AP Calculus Chapter 3 Testbank
(Mr. Surowski)
Part I. Multiple-Choice Questions (5 points each; please circle the correct
answer.)
4
1. If f (x) = 10x 3 + x, then f 0 (8) =
(A) 11
(B)
41
3
(C)
3x2 + x
, then g 0 (x) =
2
3x − x
6x2 + 1
−6
(B) 2
(C)
6x − 1
(3x − 1)2
83
3
(D)
21
3
(E) 21
2. If g(x) =
(A) 1
p √
3. If f (x) = 1 + x, then f 0 (x) =
−1
(A) √ p √
4 x 1+ x
1
(B) √ p √
2 x 1+ x
1
(C) p √
4 1+ x
(D)
1
√ p √
4 x 1+ x
−1
(E) √ p √
2 x 1+ x
(D)
−2x2
(x2 − x)2
(E)
36x2 − 2x
(x2 − x)2
4. Find
dy
given that x3 y + xy 3 = −10.
dx
(A) 3x2 + 3xy 2
(B) −(3x2 + 3xy 2 )
3x2 y + y 3
(C)
3xy 2 + x3
3x2 y + y 3
(D) −
3xy 2 + x3
x2 y + y 3
(E) − 2
xy + x3
5. If f (x) = sin2 x, find f 000 (x).
(A) − sin2 x
(B) 2 cos 2x
(C) cos 2x
6. If f (x) = 3πx , find f 0 (x) =
3πx
3πx
3πx
(A)
(B)
(C)
π ln 3
ln 3
π
(D) −4 sin 2x
(D) π 3πx−1
(E) − sin 2x
(E) π ln 3 (3πx )
7. Find the slope of the normal line to the graph of y = x + cos xy at the point
(0, 1).
(A) 1
(B) −1
(C) 0
(D) 2
(E) Undefined
8. If f (x) = 3x2 − x and g(x) = f −1 (x), then g 0 (10) could be
1
1
1
(C) 11
(D)
(E)
59
11
10
Note that f (x) = 3x2 − x =⇒ x = g(3x2 − x). Differentiate this and get
1 = (6x − 1)g 0 (3x2 − x). One of the solutions of 3x2 − x = 10 is x = 2; for this
value get 1 = 11g 0 (10).
(A) 59
(B)
(
ax3 − 6x
9. If the function f (x) is differentiable and f (x) =
bx2 + 4
a=
(A) 0
(B) 1
(C) −14
(D) −24
if x ≤ 1
then
if x > 1,
(E) 26
10. Two particles leave the origin at the same time and move along the y-axis
with their respective positions determined by the functions y1 = cos 2t and
y2 = 4 sin t for 0 < t < 6. For how many values of t do the particles have the
same acceleration?
(A) 0
(B) 1
(C) 2
(D) 3
dy
at y = 1 given that x2 y + y 2 = 5.
dx
2
3
2
(B) − only
(C) only
(D) ±
3
2
3
(E) 4
11. Find the value(s) of
3
(A) − only
2
(E) ±
3
2
√
12. If f (x) = x2 3x + 1, then f 0 (x) =
−3x2 − 2x
(A) √
3x + 1
9x2 + 2x
(B) √
3x + 1
−9x2 + 4x
(C) √
2 3x + 1
15x2 + 4x
(D) √
2 3x + 1
−9x2 − 4x
(E) √
2 3x + 1
13. What is the instantaneous rate of change at t = −1 of the function f , if
t3 + t
f (t) =
?
4t + 1
12
4
20
4
12
(A)
(B)
(C) −
(D) −
(E) −
9
9
9
9
9
14. What is the equation of the line tangent to the graph of y = sin2 x at x =
π
?
4
1
π
(A) y − = − x −
2
4
1
π
(B) y − = x −
2
4
1
π
(C) y −√ = x −
4
2
1
π
1
(D) y −√ =
x−
4
2 2
π
1 1
x−
(E) y − =
2 2
4
15. If the function f (x) =
(
3ax2 + 2bx + 1
real values of x, then b =
11
1
(A) −
(B)
4
4
if x ≤ 1
ax4 − 4bx2 − 3x
(C) −
if x > 1
7
16
is differentiable for all
(D) 0
(E) −
1
4
16. The position of a particle moving along the x-axis at time t is given by x(t) =
ecos 2t , 0 ≤ t ≤ π. For which of the following values of t will x0 (t) = 0?
