AP Calculus Chapter 4 Testbank
(Mr. Surowski)
Part I. Multiple-Choice Questions
1. Let f (x) = x3 + 3x2 − 45x + 4. Then the local extrema of f are
(A) a local minimum of −179 at x = 5 and a local maximum of
77 at x = −3.
(B)
a local minimum of −77 at x = 3 and a local maximum of
.
179 at x = −5
(C) a local minimum of −179 at x = −5 and a local maximum
of −77 at x = 3.
(D) a local minimum of −77 at x = 3 and a local maximum of
77 at x = 5.
(E) a local minimum of 77 at x = −5 and a local maximum of
179 at x = 3.
2. Consider the function f (x) = x3 + 3x2 − 45x + 4. Then
I. f is decreasing on (−5, 3) and increasing on (−∞, −5)∪(3, ∞).
II. f has a local minimum at x = 3.
III. f has a local maximum at x = −5.
(A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III
3. For what value of x does the function f (x) = x3 − 9x2 − 120x + 6
have a local minimum?
(A) 10
(B) 4
(C) 3
(D) −4
(E) −10
4. Use differentials to approximate the change in the volume of a
sphere when the radius is increased from 10 to 10.02 cm.
(A) 4213.973
(B) 1261.669
(C) 1256.637
(D) 25.233
(E) 25.133
5. The graph of y = x3 − 5x2 + 4x + 2 has a local minimum at
(A) (0.46, 2.87)
(B) (0.46, 0)
(C) (2.94, −4.05)
(D) (4.06, 2.87)
(E) (1.66, −0.59)
6. Find a positive value, c, for x, that satisfies the conclusion of the
Mean Value Theorem for Derivatives for f (x) = 3x2 − 5x + 1 on
the interval [2, 5].
(A) 1
(B)
13
6
(C)
11
6
(D)
23
6
(E)
7
2
7. The graph of y = x3 − 2x2 − 5x + 2 has a local maximum at
(A) (2.120, 0)
(B) (2.120, −8.061)
(C) (−0.786, 0)
(D) (−0.786, 4.209)
(E) (0.666, −1.926)
8. A 20-foot ladder slides down a wall at 5 ft/sec. At what speed
is the bottom sliding out (in ft/sec) when the top is 10 feet from
the floor?
(A) 0.346
(B) 2.887
(C) 0.224
(D) 5.774
(E) 4.472
9. What are the coordinates of the point of inflection on the graph
of y = x3 − 15x2 + 33x + 100?
(A) (9, 0)
(B) (5, −48)
(D) (9, −89)
(C) (1, 119)
(E) (5, 15)
10. The graph of y = x4 + 8x3 − 72x2 + 4 is concave down for
(A) −6 < x < 2
(B) x > 2
(C) x < −6
√
√
(D) x < −3 − 3 5 or x > −3 + 3 5
√
√
(E) −3 − 3 5 < x < −3 + 3 5
11. The function f is given by f (x) = x4 + 4x3 . On which of the
following intervals is f decreasing?
(A) (−3, 0)
(B) (0, ∞)
(C) (−3, ∞)
(D) (−∞, −3)
(E) (−∞, 0)
12. The value of c that satisfies the Mean Value Theorem on the interval [0, 5] for the function f (x) = x3 − 6x is
5
(A) −√
3
(B) 0
(C) 1
(D)
5
3
5
(E) √
3
13. The graph of the function y = x3 + 12x2 + 15x + 3 has a relative
maximum at x =
(A) −10.613
(B) −.248
(C) −7.317
(D) −1.138
(E) −.683
14. The side of a square is increasing at a constant rate of 0.4 cm/sec.
In terms of the perimeter, P , what is the rate of change of the
area of the square, in cm2 /sec?
(A) 0.05P
(B) 0.2P
(C) 0.4P
(D) 6.4P
(E) 51.2P
15. The second derivative of a function is given by f 00 (x) = x sin x −
2. How many points of inflection does f have on the interval
(−10, 10)?
