 = 7 f(6) = 6 and f(8) = 8 Since f is continuous on [6, 8] and since y0 = 7 is between f(6)and f(8), by the Intermediate Value Theorem, there exists a c in the interval (6, 8) with the property that f(c) = 7 Such

a c is a solution to the equation x(x - 7)2 = 7

206) 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

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

x - 7 - 14 < ε231) Let ε > 0 be given Choose δ = ε/2 Then 0 < x - 9 < δ implies that

x - 9 - 21 < ε232) Let ε > 0 be given Choose δ = min{7/2, 49ε/2} Then 0 < x - 7 < δ implies that

1

x- 1

7 = 7 - x7x

= 1

x ∙ 1

7 ∙ x - 7

< 17/2 ∙ 1

7 ∙ 49ε

2 = ε

Thus, 0 < x - 7 < δ implies that 1

x - 1

7 < ε