SSEA Math 41 Track
Final Exam
August 30, 2012
Name:
• This is the final exam for the Math 41 track at SSEA. Answer as many problems as
possible to the best of your ability; do not worry if you are not able to answer all of the
problems. Partial credit is available. No calculators, notes, or other electronic devices
are permitted.
• As with all math tests at Stanford, you are required to show your work in order to
receive credit. In particular, you should not do computations in your head; instead,
write them out on the test paper. You should also justify all conclusions that you
make, and do not be afraid to explain yourself by writing a sentence or two. The goal
is to make your thought process as clear as possible. However, you are not required to
simplify your answers.
• Please sign below to indicate your acceptance of the following statement:
“On my honor, I have neither given nor received any aid on this examination. I have
furthermore abided by all other aspects of the honor code with respect to this examination.”
Signature:
Problem
Total Points
1
10
2
10
3
10
4
10
5
10
6
10
7
10
8
10
9
10
10
10
Total
100
Score
SSEA Math 41 Track
1
Final Exam
August 30, 2012
Compute the following limits if they exist, or write DNE if they do not exist. Write
∞ or −∞ to indicate an infinite limit.
(a)
lim−
t→2
t3
.
t−2
Solution.
Since the numerator approaches 8 and the denominator approaches 0 from the
negative side, we have
t3
lim−
= −∞.
t→2
t−2
(b)
lim
x→∞
3ex − e−x
.
ex + e−x
Solution.
3ex − e−x
(3ex − e−x )/ex
=
lim
x→∞ (ex + e−x )/ex
ex + e−x
3 − e−2x
= lim
x→∞ 1 + e−2x
limx→∞ 3 − e−2x
=
limx→∞ 1 + e−2x
3
=
1
= 3.
lim
x→∞
(c)
lim x2 2 + sin(1/x) .
x→0
Solution.
Note for all x 6= 0 we have
1 ≤ 2 + sin(1/x) ≤ 3
and hence
x2 ≤ x2 2 + sin(1/x) ≤ 3x2 .
Since
lim x2 = 0 = lim 3x2 ,
x→0
x→0
the Squeeze Theorem implies that
lim x2 2 + sin(1/x) = 0.
x→0
SSEA Math 41 Track
2
Final Exam
August 30, 2012
Compute the derivative of each of the following functions:
(a)
f (x) = ex cos x .
Solution.
f 0 (x) = ex cos x (x cos x)0
by the chain rule
x cos x
=e
(cos x − x sin x)
by the product rule.
(b)
f (x) = sin2 (ln(x)).
Solution.
f 0 (x) = 2 sin(ln(x))(sin(ln x))0
by the chain rule
= 2 sin(ln(x)) cos(ln x)(ln x)0
by the chain rule again
1
= 2 sin(ln(x)) cos(ln x)
by the chain rule again.
x
(c)
Z
x
g(x) =
sin
√
t4 + 1 dt.
−8
Solution.
g 0 (x) = sin
√
x4 + 1
by the Fundamental Theorem of Calculus.
SSEA Math 41 Track
3
Final Exam
August 30, 2012
2
(a) Compute the derivative of f (x) = 3x + x using the definition of the derivative
(involving limits).
Solution.
f 0 (x) = lim
h→0
= lim
h→0
= lim
h→0
= lim
h→0
= lim
h→0
f (x + h) − f (x)
h
[3(x + h)2 + (x + h)] − [3x2 + x]
h
2
2
3x + 6xh + 3h + x + h − 3x2 − x
h
6xh + 3h2 + h
h
6x + 3h + 1
= 6x + 1.
(b) Find the absolute maximum and minimum values of
f (x) =
x2 − 4
x2 + 4
on the interval [−4, 4].
Solution.
Since f is a continuous function, we know that f achieves both an absolute maximum and an absolute minimum on this closed interval. Note
f 0 (x) =
(x2 + 4)(2x) − (x2 − 4)(2x)
16x
= 2
.
2
2
(x + 4)
(x + 4)2
Hence f 0 (x) = 0 if and only if x = 0, and this is the only critical point. The
possible absolute extrema are the endpoints ±4 and the critical point 0. We
evaluate f at these points and get f (−4) = 53 , f (4) = 35 , and f (0) = −1. So the
absolute maximum of f on this interval is f (−4) = f (4) = 35 , and the absolute
minimum of f on this interval is f (0) = −1.
SSEA Math 41 Track
4
Final Exam
August 30, 2012
(a) Sketch the graph of a function f (x) which satisfies the following conditions:
•
•
•
•
•
•
•
•
f (0) = 0,
f 0 (−2) = f 0 (1) = f 0 (9) = 0,
limx→∞ f (x) = 0,
limx→6 f (x) = −∞,
f 0 (x) < 0 on (−∞, −2), (1, 6), and (9, ∞),
f 0 (x) > 0 on (−2, 1) and (6, 9),
f 00 (x) > 0 on (−∞, 0) and (12, ∞),
f 00 (x) < 0 on (0, 6) and (6, 12).