(A) Zero
(B) Two
(C) Four
(D) Six
(E) Eight
Part II. Free-Response Questions
16. Find the extreme values of the function f (x) = x3 + 3x2 − 36x + 4
on the interval [0, 4].
The critical x-values are obtained by solving
f 0 (x) = 0. This
√
leads to 3x2 + 6x − 36 = 0 ⇒ x = −1 + 13. The extreme values
are obtained by comparing:
f (0) = 4
√
f (−1 + 13) = −51.74
f (4) = −28
Therefore the maximum of f is 4 and the minimum is −51.74.
17. Find the extreme values of the function f (x) =
interval [−3, 3].
x
on the
x2 + 1
1 − x2
From f (x) =
, we see that the critical x-values on the
1 + x2
given interval are x = ±1. We compare:
0
−3
10
1
f (−1) = −
2
1
f (1) =
2
3
f (3) =
10
f (−3) =
1
Therefore f has a maximum of (at x = 1) and a minimum of
2
1
− (at x = −1).
2
18. Find the extreme values of the function f (x) =
interval (−∞, ∞).
x
on the
x2 + 1
This is very similar to the above problem. As above, that the
critical x-values on the given interval are x = ±1. We compare:
x
f (−∞) = lim 2
=0
x→−∞ x + 1
1
f (−1) = −
2
1
f (1) =
2
x
=0
f (+∞) = lim 2
x→+∞ x + 1
1
Therefore f has a maximum of (at x = 1) and a minimum of
2
1
− (at x = −1).
2
Drawing the graph of y = f (x) will reveal the relationship between the above two problems.
√
19. Find the extreme values of the function g(x) = x 2 − x on the
interval [−2, 2].
The critical x-values are those for which either g 0 (x) = 0 or g 0 (x)
√
x
doesn’t exist. From 0 = g 0 (x) = 2 − x− √
we get 4−2x =
2 2−x
4
x ⇒ x = . Also, note that g 0 does not exist at x = 2 (an endpoint
3
of the given interval). We now compare:
g(−2) = −4
q
4
4
g
= 3 23
3
g(2) = 0
4
Therefore the maximum of g is
3
mum of g is −4 (at x = −2).
r
2
4
(at x = ), and the mini3
3
20. Find all relative extrema of the function h(x) = x2 e1/x .
We find the critical x-values:
1
1
set
h0 (x) = 2xe1/x − e1/x = 0 ⇒ x = . Since h0 (x) < 0 when x <
2
2
1
and h0 (x) > 0 when x > , we conclude that h has a relative
2
1
2
minimum of e /4 at x = .
2
√
21. Let f (x) = x 4 − x2 .
(a) Determine all relative extrema of f .
The critical x-values are obtained by solving f 0 (x) = 0. This
√
x2
2
= 0, and so x =
leads to the equation 4 − x − √
2
4
−
x
√
± 2. Since f (x) < 0 when −2 < x < 0, and since f (x) > 0
when 0√ < x < 2 we infer √
that f has a relative minimum
of √
f (− 2) = −2 (at
√ x = − 2) and a relative maximum of
f ( 2) = 2 (at x = 2).
(b) Determine all points of inflection of the graph of y = f (x).
2x(x2 − 4)
00
Here, we have (after some work) that f (x) =
=
(4 − x2 )3/2
0 ⇒ x = 0 (It’s true that x = ±2 are also solutions of the
above, but we don’t allow points of inflection to be at the
endpoints of the domain of definition of a function.) Since
f 00 (x) > 0 when x < 0 and f 00 (x) < 0 when x > 0, we see that
the inflection changes at x = 0, and so the point (0, f (0)) =
(0, 0) is a genuine point of inflection.
22. Determine all relative extrema of the function f (x) = x + cos x.
We compute the critical x-values:
π
set
f 0 (x) = 1 − sin x = 0 ⇒ sin x = 1 ⇒ x =
+ 2kπ, where
2
k is an arbitrary integer. However, as f 0 (x) is ALWAYS nonnegative, we see f is always increasing and no there can be NO
RELATIVE EXTREMA.
