Solution: APPM 1350
Review #2
Summer 2014
1. The cost (in dollars) of producing x units of a certain commodity is C(x) = 5000 + 10x + 0.05x2 .
(a) Find the average rate of change of C with respect to x when the production level is changed
i. from x = 100 to x = 105
ii. from x = 100 to x = 101
(b) Find the instantaneous rate of change of C with respect to x when x = 100.
Solution:
C(105) − C(100)
= 20.25
105 − 100
C(101) − C(100)
(ii) Average rate of change =
= 20.05
101 − 100
(b) The instantaneous rate of change is C 0 (100) where C 0 (x) = 10 + 0.01x, so C(100) = 20
2
x sin(1/x), if x 6= 0
0
2. Determine whether f (0) exists given f (x) =
0, if x = 0
(a) (i) Average rate of change =
Solution: We first need to check if f (x) is continuous at x = 0, so note that −1 ≤ sin(1/x) ≤ 1 and
so −x2 ≤ x2 sin(1/x) ≤ x2 and since lim x2 = lim −x2 = 0, we have that lim x2 sin(1/x) = 0 by the
x→0
x→0
x→0
Squeeze Theorem, that is, lim x2 sin(1/x) = f (0) and so f (x) is continuous at x = 0.
x→0
that f 0 (0)
Now we need to show
exists, i.e. that the derivative from the left and right of x = 0 exist
and are equal. Note that from the left of 0, we have
lim
h→0−
f (0 + h) − f (0)
h2 sin(1/h) − 0
= lim
= lim h sin(1/h)
h
h
h→0−
h→0−
and recall that −1 ≤ sin(1/h) ≤ 1 and so if h < 0 then −h ≥ h sin(1/h) ≥ h and now since
lim h = lim −h = 0 then lim h sin(1/h) = 0 by the Squeeze Theorem. Similarly, we can show
h→0−
h→0−
h→0−
lim h sin(1/h) = 0 by the Squeeze Theorem and thus f 0 (0) = 0.
h→0+
3. Find an equation of the tangent line to y =
2x
at the point (0, 0).
+1
x2
2 − 2x2
and f (0) = 0 and f 0 (0) = 2 so the equation of the tangent
(x2 + 1)2
line is y = f (0) + f 0 (0)(x − 0) = 2x, i.e. y = 2x.
Solution: Note that f 0 (x) =
4. If a particle has position function s(t) = t3 − 9t2 + 15t + 10, where t is measured in seconds and s in
feet. Find the total distance traveled in the first 8 seconds.
Solution: Note that v(t) = s0 (t) = 3t2 − 18t + 15 = 3(t2 − 6t + 5) = 3(t − 1)(t − 5), so v(t) = 0 when
t = 1 and t = 5, so the total distance traveled in the first 8 seconds is
Total Distance = |s(8) − s(5)| + |s(5) − s(1)| + |s(1) − s(0)| = 120ft
5. If f is a differentiable function, find the derivative of y =
1 + xf (x)
√
.
x
Solution:
√
(f (x) + xf 0 (x)) x − 21 x−1/2 (1 + xf (x))
2x2 f 0 (x) + xf (x) − 1
f (x) =
=
x
2x3/2
0
6. Find the first and second derivatives of the function:
(a) F (t) = (1 − 7t)6
(b) y = (x3 + 1)2/3
(c) y =
2x
(x − 1)2
Solution:
(a) F 0 (t) = −42(1 − 7t)5 and F 00 (t) = 1470(1 − 7t)4
2x(x3 + 2)
2x2
00
and
y
=
(x3 + 1)1/3
(x3 + 1)4/3
−2(x + 1)
4(x + 2)
(c) y 0 =
and y 00 =
3
(x − 1)
(x − 1)4
r
x
√
7. Find dy/dx: (a) y =
(b) y sin(x2 ) = x sin(y 2 ) (c) xy = 1+x2 y (d) y 5 +x2 y 3 = 1+x4 y
x−5
Solution:
sin(y 2 ) − 2xy cos(x2 )
−5
y − 4(xy)3/2
4x3 y − 2xy 3
0
0
0
(b)
y
=
(a) y 0 = √
(c)
y
=
(d)
y
=
√
2
2
2
4
sin(x ) − 2xy cos(y )
2x xy − x
5y + 3x2 y 2 − x4
2 x(x − 5)3/2
(b) y 0 =
8. A man starts walking north at 4 ft/s from a point P. Five minutes later, a woman starts walking
south at 5 ft/s from a point 500 ft due east of point P. At what rate are the people moving apart 15
minutes after the woman starts walking?
