Fall 2010 Math 113 Final Exam Solutions
1. If f (x) =
x
ln x ,
find f 0 (e3 ).
Answer: Using the quotient rule,
f 0 (x) =
ln x ·
d
dx (x)
−x·
(ln x)2
d
dx (ln x)
=
ln x · 1 − x ·
(ln x)2
1
x
=
ln x − 1
.
(ln x)2
Therefore,
f 0 (e3 ) =
3−1
2
ln(e3 ) − 1
=
= .
3
2
2
(ln(e ))
3
9
2. At what value(s) of x (if any) is the function f (x) defined below discontinuous?

2

x + 4x + 5 if x < −2
x
f (x) = 2
if − 2 ≤ x ≤ 2

√

1+ x−2
if x > 2
For x < −2, f (x) = x2 + 4x + 5, which is a polynomial and, hence, continuous. Also, for −2 <√x < 2,
f (x) = x2 , which is also a polynomial and hence continuous. Finally, for x > 2, f (x) = 1 + x − 2,
which is continuous
√ since it’s the composition of continuous functions (specifically, f (x) = g ◦ h(x)
where g(u) = 1 + u and h(x) = x − 2). Therefore, the only possible points of discontinuity for f are
at x = −2 and x = 2.
At x = −2, f (x) will be continuous if
lim f (x) = f (−2) =
x→−2
Clearly, limx→−2+ f (x) = limx→−2+
lim f (x) =
x→−2−
lim
x→−2−
−2
= −1.
2
x
2
= −1. On the other hand,
x2 + 4x + 5 = (−2)2 + 4(−2) + 5 = 4 − 8 + 5 = 1.
Therefore, since limx→−2+ f (x) = −1 and limx→−2− f (x) = 1, we can see that f (x) is not continuous
at x = −2.
Turning to x = 2, f (x) will be continuous if
lim f (x) = f (2) =
x→2
Now limx→2− f (x) = limx→2−
x
2
2
= 1.
2
= 1. On the other hand
lim+ f (x) = lim+ 1 +
x→2
√
x − 2 = 1.
x→2
Since the limits from the left and the right are both equal to 1, we conclude that
lim f (x) = 1,
x→2
so f (x) is continuous at x = 2.
Therefore, f (x) is continuous at all real numbers except x = −2.
1
3. Find the equation of the tangent line to the curve
x cos y = 1
at the point 2, π3 .
Answer: We will find the slope of the tangent line using implicit differentiation. Differentiating both
sides yields
1 · cos y + x · (− sin y · y 0 ) = 0
or
cos y − xy 0 sin y = 0.
Therefore,
cos y = xy 0 sin y
and so we have that
y0 =
cos y
.
x sin y
Therefore, at the point 2, π3 the slope of the tangent line is
y0 =
cos(π/3)
1/2
1
√ = √ .
=
3
2 · sin(π/3)
2 3
2· 2
By the point-slope formula, the equation of the tangent line is
y−
π
1
= √ (x − 2)
3
2 3
y−
x
π
1
= √ −√ .
3
2 3
3
or
Therefore, the equation of the tangent line is
x
1
π
y= √ −√ + .
3
2 3
3
4. A particle moves so that its position at time t is given by
p
s(t) = 3t2 + 4.
If v(t) is the velocity of the particle at time t, what is limt→+∞ v(t)?
Answer: First, we know that the velocity v(t) is just the derivative of the position function. In other
words,
1
3t
v(t) = s0 (t) = (3t2 + 4)−1/2 · 6t = √
.
2
3t2 + 4
Therefore,
lim v(t) = lim √
t→+∞
t→+∞
3t
.
3t2 + 4
Now, notice that the leading terms of both numerator and denominator are essentially t (since the t2 in
the denominator is under the square root). Therefore, we should simplify things by multiplying both
2
numerator and denominator by 1t :
lim √
t→+∞
3t
3t
= lim √
·
2
t→+∞
3t + 4
3t2 + 4
= lim q
t→+∞
= lim q
t→+∞
1
t
1
t2
1
t
1
t
· 3t
(3t2 + 4)
3
3+
4
t2
3
=√
3
√
= 3.