23. Let f (x) = 2xe−x , x ≥ 0 and determine the extrema of f (x) on
[0, ∞).
We have f 0 (x) = 2e−x − 2xe−x = (2 − 2x)e−x . From this we see
that the only critical x-value is x = 1. Since f 0 (x) > 0 when
0 ≤ x < 1 and f 0 (x) < 0 when x > 1, we conclude that f has a
relative maximum of 2/e (at x = 1).
24. Let f (x) = 2xe−x , x ≥ 0.
(a) Determine where f is increasing and where f is decreasing.
Using the work already done above, we conclude that f %
on the interval 0 ≤ x < 1 and that f & on the interval x > 1.
(b) Determine where f is concave up and where f is concave
down.
f 00 (x) = (2x − 4)e−x = 0 when x = 2. Also f 00 (x) < 0 when
x < 2 and f 00 (x) < 0 when x > 2. Therefore there is a change
in concavity at x = 2 forcing the point (2, 4/e2 ) to be the
unique point of inflection.
(c) Sketch the graph of y = f (x),
2
.
1 + e−x
There are at least two ways of doing this. The first, and easiest,
way is to recognize the function as a logistic function, where we
have already learned that the point of inflection occurs where
y is one-half the carrying capacity. In this case the carrying capacity is y = 2 and so the point of inflection is at (x, 1), where
2
⇒ x = 0. That is the point of inflection is (0, 1).
1=
1 + e−x
The more straightforward method is to compute the second derivative y 00 of y. We have
25. Find the points of inflection of the graph y =
−2e−x
y =
,
(1 + e−x )2
0
and so
y
00
=
=
=
=
2e−x (1 + e−x )2 − 4e−2x (1 + e−x )
(1 + e−x )4
2e−x + 2e−2x − 4e−2x
(1 + e−x )3
2e−x (1 − e−x )
(1 + e−x )3
0 when x = 0.
Furthermore, it’s easily checked that a change in concavity occurs through x = 0 and so the point of inflection is (0, 1).
26. A particle moves along the x-axis so that its position at time t ≥
0 is given by the function x(t) = t3 − 12t2 + 48t, t ≥ 0, where x
is measured in meters and t is measured in seconds.
(a) Determine those values
of t where the particle is
not moving. The particle is
not moving when v(t) =
x0 (t) = 0. This leads to the
equation 3t2 − 24t + 48 =
0 ⇒ (t − 4)2 = 0, and so
t = 4.
(b) Determine when the particle is moving to the right
and when the particle is
moving to the left. As
v(t) = x0 (t) = 3(t − 4)2 ≥ 0
for all t, we see that unless
t = 4 the particle is always
moving to the right.
(c) Determine when the particle is accelerating and
when the particle is decelerating.
We have that
00
a(t) = x (t) = 6t − 24 > 0
whe t > 4 and x00 (t) < 0
when t < 4. That is to say,
the particle is accelerating
when t > 4 and is decelerating with t < 4.
(d) At what value(s) of t does
the particle have zero acceleration? This happens
when 0 = x00 (t) = 6t − 24,
i.e, when t = 4.
27. Find all relative extrema of the function f (x) = e−x sin x, x > 0.
The critical x values are those for which f 0 (x) = 0, i.e., where
e−x cos x − e−x sin x = 0. This happens when cos x = sin x, i.e.,
π
when x = + kπ, where k is any non-negative integer. Next,
4
note that f 00 (x) = −e−x cos x − e−x sin x + e−x sin x − e−x cos x =
−2e−x cos x. We have
π
+ kπ, k even
4
f 00
+ kπ

4
 > 0 if x = π + kπ, k odd.
4
√
− π4 +kπ
and so f has relative maxima
of
e
/
2, k even and ≥ 0 and
√
− π4 +kπ
relative minima of e
/ 2, k odd and > 0.
π

< 0

if x =
√
28. Find the critical x-values of the function f (x) = x2 3 x − 1.
These are the solutions of f 0 (x) = 0 or where the derivative fails
to exist. We have
√
1
f 0 (x) = 2x 3 x − 1 + x2 (x − 1)−2/3
3
6x(x − 1) + x2
=
3(x − 1)2/3
x(7x − 6)
.