Solution:
x
x
h
P
500
h
P
y
y
500
Figure 1: Man and womans’ path
Figure 2: Man and womans’ path organized as a triangle
Let:
dx
= 4 ft/s
dt
dy
• y = the distance the woman has walked, then
= 5 ft/s
dt
• h = the distance between the man and the woman
• x = the distance the man has walked, then
dh
Here we have h2 = (x + y)2 + 5002 , and we wish to find
when t = 15 min = 900 sec. Recall that
dt
distance = rate × time and so when t = 900 sec, we have
• x = 4(|{z}
300 +900) = 4800 ft
5 min headstart
• y = 5(900) = 4500 ft
√
√
• h2 = (x + y)2 + 5002 = 93002 + 5002 = 1002 (932 + 52 ), so h = 100 932 + 52 = 100 932 + 25
dh
dx dy
And note,
= (x +
+
implies 2h
and so
= 2(x + y)
+
dt
dt
dt
dy
dx
2(x
+
y)
+
dt
dt
93(9)
837
dh
9300(4 + 5)
√
=√
=√
ft/sec
=
=
2
2
dt
2h
100 93 + 25
93 + 25
932 + 25
h2
y)2
5002
9. Water is leaking out of an inverted conical tank at a rate of 10,000 cm3 /min at the same time that
water is being pumped into the tank at a constant rate. The tank has height 600 cm and the diameter
of the top is 400 cm. If the water level is rising at a rate of 20 cm/min when the height of the water
is 200 cm, find the rate at which water is being pumped into the tank.
Solution:
Let:
• K = rate at which water is pumped in (cm3 /min)
• r = radius of water
• h = height of water
•
dh
= 20 cm/min when h = 200 cm
dt
• leak = 10, 000 cm3 /min
1
We wish to find K, note that V = πr2 h and since we have similar triangles we have
3
r
200
h
1
=
=⇒ r = so V = πh3
h
600
3
27
dV
1
dV
1
dh
Now note that
= rate in−rate out = K−10000, also note that V = πh3 implies
= πh2
dt
27
dt
9
dt
1 2 dh
And so K − 10, 000 = πh
and thus
9
dt
dh
π
800, 000π
1
+ 10, 000 = (200)2 (20) + 10, 000 =
+ 10, 000 cm3 /min
K = πh2
9
dt
9
9
√
√
10. Find the linearizaton L(x) of f (x) = 3 1 + x at x = 0 and use it to approximate 3 0.95
Solution:
1
Note that L(x) = f (0) + f 0 (0)x = 1 + x, and
3
√
1
1 1
3
0.95 = f (−0.05) ≈ L(−0.05) = L(− ) = 1 − ·
= 59/60
20
3 20
11. Find the absolute maximum and minimum values of f on the given interval:
(a) f (x) = x3 − 6x2 + 9x + 2, [−1, 4]
(b) f (x) = sin(x) + cos(x), [−π/4, π/2]
Solution:
(a) Note that f 0 (x) = 3x2 − 12x + 9 = 3(x − 1)(x − 3) so the critcal points are x = 1 and x = 3 and
f (1) = 6, f (3) = 2 and f (−1) = −14, f (4) = 6. So, the absolute maximum value is 6 at x = 1 and
x = 4 and the absolute minimum value is -14 at x = −1.