π
I. t = 0
II. t =
III. t = π
2
(A) I only
(B) II only
(C) I and III only
(D) I and II only
(E) I, II, and III
17. If f (x) = (3x)3x , then f 0 (x) =
(A) (3x)3x (3 ln(3x) + 3)
(B) (3x)3x (3 ln(3x) + 3x)
(C) (9x)3x (ln(3x) + 1)
(D) (3x)3x−1 (3x)
(E) (3x)3x−1 (9x)
18. Given that f (x) = 2x2 + 4, which of the following will calculate the derivative
of f (x)?
[2(x + ∆x)2 + 4] − (2x2 + 4)
∆x
2
(2x + 4 + ∆x) − (2x2 + 4)
(B) lim
∆x→0
∆x
(A)
[2(x + ∆x)2 + 4] − (2x2 + 4)
(C) lim
∆x→0
∆x
(2x2 + 4 + ∆x) − (2x2 + 4)
∆x
(E) None of the above.
(D)
19. Given that g(x) =
1
, which of the following will calculate the derivative
x+1
of g(x)?
1
1
−
x + ∆x + 1 x + 1
1
1
1
(B) lim
−
∆x→0 ∆x
x + ∆x + 1 x + 1
1
1
1
−
(C) lim
lim
∆x→0 ∆x ∆x→0 x + ∆x + 1
x+1
1
1
(D) lim
−
∆x→0 x + ∆x + 1
x+1
(E) None of the above.
1
(A)
∆x
The next two questions pertain to the function f , whose graph is given below:
20. For the function f ,
10 y
I. f 0 (−3) > 0
y=f(x)
II. f 0 (0) < 0
III. f is differentiable
on the interval
(0, 1)
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I, II, and III
5
x
-6
-4
-2
2
-5
-10
21. For the function f
I. f 0 (x) > 0 on the interval (−5, −4)
II. f 0 (x) is constant on the interval (4, 6)
III. f 0 is not defined at all points of the interval (1, 5)
(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III
4
6
22. Given the graph of the rational function f below, give a sketch of the graph
of y = f 0 (x) on the same coordinate axes. (Note: the graph of y = f (x)
has a vertical asymptote at x = 1.) The graph of y = f 0 (x) is in blue .
8
Y
6
4
2
-4 -2
2
-2
4
X
-4
23. The following graph represents a function g defined on the interval [−4, 4] and
differentiable on (−4, 4). On the same coordinate axes, graph y = g 0 (x) over
the interval (−3, 3). The graph of y = g 0 (x) is in blue .
4
Y
2
-4 -2
2
-2
-4
4
X
24. The following graph is that of y = h0 (x). On the same coordinate axes, give
a sketch of y = h(x), assuming that h(0) = 1. The graph of y = h(x) is in
blue .
Y
4
3
2
1
-4 -2
-1
-2
2
4
X
25. The following graph is that of y = h0 (x). On the same coordinate axes, give
a sketch of y = h(x), assuming that h(0) = 0. The graph of y = h(x) is in
blue .
Y
X
26. Using the definition of the derivative of a function, find f 0 (x), where f (x) =
x − x4 . Then find f 0 (1).
f (x + h) − f (x)
h→0
h
(x + h) − (x + h)4 − (x − x4 )
lim
h→0
h
4
(x + h) − (x + 4x3 h + 6x2 h2 + 4xh3 + h4 ) − (x − x4 )
lim
h→0
h
3
2 2
3
h − (4x h + 6x h + 4xh + h4 )
lim
h→0
h
3
2
lim [1 − (4x + 6x h + 4xh2 + h3 )] = 1 − 4x3 .
f 0 (x) = lim
=
=
=
=
h→0
That is to say, f 0 (x) = 1 − 4x3 , and so f 0 (1) = 1 − 4 · 13 = −3.
√
dy
where y = x.