=
3(x − 1)2/3
The above shows that the critical x-values are x = 0,
f 0 (x) = 0) and x = 1 (where f 0 (x) is not defined).
6
7
(where
29. Find the open intervals on which the function f (x) = 41 x3 − 3x is
increasing or decreasing.
We know that f% precisely where f 0 (x) > 0 and that f &
precisely where f 0 (x) < 0. From f 0 (x) = 43 x2 − 3 = 43 (x2 − 4).
From this it follows quickly that f 0 (x) > 0 on the open intervals
(−∞, −2) and (2, ∞) and that f 0 (x) < 0 on the open interval
(−2, 2). Therefore
f % on (−∞, −2) and on (2, ∞), and
f & on (−2, 2).
30. Use the second derivative test to find the relative extrema of
f (x) = x4 − 4x3 + 3. As part of your work, be sure to identify all points of inflection. Sketch a graph, indicating all points
of inflection.
We have f 0 (x) = 4x3 − 12x2 = 4x2 (x − 3) revealing critical xvalues at x = 0, 3. Next, f 00 (x) = 12x2 − 24x = 12x(x − 2). Note
that f 00 (0) = 0 and so the second derivative test is INCONCLUSIVE at the critical point x = 0. Next, f 00 (3) = 36·1 > 0 and so we
see that f has a relative minimum of f (3) = 34 − 4 · 33 + 3 = −24
(at x = 3). Finally note that f 00 (x) > 0 on the intervals (−∞, 0)
and (2, ∞) and f 00 (x) < 0 on the interval (0, 2). Therefore we
obtain points of inflection at x = 0, 2; the corresponding points
are P and Q, with coordinates P (0, 3) and Q(2, −13)
y6
y = f (x)
6
x0 x1 x2
x3 x4
x5 x6
-
x
?
?
31. The graph above depicts a function f defined on the interval
(−∞, ∞). In terms of the notation given in the graph, determine
(i) the critical x values of f ; These are x = x0 , x2 , x4 , and x6 .
(ii) the open intervals on which f is increasing and decreasing; f % on (−∞, x0 ), (x2 , x4 ), (x6 , ∞); f & on (x0 , x2 ), and on
(x4 , x6 ).
(iii) the open intervals on which f is concave up and concave
down. f ∪ on the intervals (x1 , x3 ) and on (x5 , ∞); f ∩ on the
intervals (−∞, x1 ) and on (x3 , x5 ).
(iv) Find the values of x at which f has points of inflection.
These are the values x = x1 , x3 , and x5 .
32. Find the production level that produces the maximum profit for
hamburgers in a fast-food restaurant whose profit function (P
dollars) in terms of hamburgers (x) is described as
x2
P = 2.44x −
− 5, 000 ,
20, 000
0 ≤ x ≤ 50, 000.
x
set
= 0 we infer the critical value
10, 000
x = 24, 400. We now compare:
From P 0 (x) = 2.44 −
P (0) = −5000 (certainly not a maximum!);
244002
P (24400) = 2.44 · 24400 −
− 5000 = 24768;
20000
500002
− 5000 = −8000
P (50000) = 2.44 · 50000 −
20000
Therefore the production level which maximizes profit is x =
24, 400 hamburgers.
√
33. Suppose that the semicircle y = 1 − x2 is drawn. Now draw
a rectangle whose base is on the x-axis and height is such that
the rectangle is incribed within the semicircle. What is the maximum area of this rectangle?
If the base of the rectangle is drawn√between the points (−x, 0)
and (x, 0), then the height would be 1 − x2 . Therefore,
the area
√
to be maximized is given by the√
function A(x) = 2x 1 − x2 , −1 ≤
x ≤ 1. Solving 0 = A0 (x) = 2 1 − x2 − 2x2 (1 − x2 )−1/2 yields
1
1 − x2 = x2 and so x = ± √ The resulting maximum area is
2
q
√
1
then A ± √
= 2 1 − 12 = 1.