Solution.
SSEA Math 41 Track
Final Exam
3
August 30, 2012
2
(b) Sketch the graph of f (x) = 2x + 3x − 36x. You should find the intervals of
increase and decrease, find the intervals of concavity, label all local and absolute
maxima and minima, and label all inflection points.
Solution.
Note
f 0 (x) = 6x2 + 6x − 36 = 6(x2 + x − 6) = 6(x + 3)(x − 2).
So f is increasing on (−∞, 3) and (2, ∞) and f is decreasing on (−3, 2). Also, f
has a local maxima at x = −3 and a local minima at x = 2. Function f has no
absolute maxima or absolute minima.
Note
1
f 00 (x) = 12x + 6 = 6(x + ).
2
1
So f is concave down on (−∞, − 2 ) and concave up on (− 12 , ∞). Also, f has an
inflection point at x = − 12 .
SSEA Math 41 Track
Final Exam
August 30, 2012
√
5 (a) Let f (x) = 3 + x. Find the linear approximation L(x) to f (x) at x = 1. This
is also the equation of the tangent line to f (x) at x = 1.
Solution.
√
1
. Let a = 1. Note f (1) = 3 + 1 = 2 and f 0 (1) = 2√13+1 = 14 .
Note f 0 (x) = 2√3+x
L(x) = f (a) + f 0 (a)(x − a)
= f (1) + f 0 (1)(x − 1)
1
= 2 + (x − 1).
4
(b) Use the linear approximation from part (a) to estimate
Solution.
√
√
3.99 = 3 + 0.99
= f (0.99)
≈ L(0.99)
1
= 2 + (0.99 − 1)
4
= 1.9975.
√
3.99.
SSEA Math 41 Track
6
Final Exam
August 30, 2012
A rectangular storage container with an open top is to have a volume of 10 m3 . The
length of the base is twice the width of the base. Material for the base of the container
costs $10 per square meter. Material for each of the four sides costs $6 per square
meter. What is the cost of materials for the cheapest such container? You do not need
to simplify the expression you obtain.
Solution.
Let w be the width of the base and let h be the height of the box. Since the length of
the base is 2w, the volume of the box is
10 = (2w)wh = 2w2 h.
Hence
h=
5
10
= 2.
2
2w
w
Now let’s find the cost of the box. The surface area of the base is 2w2 and the surface
area of the four sides is 2(2w)h + 2wh = 6wh. Hence the cost of the box is
C = 10(2w2 ) + 6(6wh)
= 20w2 + 36wh.
We substitute h =
5
w2
to write the cost as a function of w alone.
C(w) = 20w2 + 36wh
= 20w2 + 36w(
= 20w2 +
5
)
w2
180
.
w
We take the derivative to get
40w3 − 180
180
.
C (w) = 20w − 2 =
w
w2
0
p
3
3
The derivative is zero when
w
=
9/2,
that
is,
when
w
=
9/2. The derivative
is
p
p
p
3
3
3
negative when 0 < w < 9/2 and positive when w > 9/2. Hence w = 9/2 is
an absolute minimum of the cost function. Therefore the cost of the materials for the
cheapest such container is
p
p
180
C( 3 9/2) = 20( 3 9/2)2 + p
.
3
9/2
SSEA Math 41 Track
7
Final Exam
August 30, 2012
Consider the integral
Z
9
ln(x) dx.
1
(a) Write an approximation to this integral, using left endpoints and 4 subintervals.
You do not need to simplify. Is this an overestimate or underestimate?
Solution.
Note ∆x = 9−1
= 2. Hence
4
Z
9
ln(x) dx ≈ ∆x ln(1) + ln(3) + ln(5) + ln(7)
= 2 ln(1) + ln(3) + ln(5) + ln(7) .
1
This is an underestimate because ln(x) is an increasing function.
(b) Write an approximation to this integral, using right endpoints and 4 subintervals.
You do not need to simplify. Is this an overestimate or underestimate?
Solution.
= 2. Hence
Note ∆x = 9−1
4
Z
9
ln(x) dx ≈ ∆x ln(3) + ln(5) + ln(7) + ln(9)
= 2 ln(3) + ln(5) + ln(7) + ln(9) .
1
This is an overestimate because ln(x) is an increasing function.
(c) Write an approximation to this integral, using midpoints and 4 subintervals.
Solution.
Note ∆x = 9−1
= 2. Hence
4
Z
1
9
ln(x) dx ≈ ∆x ln(2) + ln(4) + ln(6) + ln(8)
= 2 ln(2) + ln(4) + ln(6) + ln(8) .
SSEA Math 41 Track
8
Final Exam
August 30, 2012
Evaluate the following integrals:
(a)
Z
π
(2x3 + 5 + sin x) dx.
0
Solution.
Z π
π
1
(2x3 + 5 + sin x) dx = x4 + 5x − cos x 0
2
0
1
1
= ( (π)4 + 5(π) − cos(π) − ( (0)4 + 5(0) − cos(0)
2
2
1 4
= π + 5π + 2.