Therefore, we can conclude that limt→+∞ v(t) =
√
3.
5. A technical writer is producing a book which must have 1-inch side margins and 2-inch top and bottom
margins. If the area of the page can be at most 50 in2 , what dimensions give the most printed area
per page?
Answer: If we call the width of the page x and the height of the page y, then x consists of two 1-inch
margins plus the width of the text, so the actual width of the text is x − 2. Likewise, y consists of two
2-inch margins plus the height of the text, so the height of the text is y − 4.
Now, we know that the area of the page is 50 in2 , meaning that
xy = 50.
Solving for y, we see that
y=
50
.
x
The quantity we want to maximize is the area of the printed text, which, by the discussion above is
100
100
50
− 4 = 50 −
− 4x + 8 = 58 −
− 4x.
(x − 2)(y − 4) = (x − 2)
x
x
x
Hence, we’re looking to maximize the function A(x) = 58 − 100
x − 4x. Of course, it must be the case
that x > 0, and in fact this is the only constraint (since x > 0, y = 50
x is automatically bigger than
zero).
To find the maximum, let’s determine the critical points:
A0 (x) =
100
− 4.
x2
Therefore, A0 (x) = 0 when
100
− 4 = 0,
x2
meaning that
100
= 4,
x2
and so
100 = 4x2 .
3
Hence x2 = 25 and so x = ±5. Clearly −5 is not greater than zero, so the only critical point we care
about will be x = 5.
Since A00 (x) = − 200
x3 , which is negative for all x > 0, we see that x = 5 is a local maximum. Since it’s
the only critical point in the interval we’re interested, that means it must be the absolute maximum.
Therefore, the printed area will be maximized when the width of the page is x = 5 and when the height
50
of the page is y = 50
x = 5 = 10.
6. Suppose f (t) = 2t + cos t. Evaluate the definite integral
Z
π/2
f (t)dt.
−π/2
Answer: Since t2 + sin t is an antiderivative of f (t), the Fundamental Theorem of Calculus tells us
that
Z π/2
π/2
(2t + cos t)dt = t2 + sin t −π/2
−π/2
π 2
π 2
=
+ sin(π/2) −
−
+ sin(−π/2)
2
2
π2
π2
=
+1−
− (−1)
4
4
= 2.
7. Evaluate the limit
ln(3 + 2e3x )
.
x→∞
6x
lim
Answer: Notice that
lim ln(3 + 2e3x ) = ∞
x→∞
and
lim 6x = ∞.
x→∞
Therefore, we can apply L’Hôpital’s Rule, so the above limit is equal to
lim
x→∞
1
3+2e3x
· 2e3x · 3
6
= lim
x→∞
e3x
3+2e3x
1
e3x
.
x→∞ 3 + 2e3x
= lim
Again, both numerator and denominator are going to ∞, so we can apply L’ôpital’s Rule again to get
e3x · 3
1
1
= lim
= .
x→∞ 2e3x · 3
x→∞ 2
2
lim
Therefore, we conclude that
ln(3 + 2e3x )
1
= .
x→∞
6x
2
lim
8. Suppose a rabbit colony starts with 20 members and grows (at least for the first few years) at a rate
proportional to its size. If there are 45 rabbits after two years, how many rabbits were there after 1
year? (Note: Your answer should be an integer)
Answer: Since the rate of growth of the rabbit colony is proportional to the size of the population,
the population function is given by
P (t)
=
Cekt
4
for some constants C and k. To determine C, we plug in t = 0:
(
(k·0
P (0)
=(
Ce
((
(
20
=
Ce0
20
=
C.