2
34. Suppose that on one side of a 2 km-wide river is an electricitygenerating plant. On the opposite side, and 10 km down the
river is a small town that will be consuming the electricity. If it
costs $80/m to lay cable under the river and $40/m to lay cable
over land, find the most strategic method for laying the cable.
Power Plant
•HH
H
2 km
|
HH
H
HH
H
HH
H
H
{z
x
}
z
10 − x
}|
{
•
Town
From the above picture, the total cost of running the cable from
the power plant to the town is
p
C(x) = 80000 4 + x2 + 40000(10 − x) dollars.
Differentiating and setting equal to 0 give the equation
80000x
√
− 40000 = 0,
4 + x2
and so
2
4x2 = 4 + x2 ⇒ x = √ km.
3
Comparing the cost at the critical x-value versus that at the endpoints yields
C(0) = 560, 000 (dollars)
C(10) = 815, 843 (dollars)
√
C(2/ 3) = 538, 564 (dollars)
from which we see that the most economical method is
√ to run
the cable to the opposite side of the river where x = 2/ 3 (km)
and then run the cable along the ground to the town. The resulting cost of doing this is $538,564.
35. Find the minimum distance from the point (2, 0) to a point on
the parabola whose graph is y = 2x2 .
We shall minimize the distance-squared from the point P (2, 0)
to the parabola whose equation is y = 2x2 . Setting D equal to
this distance-squared, then we have D(x) = (x − 2)2 + (2x2 )2 .
Differentiating and setting equal to 0 produces
0 = D0 (x) = 2(x − 2) + 8x3 .
This cannot be solved (easily) in closed form; using a graphics
calculator gives the approximate solution x ≈ 0.689 Therefore,
the minimum distance is approximated as
p
p
D(0.689) = (0.689 − 2)2 + 4(0.689)4 ≈ 1.62
√
36. Write the linearization of the function f (x) =√ 3 1 + x valid near
x = 0. Use this linearization to approximate 3 1.1.
We have that f (x) ≈ f (0) + f 0 (0)x = 1 + 13 x.
From this we get the approximation
√
3
1
1.1 = f (.1) ≈ 1 + (0.1) ≈ 1.033.
3
(Note that this is pretty good, as a calculator yields
1.032.)
√
3
1.1 ≈
37. Write the linearization of the function g(x) = cos x valid near
x = 0. Use this to approximate cos(.01). (Are you surprised by
this linearization?)
Setting f (x) = cos x we have cos x ≈ f (0) + f 0 (0)x = 1 + 0, since
d
dx cos x = − sin x and sin 0 = 0. This forces the approximation
cos(.01) ≈ 1.
38. Write the linearization of f (x) = ln x valid near x = 1. Use this
approximate ln(1.2).
We have f (x) ≈ f (1)+f 0 (1)(x−1) = 0+(x−1) = x−1. Therefore,
ln(1.2) = f (1.2) ≈ 1.2 − 1 = .2.
39. Assume that y = x2 ln x.
(a) Compute dy in terms of x and dx.
dy
From
= 2x ln x + x we get dy = (2x ln x + x) dx.
dx
(b) Compute dy given that x = 1 and that dx = 0.1.
dy = dx = 0.1
40. Let V be the volume of a sphere of radius r and let dV be its differential value (in terms of r and dr). For fixed dr, what would
you say about dV :
(a) that it is an increasing function of r,
(b) that it is a decreasing function of r, or
(c) that it doesn’t depend on r?
From V = 43 πr3 we get dV = 4πr2 dr. Therefore, for fixed dr, dV
is an increasing function of r .
41. Let Q be a quantity
√ that depends on a measurable quantity, x,
by the rule Q = 1 + x2 . Suppose that we measure x to be 5.2 ±
0.015.
(a) Estimate the resulting range in computed values for Q.
x dx
We have dQ = √
; we take x = 5.2 and dx = 0.015. The
1 + x2
resulting range of values of Q is approximately Q ± dQ =
√
5.2 · 0.015
1 + 5.22 ± p
= 5.30 ± 0.015.
