Honors Calculus
1st Semester Final Review WS
Name: ______________________________________________
Block: _____
SHOW ALL WORK !!!
Calculate the derivative of the function using the limit
definition of derivative. Then find the value of the
derivative as specified.
′
3) f(x) = 5x + 9; f (2)
Use the definition f'(a) = lim f(a + h) - f(a) to find the
h
h->0
derivative of the given function at the indicated point.
1) f(x) = 17 - 10x, a = 1
′
4) g(x) = 3x 2 - 4x; g (3)
2) f(x) = 3x2, a = 1
1
For quions 5 - 13:
(a) Find all points where the function is discontinuous
and give the type of discontinuity.
(b)List all points where the function is not differentiable.
5)
9)
10)
6)
11)
7)
12)
8)
13)
2
Calculate the derivative of the function. Then find the
value of the derivative as specified.
′
14) f(x) = 5x + 9; f (2)
23) y = 13x-2 - 6x3 - 4x
24) y = 5x2 + 7x + 5x-3
′
15) g(x) = 3x 2 - 4x; g (3)
25) w = z-3 - 1
z
′
16) f(x) = x 2 + 7x - 2; f (0)
26) r = 4 - 8
s
s3
′
17) g(x) = x 3 + 5x; g (1)
27) y = 1 + 1
7x
5x2
′
18) f(x) = 8 ; f (-1)
x
Find the derivative.
19) y = 8 - 8x2
Find the second derivative.
28) y = 6x2 + 12x + 2x-3
29) w = z-6 - 1
z
20) y = 4 - 4x3
21) y = 2x4 - 8x3 - 7
30) r = 2 - 5
s
s3
22) s = 3t2 + 8t + 5
31) y = 1 + 1
9x
5x2
3
′
Using the Product Rule, Find y .
You do not need to simplify.
32) y = (4x - 4)(5x + 1)
2
38) g(x) = x + 5
x2 + 6x
33) y = (5x - 4)(2x3 - x 2 + 1)
2
39) y = x + 8x + 3
x
34) y = (x2 - 4x + 2)(4x3 - x 2 + 5)
2
40) y = x + 2x - 2
x2 - 2x + 2
35) y = (5x3 + 7)(3x7 - 4)
Find the second derivative of the function.
4
41) y = x + 2
x2
Find the derivative of the function.
2
36) y = x - 3x + 2
x7 - 2
7
42) s = t + 7t + 4
t2
3
37) y = x
x-1
4
Suppose u and v are differentiable functions of x. Use
the given values of the functions and their derivatives to
find the value of the indicated derivative.
′
′
43) u(1) = 2, u (1) = -6, v(1) = 6, v (1) = -4.
d (uv) at x = 1
dx
′
Solve the problem.
49) Find an equation of the tangent line to the graph
of y = x2 - x, at the point (-2, 6).
′
44) u(1) = 5, u (1) = -5, v(1) = 6, v (1) = -2.
d u at x = 1
dx v
′
50) Find an equation of the tangent line to the graph
of y = x - x2 at the point (-1, -2).
′
45) u(1) = 3, u (1) = -5, v(1) = 7, v (1) = -4.
d v at x = 1
dx u
51) If y = x3 - 4x - 3, find an equation of the tangent
line to the graph of y at x = 2.
′
′
′
′
46) u(1) = 2, u (1) = -6, v(1) = 6, v (1) = -4.
d (2u - 4v) at x = 1
dx
47) u(2) = 8, u (2) = 2, v(2) = -3, v (2) = -5.
d (uv) at x = 2
dx
′
52) Find an equation of the tangent line to the graph
of y = 10 x - x + 1 at the point
(100, 1).
′
48) u(2) = 9, u (2) = 2, v(2) = -3, v (2) = -4.
d u at x = 2
dx v
5
The function s = f(t) gives the position of a body moving
on a coordinate line, with s in meters and t in seconds.
53) s = 5t2 + 3t + 7, 0 ≤ t ≤ 2
58) s = - t3 + 2t2 - 2t, 0 ≤ t ≤ 2
Find the body's speed and acceleration at the
end of the time interval.
Find the body's displacement and average
velocity for the given time interval.
54) s = 7t - t2, 0 ≤ t ≤ 7
Find the body's displacement and average
velocity for the given time interval.
The figure shows the velocity v or position s of a body
moving along a coordinate line as a function of time t .
Use the figure to answer the question.
59)
v (ft/sec)
t (sec)
55) s = - t3 + 6t2 - 6t, 0 ≤ t ≤ 6
Find the body's displacement and average
velocity for the given time interval.
When is the body's acceleration equal to zero?
56) s = 6t2 + 3t + 2, 0 ≤ t ≤ 2
Find the body's speed and acceleration at the
end of the time interval.
60)
v (ft/sec)
t (sec)
57) s = 3t - t2, 0 ≤ t ≤ 3
Find the body's speed and acceleration at the
end of the time interval.
What is the body's greatest velocity?