(b) Here f 0 (x) = cos(x) − sin(x) and f 0 (x) = 0 implies cos(x)
√ = sin(x) which implies tan(x) = 1
so x = π/4 is a critical
point,
and
f
(−π/4)
=
0,
f
(π/4)
=
2 and f (π/2) = 1, so the absolute
√
maximum value is 2 at x = π/4 and the absolute minimum value is 0 at x = −π/4.
12. State Rolle’s Theorem. Now, let f (x) = 1 − x2/3 . Show that f (−1) = f (1) but there is no number c
in (−1, 1) which satisfies Rolle’s Theorem. Does this contradict Rolle’s Theorem? Why or why not?
Solution:
Rolle’s Theorem: If f (x) is continuous on [a, b] and differentiable on (a, b) and if f (a) = f (b) then
there exist a c in (a, b) such that f 0 (c) = 0.
−2
Note that here clearly f (−1) = 0 = f (1) and f 0 (x) = 1/3 and clearly there is no number x = c for
3x
which f 0 (c) = 0.
This does not contradict Rolle’s Theorem since f (x) is not differentiable on (−1, 1).
√
13. Verify that f (x) = 3 x satisfies the hypothesis of the Mean Value Theorem on the interval [0, 1].
Find all values of c that satisfy the conclusion of the Mean Value Theorem.
Solution:
√
3
1
and so f (x) is differentiable on (0, 1).
3x
r2/3
f (1) − f (0)
1
Now note that
= 1 and f 0 (c) = 1 implies c =
.
1−0
27
Note that f (x) =
x is continuous on [0, 1] and f 0 (x) =
14. At 2:00 p.m., a car’s speedometer reads 30 mi/h. At 2:10 p.m., it reads 50 mi/h. Assuming the
car has been driven the whole time, show that at some time between 2:00 and 2:10, the acceleration
is 120 mi/h2 .
Solution:
Let v(t) be the speed of the vehicle as a function of time (in hours). Then v(0) = 30 and v(1/6) =
50, and so by the Mean Value Theorem there exists a time t0 between t = 0 (2 o’clock) and
t = 1/6 (2:10 p.m.) such that
a(t0 ) = v 0 (t0 ) =
v(1/6) − v(0)
50 − 30
=
= 120 mi/h2
1/6
1/6
15. Show that the equation 2x − 1 − sin(x) = 0 has exactly one real root.
Solution:
Existence: Let f (x) = 2x − 1 − sin(x), note that f (0) = −1 < 0 and f (π/2) = π − 2 > 0. Since f (x) is
the combination of a polynomial and the sine function, f (x) is continuous. Now, by the Intermediate
Value Theorem, there exists at least one point c in the interval (0, π/2) such that f (c) = 0, that is,
there exists at least one root of f (x).
Uniqueness: By way of contradiction, assume there is more than one root, assume there are two roots
a and b of f (x) such that a 6= b. Without loss of generality, we can assume a < b, now we apply the
Mean Value Theorem on f (x) on the interval [a, b]. Note that that f (x) is continuous on [a, b] and
f 0 (x) = 2 − cos(x) which is defined on (a, b) and so f (x) is differentiable on (a, b), thus by the Mean
Value theorem there is a number z between a and b such that
f 0 (z) =
f (b) − f (a)
=0
b−a
Now note that −1 ≤ cos(x) ≤ 1 implies −1 ≤ − cos(x) ≤ 1, and so 1 ≤ 2 − cos(x) ≤ 3, that is,
f 0 (x) = 2 − cos(x) ≥ 1 which contradicts f 0 (z) = 0. This contradiction shows that the given equation
can’t have two distinct roots and so it has exactly one real root.
16. For the functions given below
(a) Find the intervals on which f is increasing or decreasing
(b) Find the local maximum and minimum values of f
(c) Find the inflection points and state where the function is concave up/down.