So now we know that P (t) = 20ekt for some k. To determine k, we plug in our other known value,
which is the population after 2 years:
(
(
(k·2
P (2)
=(
20e
((
(
((
(
45(
=(
20e2k
45 = e2k ,
20
where the last line follows from dividing both sides by 20. Notice that
Taking the natural log of both sides yields
45
20
= 94 , so we have
9
4
= e2k .
9
ln = 2k,
4
so
1/2 ! ((
(
((
(
(
1
9
9
3
(
(
k = ln ((
=(
ln((
= ln
.
2 (4
4
2
((((
Hence, the population function is
((
(t(
ln( 23 )
P (t)
=(
20e
.
((
(
Therefore, the population at time t = 1 must be
(
(
((((
(
(
(
3
60
ln((
)(
((
P (1) = 201·ln(()(
=(20e
= 20 · =
= 30.
(
2
2
(
(
(
(
(
3
2
3
2
So there were 30 rabbits after 1 year.
3
9. Consider the function f (x) = 1 − e−x .
(a) What are the domain and range of f ?
Answer: The domain of the function ex consists of all real numbers. In turn, the domain of the
function −x3 is also all real numbers, so the domain of the function f is all real numbers.
3
As for the range, notice that e−x will hit all positive numbers. Since this is being subtracted
from 1 in the definition of f , we see that the range of f consists of all numbers less than 1; i.e.,
the range of f is the set {x : x < 1}.
(b) What is the inverse of f (i.e., what is the function f −1 )?
3
Answer: If y = f (x), then y = 1 − e−x . Then we can determine the inverse of f by swapping x
and y in this expression and solving for y. Swapping x and y yields
3
x = 1 − e−y .
3
Our first step in solving for y is to add e−y to both sides and subtract x from both sides, yielding
3
e−y = 1 − x.
5
Now take the natural log of both sides:
−y 3 = ln(1 − x)
or
y 3 = − ln(1 − x).
Hence,
y=
p
3
− ln(1 − x),
so
f −1 (x) =
p
3
− ln(1 − x).
Notice that the domain of this function is {x : x < 1}, which is reassuring, since that’s the range
of f .
10. As a spherical raindrop falls, it evaporates (i.e. loses volume) at a rate proportional to its surface area.
Show that the radius of the raindrop decreases at a constant rate. (Hint: the volume of a sphere of
radius r is 34 πr3 , and the surface area is 4πr2 )
Answer: First, let’s record what we know. We know that the volume is given by V (t) = 34 π(r(t))3
and that the surface area is given by A(t) = 4π(r(t))2 . Moreover, since the volume changes at a rate
proportional to the surface area, we know that
V 0 (t) = C · A(t)
for some constant C.
Now, what we’re trying to show is that the radius decreases at a constant rate; in other words, that
r0 (t) is constant.
To get at r0 (t), let’s differentiate V (t):
V 0 (t) =
4
π · 3(r(t))2 · r0 (t) = 4π(r(t))2 r0 (t)
3
by the Chain Rule. On the other hand, we know that V 0 (t) = C · A(t), so we have that
4π(r(t))2 r0 (t) = C · A(t)
4π(r(t))2 r0 (t) = C · 4π(r(t))2 .
Dividing both sides by 4π(r(t))2 yields that
r0 (t) = C,
so indeed the rate of change of the radius is constant.
11. Suppose f 0 (t) = 3et − 2 sec t tan t and that f (0) = 4. Give a formula for f (t).
Answer: First, let’s take the indefinite integral of f 0 (t):
Z
Z
0
f (t)dt =
3et − 2 sec t tan t)dt = 3et − 2 sec t + C
for some constant C. In other words,
f (t) = 3et − 2 sec t + C
6
for some C. To determine C, we want to plug in the initial value:
f (0) = 3e0 − 2 sec(0) + C
4=3−2·1+C
4 = 1 + C,
so we see that C = 3. Therefore, the formula for f (t) is
f (t) = 3et − 2 sec t + 3.
7