1 + (5.2)2
(b) Estimate the resulting relative error in the computed value
dQ
= 0.015/5.30 ≈ 0.003, or a .3%
for Q. The relative error is
Q
relative error.
42. Assume that ball bearings are to be made whose volume must be
within a 1% error. Use differentials to determine the necessary
tolerance of the radius of the given ball bearings.
dV
dV
We require that
≤ .01. Since dV = 4πr2 dr, we get
=
V
V
4πr2 dr
3dr
=
. Therefore the relative error on the measure(4/3)πr3
r
ments of the radius cannot exceed 1/3 of 1%.
43. Suppose that a sphere is expanding at a rate given by dr/dt =
5m/sec. If S denotes the surface area of this sphere and V denotes its volume, compute dV /dt and dS/dt when r = 2. Do the
same for r = 4. Does this make sense?
Using the chain rule and the facts that V = 43 πr3 , S = 4πr2 , we
dV
dV dr
r=2
get
=
·
= (4πr2 ) · (5 m/sec) = 80πm3 /sec. Similarly
dt
dr dt
dS dr
dS
r=2
=
·
= (8πr) · (5 m/sec) = 80πm2 /sec.
dt
dr dt
The corresponding results for r = 4 are similarly obtained.
44. Suppose that you are standing exactly 1km from the Pearl Tower
in Shanghai. Now suppose that someone releases a heliumfilled balloon from the top of the Pearl Tower, and that this balloon is rising at a constant rate of 1m/sec. Let D be the distance
between you (standing on the ground) and the balloon. Let h(t)
be the height of the balloon as a function of time, where we assume that h(0) = 400m.
(a) Compute dD/dt when h = 1000m. We have that D2 = h2 +1,
dD
dh
dD
which implies that 2D
= 2h . This implies that
=
dt
dt
dt
h dh
h
1000
=
=√
≈ 1m/sec
D dt
D
1 + 10002
(b) Compute dD/dt when h = 2000m.
dD
h dh
h
2000
≈ 1m/sec.
Here,
=
=
=√
dt
D dt
D
1 + 20002
(c) Compute lim dD/dt. (Does this make sense?)
h→∞
The above work shows that
dD
h dh
h
h
=
=
=√
≈
dt
D dt
D
1 + h2
1m/sec → 1 as h → ∞.
This makes sense for as the balloon continues to rise, it becomes almost directly overhead, rising at very nearly 1 m/sec
relative to the observer.
(d) Show that dD/dt is a increasing function of both h and t.
h
This is almost obvious for the function F (h) = √
sat1 + h2
dD
isfies F (h) < 1, but that F (h) → 1 as h → ∞. Therefore,
dt
is an increasing function of h. As h increases as t increases,
dD
is also an increasing function of t.
dt
Of course, if you’re nervous about this, you can show that
F (h) is increasing by computing its derivative with respect
to h and showing that it’s positive.
45. Let h be a function defined for all x 6= 0 such that h(4) = −3 and
x2 − 2
0
the derivative of h is given by h (x) =
for all x 6= 0
x
(a) Find all values of x for which the graph of h has a horizontal
tangent, and determine whether h has a local maximum, a
local minimum, or neither at each of these values. Justify
your answers. From what’s given we see immediately that
√
2
h0 (x) = 0 when x = ± 2. We have that h0 (x) = x − and
x
√
2
so h00 (x) = 1 + 2 . Therefore, h00 (± 2) = 1 + 1 = 2 > 0,
x
and so√h will have a relative minimum at each of the values
x = ± 2.
(b) On what intervals, if any, is the graph of h concave up? Justify your answer. Since h00 (x) > 0 for all values of x in the
domain of h, we infer that h is concave up on the intervals
(−∞, 0) and (0, ∞).
(c) Write an equation for the line tangent to the graph of h at
42 − 2
7
0
x = 4. We have that h (4) =
= . Therefore, since
4
2
we’re already given that h(4) = −3, the equation of the line
tangent to the graph of y = h(x) at the point (4, −3) is
7
y + 3 = (x − 4).