(i) f (x) = x4 − 4x − 1
(ii) f (x) =
x2
x2 + 3
(iii) f (x) = x − 2 sin(x), 0 ≤ x ≤ 2π
Solution:
(i) Note that f 0 (x) = 4x3 − 4 = 4(x3 − 1) = 4(x − 1)(x2 + x + 1) and f 00 (x) = 12x2 so
(a) f (x) decreases on (−∞, 1) and increases on (1, ∞) (Note, the only real root of f 0 (x) is x = 1)
(b) There is a local minimum at x = 1 with value f (1) = −4 (Note x = 1 is in the domain of f (x))
(c) f (x) is concave up on (−∞, 0) and (0, ∞), there is no inflection point.
First derivative test for (i)
&
Concavity test for (i)
^
%
|
1
(ii) Note that f 0 (x) =
(x2
^
|
0
6x
−18(x2 − 1)
and f 00 (x) =
2
+ 3)
(x2 + 3)3
(a) f (x) decreases on (−∞, 0) and increases on (0, ∞)
(b) There is a local minimum at x = 0 with value f (0) = 0.
(c) f (x) is concave up on (−1, 1) and concave down on (−∞, −1) and (1, ∞). There are inflections
points at (−1, 1/4) and (1, 1/4).
First derivative test for (ii)
&
Concavity test for (ii)
_
%
|
0
^
|
-1
_
|
1
(iii) Note that f 0 (x) = 1 − 2 cos(x) and f 00 (x) = 2 sin(x)
(a) f (x) decreases on (0, π/3) ∪ (5π/3, 2π) and increases on (π/3, 5π/3).
√
(b) There is a local minimum at x = π/3√
with value f (π/3) = π/3 − 3. There is a local maximum
at x = 5π/3 with value f (5π/3) = 5π/3 + 3
(c) f (x) is concave up on (0, π) and concave down on (π, 2π). There is an inflection point at (π, π).
First derivative test for (iii)
&
0
%
|
π/3
Concavity test for (iii)
^
&
|
5π/3
2π
0
_
|
π
2π/
17. For the functions given below, state the domain, find all horizontal and vertical asymptotes, find all
local maxima, minima, and inflection points and then sketch the graph.
√
r
2x
1 − x2
x
(a) y =
(b) y =
(c) y =
2
(x − 1)
x
x−5
Solution:
4(x + 2)
−2(x + 1)
and f 00 (x) =
(a) Note that f 0 (x) =
3
(x − 1)
(x − 1)4
2.4
1.6
0.8
-4.8
-4
-3.2
-2.4
-1.6
-0.8
0
0.8
1.6
2.4
3.2
4
4.8
-0.8
-1.6
-2.4
Domain: x 6= 1,
Horizontal Asymptote: y = 0,
Local Maximum Point: None,
Vertical Asymptote: x = 1
Local Minimum point at (x, y) = (−1, −1/2),
Inflection Point at (−2, −4/9)
(b) Note here f 0 (x) =
−1
−(3x2 − 2)
√
and f 00 (x) = 3
x (1 − x2 )3/2
x2 1 − x2
2.4
1.6
0.8
-4.8
-4
-3.2
-2.4
-1.6
-0.8
0
0.8
1.6
2.4
3.2
4
4.8
-0.8
-1.6
-2.4
Domain: [−1, 0) ∪ (0, 1],
Horizontal Asymptote: None,
Local Maximum point: None,
Vertical Asymptote: x = 0
Local Minimum point: None,
r
Inflection Points at
2
,
3
r !
1
and
2
r
−
r !
2
1
,−
3
2
−5
5(4x − 5)
(c) Note here f 0 (x) = p
and f 00 (x) = p
2 x(x − 5)3
4 x3 (x − 5)5
2.4
1.6
0.8
-5
-2.5
0
2.5
5
7.5
10
12.5
15
-0.8
-1.6
-2.4
Domain: (−∞, 0] ∪ (5, ∞),
Horizontal Asymptote: y = 1,
Local Maximum point: None,
Local Minimum point: None,
Vertical Asymptote: x = 5,
Inflection Points: None