2
(b)
Z
x3/2 ln x dx.
Solution.
We’ll use integration by parts. Let u = ln x and let dv = x3/2 dx. Then du =
and v = 25 x5/2 .
Z
x
3/2
Z
ln x dx = uv −
v du
Z
2 5/2
2 5/2 1
= x ln x −
x
dx
5
5
x
Z
2 5/2
2 3/2
= x ln x −
x dx
5
5
2
4
= x5/2 ln x − x5/2 + C.
5
25
1
x
dx
SSEA Math 41 Track
Final Exam
August 30, 2012
(c)
Z
1
(x2 + 1)(x3 + 3x)4 dx.
0
Solution.
We’ll use a u-substitution. Let u = x3 + 3x. Then du = 3x2 + 3 dx = 3(x2 + 1) dx.
We have
Z 1
Z
1 1
2
3
4
(x + 1)(x + 3x) dx =
3(x2 + 1)(x3 + 3x)4 dx
3
0
0
Z
1 ? 4
=
(u) du
3 ?
?
1
= (u)5 ?
15
1
1
= (x3 + 3x)5 0
15
5
5
1
1
(1)3 + 3(1) −
(0)3 + 3(0)
=
15
15
1 5
= 4.
15
The question marks above are just place-holders for the limits of the integral in
terms of u, which we didn’t need to find because we substituted u = x3 + 3x back
in. Or, we could have evaluated using the limits in terms of u, which would have
been (0)3 + 3(0) = 0 and (1)3 + 3(1) = 4.
(d) If
F (x) = x3/2 ln x,
evaluate
Z
3
F 0 (x) dx.
1
Solution.
By the Fundamental Theorem of Calculus, we have
Z 3
F 0 (x) dx = F (3) − F (1)
1
= (3)3/2 ln(3) − (1)3/2 ln(1)
= 33/2 ln 3.
SSEA Math 41 Track
9
Final Exam
August 30, 2012
For the following true and false questions, you do not need to explain your answer at
all. Just write “True” or “False”.
(a) True or false: If a function f (x) is differentiable at all real numbers x, then it is
also continuous at all real numbers x.
Solution.
True. This follows from Theorem 4 in Section 2.7 of Stewart, which says that if
f is differentiable at a then f is continuous at a. We also proved this theorem in
class on August 13.
(b) True or false: If function f (x) is not defined at x = a, then limx→a f (x) does not
exist.
Solution.
False. For example, consider the function f defined by
(
3
if x 6= 1
f (x) =
not defined if x = 1.
Then f (x) is not defined at x = 1, but limx→1 f (x) = 3 does exist.
(c) True or false: If a function f (x) is continuous on [2, 5] then it attains an absolute
maximum on [2, 5].
Solution.
True. This is by the Extreme Value Theorem in Section 4.2 of Stewart.
(d) True or false: If a function f (x) has an antiderivative function, then it has many
different antiderivative functions.
Solution.
True. Remember that F (x) is an antiderivative of f (x) if F 0 (x) = f (x). So if
F (x) is an antiderivative of f (x), then so is F (x) + C for any real number C.
(e) True or false: Suppose f 00 is continuous near c. If f 0 (c) = 0 and f 00 (c) < 0, then
f has a local minimum at c.
Solution.
False. By the Second Derivative Test in Section 4.3 of Stewart, f has a local
maximum at c.
SSEA Math 41 Track
Final Exam
August 30, 2012
10 Two athletes run a 10 mile race. It’s a tie! They both finish in exactly one hour. In
part (b), we will prove that at some time during the race the two athletes have the
exact same speed.
(a) What is the name of the theorem you want to use for this proof? What is the
statement of this theorem, as best as you can remember?
Solution.
The theorem that you want to use for this proof is the Mean Value Theorem in
Section 4.3 of Stewart. It says that if f is a differentable function on the interval
[a, b], then there exists a number c between a and b such that
f 0 (c) =
f (b) − f (a)
.
b−a
(b) Prove that at some time during the race the two athletes have the exact same
speed. For notational purposes, you may want to let g(t) be the position in miles
of the first athlete at time t and let h(t) be the position in miles of the second
athlete at time t. You may want to let 0 ≤ t ≤ 1 be measured in hours. If so,
then g(0) = h(0) = 0 miles and g(1) = h(1) = 10 miles.
Solution. Using the suggested notation above, let f (t) = g(t) − h(t). Then
f (0) = g(0) − h(0) = 0 − 0 = 0
and
f (1) = g(1) − h(1) = 10 − 10 = 0.
By the Mean Value Theorem, there exists a time t = c between 0 and 1 such that
f 0 (c) =
0
f (10) − f (0)
=
= 0.
10 − 0
10
So g 0 (c) − h0 (c) = f 0 (c) = 0, which means that g 0 (c) = h0 (c). So at time t = c, the
two athletes have the exact same speed.