27. Using the definition of the derivative of a function, find
dx
dy Then find
.
dx x=4
dy
f (x + h) − f (x)
= lim
h→0
dx
h
√
√
x+h− x
= lim
h→0
√ h
√ √
√
( x + h − x)( x + h − x)
√
= lim
√
h→0
h( x + h + x)
x+h−x
= lim √
√
h→0 h( x + h +
x)
h
= lim √
√
h→0 h( x + h +
x)
1
1
= lim √
√ = √
h→0
x+h+ x 2 x
1
dy 1
Therefore,
= √ = .
dx x=4 2 4 4
28. Let f (x) = 4x3 − 21x2 − 24x + 23.
(a) Compute f 0 (x). f 0 (x) = 12x2 − 42x − 24.
(b) Find all values of x satisfying f 0 (x) = 0. f 0 (x) = 0 ⇒ 12x2 − 42x − 24 =
0 ⇒ 2x2 − 7x − 4 = 0 ⇒ (2x + 1)(x − 4) = 0 ⇒ x = − 12 , 4.
1
29. Let f (x) = x + .
x
(a) Compute f 0 (x). f 0 (x) = 1 −
1
x2
(b) Find all values of x satisfying f 0 (x) = 0.
1
x2 − 1
We have 0 = f 0 (x) = 1 − 2 =
⇒ x = ±1.
x
x2
30. Let y =
x
.
1 + x2
dy
.
dx
Using the quotient rule, one has
(a) Compute
d
+ x2 ) − x dx
(1 + x2 )
(1 + x2 )2
(1 + x2 ) − x(2x)
1 − x2
=
.
=
(1 + x2 )2
(1 + x2 )2
dy
=
dx
d
dx (x)(1
(b) Compute all values of x for which
From the above, it’s clear that
dy
= 0.
dx
dy
= 0 ⇒ x = ±1.
dx
31. Let s(x) =
sin x
and compute lim s0 (x).
x→∞
x
x cos x − sin x
. Therefore,
x2
cos
x
sin
x
lim s0 (x) = lim
− 2
= 0 − 0 = 0.
x→∞
x→∞
x
x
We have, using the quotient rule, that s0 (x) =
32. The graph below depicts the velocity v = s0 (t) of a particle moving along a
straight line, where on this straight line positive direction is to the right.
v (velocity)
4
2
t (time)
2
4
6
8
10
-2
-4
(a) Would you say that at time t = 1 the particle is moving to the left,
moving to the right, or not moving at all? Please explain. Particle
is moving to the right, as v(1) > 0.
(b) Would you say that at time t = 3 the particle is moving to the left,
moving to the right, or not moving at all? Please explain. Particle
is moving to the left, as v(3) < 0.
(c) Would you say that at time t = 4 the particle is moving to the left,
moving to the right, or not moving at all? Please explain. Particle
is not moving, as v(4) = 0.
(d) Find (estimate) two values of t at which time the particle is not accelerating. It appears that a(t) = v 0 (t) = 0 where t ≈ 1 or 3. At such values
of t the particle will not be accelerating.
(e) Find (estimate) a value of t at which time the particle is moving to the
left, but with zero acceleration. This would happen at t ≈ 3 as a(3) ≈ 0
and v(3) < 0.
(f) According to this graph, at how many distinct times is the particle at
rest? The particle is at rest when it’s not moving; this happens for FIVE
values of t.
(g) For which values of t is the particle not only at rest, but is not accelerating
(i.e., has no forces acting on it)? There are NO SUCH values of t.
(h) According to this graph, at how many distinct times is the particle not
accelerating? We have a(t) = 0 for FOUR values of t.
33. Using logarithmic differentiation compute f 0 (x) where f (x) = xx , x > 0.
Starting with ln f (x) = ln xx = x ln x, and differentiating both sides, we get
f 0 (x)
= ln x + 1, and so f 0 (x) = f (x)(ln x + 1) = xx (ln x + 1).
f (x)
1
, where k is a real number.
1 + e−kt
dP
(a) Show that
= kP (1 − P ).
dt
Using the chain rule, we have
34. Let P (t) =
−ke−kt
dP
=
dt
(1 + e−kt )2
k
e−kt
=
·
1 + e−kt 1 + e−kt
= kP (1 − P ).
d2 P
(b) Show that 2 = 0 when P = 1/2.
dt
Using implicit differentiation, together with the result of part (a), we have
that
d2 P
= kP 0 (1 − P ) − kP P 0 = kP 0 (1 − 2P ).
