TEST BANK AND SOLUTION MANUAL FOR CALCULUS EARLY TRANSCENDENTALS 9TH EDITION BY JAMES STEWART, DANIEL & SALEEM WATSON –CALCULUS PROBLEM-SOLVING RESOURCE

TEST BANK AND SOLUTION MANUAL FOR CALCULUS EARLY TRANSCENDENTALS 9TH EDITION BY
JAMES STEWART, DANIEL & SALEEM WATSON –CALCULUS PROBLEM-SOLVING RESOURCE




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

Find the average velocity of an object which follows the position function s(t) over the given interval.
1) s(t) = t2 + 2t, [2, 9] 1)

A) 13 B) C) D) 11

2) s(t) = 4t3 + 8t2 - 5, [-8, -2] 2)


A) 256 B) - 768 C) - D)

3) s(t) = 2t, [2, 8] 3)

A) 2 B) 7 C) D) -



4) s(t) = 3 , [4, 7] 4)
t-2

A) B) 7 C) - D) 2



5) s(t) = 4t2, 0, 7 5)
4

A) 7 B) C) - D) 2

6) s(t) =-3t2 - t, [5, 6] 6)




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A) -34 B) -2 C) - D)



7) s(t) = sin (3t), 0, π 7)
6

A) - B) C) D)



8) s(t) = 3 + tan t, - π, π 8)
4 4

A) 0 B) - C) D) -
Use the table to find the instantaneous velocity of y at the specified value of x.
9) x = 1. 9)
x y
0 0
0.2 0.02
0.4 0.08
0.6 0.18
0.8 0.32
1.0 0.5
1.2 0.72
1.4 0.98
A) 1 B) 0.5 C) 2 D) 1.5


10) x = 1. 10)
x y
0 0
0.2 0.01
0.4 0.04
0.6 0.09
0.8 0.16
1.0 0.25
1.2 0.36
1.4 0.49
A) 1 B) 0.5 C) 2 D) 1.5


11) x = 1. 11)




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,TEST BANK AND SOLUTION MANUAL FOR CALCULUS EARLY TRANSCENDENTALS 9TH EDITION BY
JAMES STEWART, DANIEL & SALEEM WATSON –CALCULUS PROBLEM-SOLVING RESOURCE




x y
0 0
0.2 0.12
0.4 0.48
0.6 1.08
0.8 1.92
1.0 3
1.2 4.32
1.4 5.88
A) 2 B) 8 C) 6 D) 4
12) x = 2. 12)
x y
0 10
0.5 38
1.0 58
1.5 70
2.0 74
2.5
70
3.0
58
3.5
4.0 38
10
A) -8 B) 0 C) 4 D) 8




x y




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13) x = 1. 0.900 -0.05263 13)
0.990 -0.00503
0.999 -0.0005
1.000 A) slope is - B) slope is C) slope is 13 D) slope is -39
0.0000
1.001
15) y = x2 0.0005 + 11x - 15, x = 1 15)
1.010
1.100 0.00498
A) slope 0.04762 is - B) slope is 13 C) slope is D) slope is -39


A) 0 B) -0.5 C) 0.5 D) 1

Find the slope of the curve for the given value of x.
14) y = x2 + 5x, x = 4 14)

16) y = x3 - 8x, x = 1 16)

A) slope is -5 B) slope is 1 C) slope is -3 D) slope is 3

17) y = x3 - 2x2 + 4, x = 3 17)

A) slope is 1 B) slope is 0 C) slope is -15 D) slope is 15

18) y = 4 - x3, x =-1 18)

A) slope is 3 B) slope is -1 C) slope is -3 D) slope is 0

Solve the problem.




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