6
61)
v (ft/sec)
64)
v (ft/sec)
t (sec)
t (sec)
When is the body moving backward?
When does the body reverse direction?
62)
65)
v (ft/sec)
s (m)
t (sec)
t (sec)
When is the body moving forward?
What is the body's acceleration when t = 1 sec?
63)
66)
v (ft/sec)
s (m)
t (sec)
t (sec)
When is the body standing still?
What is the body's speed when t = 1 sec?
7
67)
s (m)
Solve the problem.
69) The position of a body moving on a coordinate
line is given by s = t2 - 9t + 7, with s in meters
and t in seconds. When, if ever, during the
interval 0 ≤ t ≤ 9 does the body change
direction?
t (sec)
When is the body moving backward?
68)
s (m)
70) At time t, the position of a body moving along
the s-axis is s = t3 - 21t2 + 120t m. Find the
body's acceleration each time the velocity is
zero.
t (sec)
What is the body's velocity when t = 8 sec?
8
71) At time t, the position of a body moving along
the s-axis is s = t3 - 18t2 + 60t m. Find the total
74) A ball dropped from the top of a building has a
height of s = 400 - 16t2 meters after t seconds.
distance traveled by the body from t = 0 to t = 3.
How long does it take the ball to reach the
ground? What is the ball's velocity at the
moment of impact?
72) At time t ≥ 0, the velocity of a body moving
along the s-axis is v = t2 - 8t + 7. When is the
body moving backward?
75) A rock is thrown vertically upward from the
surface of an airless planet. It reaches a height of
s = 120t - 12t2 meters in t seconds. How high
does the rock go? How long does it take the
rock to reach its highest point?
73) At time t ≥ 0, the velocity of a body moving
along the s-axis is v = t2 - 5t + 4. When is the
body's velocity increasing?
9
Answer Key
Testname: FINAL (UNIT 3) REVIEW WS
1) - 10
2) 6
′
′
3) f (x) = 5; f (2) = 5
′
′
4) g (x) = 6x - 4; g (3) = 14
5) x = 4
6) x = 1
7) x = -2, x = 0, x = 2
8) x = 6
9) None
10) None
11) x = 3
12) x = -2, x = 2
13) x = 0
′
′
14) f (x) = 5; f (2) = 5
′
′
15) g (x) = 6x - 4; g (3) = 14
′
′
16) f (x) = 2x + 7; f (0) = 7
′
′
17) g (x) = 3x 2 + 5; g (1) = 8
′
′
18) f (x) = - 8 ; f (-1) = -8
2
x
19) -16x
20) -12x2
21) 8x3 - 24x2
22) 6t + 8
23) -26x-3 - 18x2 - 4
24) 10x + 7 - 15x-4
25) -3z-4 + 1
z2
26) - 12 + 8
s4
s2
27) - 2 - 1
5x3
7x2
28) 12 + 24x-5
29) 42z-8 - 2
z3
30)
31)
24 - 10
s5
s3
6
5x4
+
2
9x3
32) 40x - 16
33) 40x3 - 39x2 + 8x + 5
34) 20x4 - 68x3 + 36x2 + 6x - 20
35) 150x9 + 147x6 - 60x2
1
Answer Key
Testname: FINAL (UNIT 3) REVIEW WS
8
7
6
′
36) y = -5x + 18x - 14x - 4x + 6
(x7 - 2)2
3
2
′
37) y = 2x - 3x
(x - 1)2
2
′
38) g (x) = 6x - 10x - 30
x2(x + 6)2
2
′
39) y = 3x + 8x - 3
2x3/2
′
40) y =
-4x2 + 8x
(x2 - 2x + 2)2
41)
d2y = 2 + 12
dx2
x4
42)
d2s = 20t3 + 14 + 24
dt2
t3
t4
43) -44
5
44) 9
45)
23
9
46) 4
47) -46
10
48)
3
49) y = -5x - 4
50) y = 3x + 1
51) y = 8x - 19
52) y = - 1 x + 51
2
53) 26 m, 13 m/sec
54) 0 m, 0 m/sec
55) -36 m, -6 m/sec
56) 27 m/sec, 12 m/sec2
57) 3 m/sec, - 2 m/sec2
58) 6 m/sec, -8 m/sec2
59) 2 < t < 3, 5 < t < 6
60) 5 ft/sec
61) t = 4 sec
62) -1 ft/sec2
63) 1 ft/sec
64) 0 < t < 4
2
Answer Key
Testname: FINAL (UNIT 3) REVIEW WS
65) 0 < t < 1, 3 < t < 4, 5 < t < 7, 9 < t < 10
66) 1 < t < 2
67) 2 < t < 3, 4 < t < 5, 7 < t < 9
68) -4 m/sec
69) t = 4.5 sec
70) a(10) = 18 m/sec2, a(4) = -18 m/sec2
71) 101 m
72) 1 < t < 7
73) t > 2.5
74) 5 sec, -160 m/sec
75) 300 m, 5 sec
3