2
(d) Does the line tangent to the graph of h at x = 4 lie above or
below the graph of h for x > 4. Why? The line tangent to the
graph of h at x = 4 will lie below the graph of h since the
graph of h is everywhere concave up.
46. A cubic polynomial function g is defined by
f (x) = x3 + ax2 + bx + c
where a, b, and c are constants. The function has a critical value
at x = −1, and the graph of f has a point of inflection at the
point (−2, 0). Find a, b, and c.
We have that f 0 (x) = 3x2 +2ax+b and that 0 = f 0 (−1) = 3−2a+b.
Next, f 00 (x) = 6x + 2a and so 0 = f 00 (−2) = −12 + 2a. Solving
this latter equation results in a = 6 . From this together with the
equation 3 − 2a + b = 0 we obtain b = 9 . Finally, since the point
(−2, 0) is on the graph, we know that f (−2) = 0, which forces
0 = (−2)3 + a(−2)2 + b(−2) + c = −8 + 6 · 4 + 9 · (−2) + c =
−2 + c ⇒ c = 2 .
x
f (x)
f 0 (x)
0
0<x<1 1 1<x<2
−1 Negative 0
Positive
2
2<x<3
3
3<x<4
2
Positive
0
Negative
0
Positive
DNE Negative −3 Negative
f 00 (x) −2 Negative 0
Positive
DNE Negative
4
Positive
0
Positive
47. Let f be a function that is continuous on the interval [0, 4). The
function f is twice differentiable except at x = 2. The function
f and its derivatives have the properties indicated in the table
above, where DNE indicates that the derivatives of f do not exist
at x = 2.
(a) For 0 < x < 4, find all values of x at which f has a relative
extremum. Determine whether f has a relative maximum
or a relative minimum at each of these values. Justify your
answer. f has a relative maximum at x = 2 for the derivative of f changes from positive to negative through x = 2.
(The fact that the derivative of f does not exist at x = 2 is
unimportant.) There are no other relative extrema.
y
(b) On the axes provided, sketch the graph of a function that
has all the characteristics of f .
6
x
48. An airplane is flying at a constant speed at a constant altitude of
3 km in a straight line that will take it directly over an observer
at ground level. At a given instant the observer notes that the
angle θ is π/3 and is increasing at 1/20 radians per second. Find
the speed of the airplane, in kilometers per hour.
x
}|
z
3 km
•
{
•
Airplane
θ
Observer
x
, where x is as indicated in
3
dx the diagram above. Note that the speed of the airplane is .
dt
Differentiating the above equation gives
We start with the equation cot θ =
− csc2 θ
dθ
1 dx
=
.
dt
3 dt
dx
dθ
3
= −3 csc2 θ = − csc2 θ, so that when
dt
dt
20
4
dx
3
4
π
=−
=
θ = , csc2 θ = , all of which implies that
3
3
dt
20
3
1
1
km/sec. Therefore, we see that the speed of the plane is ×
5
5
3600 = 720 km/hr.
This implies that
49. Continue that we have, as in the above problem, an airplane
flying toward the observer at an altitude of 3 km. This time,
dθ
1
however, assume that angular rate of change,
=
, and is
dt
20
constant.
(i) What is the speed of the airplane when it is directly overx
head? Again, we have the equation cot θ = , which im3
dθ
3
dx
= −3 csc2 θ = − csc2 θ, so that when the
dt
dt
20
π
plane is directly overhead we have θ = , and so
2
plies that
3
3
dx
= − csc2 θ = − .
dt
20
20
This says that the speed of the plane is
dx 3
speed = =
× 3600 = 540 km/hr.
dt
20
(ii) Is the airplane slowing down or speeding up as it approaches
the observer? Justify your conclusion.
dx 3
From =
csc2 θ, we conclude that as θ increases to
dt
20
π
2
, csc θ will decrease to 1 from values greater than 1. There2
fore the speed of the plane is decreasing .