2
dP
d2 P
1
From the above, it’s now obvious that
=
0
when
P
=
.
dP 2
2
35. Let f and g be differentiable functions and assume that f (1) = 2, f 0 (1) =
1, g(1) = −1, g 0 (1) = 0. Compute h0 (1), given that h(x) = x2 f (x)g(x).
This uses only the product rule:
h0 (x) = 2xf (x)g(x) + x2 f 0 (x)g(x) + x2 f (x)g 0 (x).
Substituting x = 1 yields
h0 (1) = 2f (1)g(1) + f 0 (1)g(1) + f (1)g 0 (1) = 2 · 2 · (−1) + 1 · (−1) + 2 · 0 = −5.
36. In your text1 it was given as a exercise that the dollar cost of producing x
washing machines is c(x) = 2000 + 100x − 0.1x2 . Why is this an absolutely
rediculous cost model? (What is lim c(x)? Is this reasonable?)
x→∞
The above model says that the cost of producing x washing machines eventually becomes NEGATIVE. This is clearly preposterous!
37. The volume V is a sphere of radius r is given by the formula V = (4/3)πr3 .
Suppose that you know that the radius r is an increasing function of t, and
dr
dV
that when r = 3,
= 2. Compute
when r = 3.
dt
dt
dV
dr r=3
Using the chain rule,
= 4πr2
= 4π · 32 · 2 = 72π.
dt
dt
d
cos x
38. Compute
, simplifying as much as possible.
dx 1 + sin x
Using the quotient rule,
d
dx
1
cos x
1 + sin x
Exercise 10, page 130.
=
− sin x(1 + sin x) − cos2 x −(1 + sin x)
−1
=
=
.
(1 + sin x)2
(1 + sin x)2
1 + sin x
39. Note that the point (1, 2) is on the curve defined by y 3 − xy 2 − x2 y − 2 = 0.
dy
at the point (1, 2). Using implicit differentiation, one has
dx
3y 2 y 0 − y 2 − 2xyy 0 − 2xy − x2 y 0 = 0, and so
(a) Compute
dy
y 2 + 2xy
y =
=
,
dx 3y 2 − 2xy − x2
0
and so
y 0 (1, 2) =
22 + 2 · 1 · 2
8
=
3 · 22 − 2 · 1 · 2 − 12
7
(b) Find an equation of the straight line tangent to the above curve at the
point (1, 2). Such an equation can be written as
y − 2 = y 0 (1, 2)(x − 1),
which becomes y − 2 = 87 (x − 1).
(c) Find an equation of the straight line normal to the above curve at the
point (1, 2). Such an equation can be written as
y−2=
−1
(x − 1),
y 0 (1, 2)
which becomes y − 2 = − 87 (x − 1).
40. Suppose that y = e−x cos x. Show that y 00 + 2y 0 + 2y = 0
This is routine: y 0 = −e−x cos x − e−x sin x, and so y 00 = e−x cos x + e−x sin x +
e−x sin x − e−x cos x = 2e−x sin x. Therefore,
y 00 + 2y 0 + 2y = 2e−x sin x + 2(−e−x cos x − e−x sin x) + 2(e−x cos x) = 0.
41. Find f 0 (x), given that f (x) = √
x
and simplify your result as much as
x2 + 9
possible.
Using the quotient and chain rules,
√
0
f (x) =
x2 + 9 − 2x2 (x2 + 9)−1/2
9 − x2
= 2
.
x2 + 9
(x + 9)−3/2
√
42. Compute f 0 (1), given that f (x) = sin π x2 + 3 .
√
2+3
x
x
cos
π
√
f 0 (x) =
; therefore, f 0 (1) = 1/2.
2
x +3
43. Let x = t2 − 1, y =
dy
in terms of t.
dx
dy
dy dx
2
4t
= /
=
·
2t
=
.
dx
dt dt
(t + 1)2
(t + 1)2
dy
Find all values of t for which
fails to exist.
dx
dy
From part (a)
fails to exist when t = −1.
dx
dy
Find all values of t for which
= 0.
dx
dy
From part (a)
= 0 when t = 0.
dx
dy
Compute lim x, lim y, lim
.
t→∞
t→∞
t→∞ dx
dy
= 0.
lim = +∞; lim y = 1; lim
t→∞
t→∞
t→∞ dx
Suppose that the graph of y = f (x), where f is a differentiable function,
has a horizontal asymptote (say with lim y = c, for some real number c.
x→∞
dy
Would you expect that lim
= 0.
x→∞ dx
dy
This is a bit subtle, but we cannot infer that lim
= 0 (even though
x→∞ dx
sin(x2 )
we might expect this to happen!). A counterexample is y =
. We
x
(a) Compute
(b)
(c)
(d)
(e)
t−1
give x and y parametrically in terms of t.
t+1
have that lim y = 0 (and so y has y = 0 as a horizontal asymptote), but
x→∞
dy
− sin x2
2
that lim
= lim
+ 2 cos x , which does not exist.
x→∞ dx
x→∞
x2
44. Let f (x) =
sin x2
.
x
(a) Compute lim f (x).
x→∞
(b) Compute lim f 0 (x). (If this limit does not exist, say so.)
x→∞
(c) Compute lim |f 0 (x)|. (If this limit does not exist, say so.)
x→∞
This was anticipated in the previous exercise.
45. Let x = t2 + t and let y = cos t.
dy
dy dx − sin t
(a) Find dy/dx as a function of t.
= /
=
.
dx
dt dt
2t + 1
d dy
(b) Find
as a function of t.
dt dx
Using the quotient rule,
d
dt
d
(c) Find
dx
dy
dx
dy
dx
d − sin t
=
dt 2t + 1
−(2t + 1) cos t + 2 sin t
.
=
(2t + 1)2
as a function of t.
d
dx
dy
dx
d dy
dx
=
/
dt dx
dt
−(2t + 1) cos t + 2 sin t
=
/(2t + 1)
(2t + 1)2
−(2t + 1) cos t + 2 sin t
=
(2t + 1)3
46. Given the equation y 2 + x2 = xy, compute both dy/dx and d2 y/dx2 .
dy
Computing
is routine; using implicit differentiation one arrives at
dx
dy
y − 2x
=
.
dx 2y − x
Computing
d2 y
is rather more complicated:
dx2
d2 y
(y 0 − 2)(2y − x) − (y − 2x)(2y 0 − 1)
=
dx2
(2y − x)2
2(y − 2x)2
(y − 2x) − 2(2y − x) −
+ (y − 2x)
2y − x
=
(2y − x)2
(y − 2x)(2y − x) − 2(2y − x)2 − 2(y − 2x)2 + (y − 2x)(2y − x)
=
(2y − x)3
−2(3y 2 − 3xy + x2 )
=
.
(2y − x)3
47. Compute dy/dx, given that
(a) y = e2x cos x
(b) y = eln x
dy
= 2e2x cos x − e2x sin x = e2x (2 cos x − sin x)
dx
dy
=1
dx
√
x
dy
(c) y = ln( 1 + x2 )
=
.
dx 1 + x2
48. Consider the curves defined by the equations y = f (x)= − 21 x2 + 4 and y =
g(x) = ln x. Show that at the point of intersection of these two curves, the
tangent lines are perpendicular. (Hint: what is f 0 (x)g 0 (x)? What does this
mean?)
1
f 0 (x) = −x and g 0 (x) = . Therefore, at the point of intersection the tangent
x
lines have negative reciprocal slopes.
√
x x2 + 1
49. Let y = √
and compute dy/dx using logarithmic differentiation.
3
x+2
Starting with ln y = ln x + 21 ln(x2 + 1) − 31 ln(x + 3), we now differentiate both
sides and get:
y0
1
x
1
= + 2
−
,
y
x x + 1 3(x + 3)
and so
y
xy
y
+ 2
−
x x + 1 3(x + 3)
√
√
x2 + 1
x2
x x2 + 1
= √
+√
−
.
√
3
x+2
x2 + 1 3 x + 2 (x + 2)4/3
y0 =
There’s not much point in trying to simplify this further!