COLUMBIA UNIVERSITY
Math V1101
Calculus I
Fall 2013
1st Midterm Exam
10.01.2013
Instructor: S. Ali Altuğ
A
Name & UNI:
Question:
1
2
3
4
5
6
7
Total
Points:
4
8
9
8
4
3
0
36
Score:
Instructions:
• Please write your NAME and UNI on EVERY PAGE.
• There are 7 questions on this exam. The last question is a bonus question and you do not have to
answer it to get full credit.
• Unless otherwise is explicitly stated (in other words, except for questions 1 and 4) JUSTIFY YOUR
WORK.
• Put your final answer in a box.
• No calculators, cell phones, books, notebooks, notes or cheat sheets are allowed.
10.01.2013
Midterm I
Name & UNI:
1. Determine whether the following statements are true (T) or false (F). You do not need to justify
your answer for this question.
(a) (1 point) If a function f has a horizontal asymptote y = L, then there is no number c in the domain
of f such that f (c) = L.
(b) (1 point) There is an even function f that is increasing on R.
(c) (1 point) The graph of f (|x|) is the graph of f reflected with respect to the x-axis.
(d) (1 point) If f (x) is not one-to-one then f (g(x)) is also not one-to-one for any function g.
Solution:
(a) False. Here is an example:
(
f (x) =
arctan(x) x 6= 0
π
x=0
2
(b) False. Let x > 0, then −x < x. So if f is even we get f (−x) = f (x), which means f can not be
increasing.
(c) False. It is the graph of f reflected with respect to the y-axis.
(d) False. Here is a counter example:
f (x) = x2
√
g(x) = x
Then the domain of the composition f (g(x)) is [, ∞). And the f (g(x)) (= x) is one-to-one in
this domain.
Page 2
10.01.2013
Midterm I
2. Let f (x) =
Name & UNI:
√
x x2 −5
x2 −9 .
(a) (2 points) What is the domain of f (x)?
(b) (2 points) Find the vertical asymptote(s) of f , if there are any.
(c) (4 points) Find the horizontal asymptote(s) of f , if there are any.
Solution:
√
(a) Note that the function is the product of three functions: x, x2 − 5, x21−9 , therefore the domain
is the domain of the intersection of the domains of these. The first function, x, has domain
R. The third, x21−9 , has domain R\{±3} because the denominator can’t be 0. For the third,
√
2
2
x2 − 5,√ we need
√ to have the inner function positive, i.e. x − 5 ≥ 5 ⇔ x ≥ 5 ⇔ x ∈
(−∞, − 5] ∪ [ 5, ∞). Therefore the domain of the function is:
√
√
Dom(f ) = (−∞, 5] ∪ [ 5, ∞)\{±3}
(b) Vertical asymptotes are the lines x = a where the function approaches nifty from either side. At
these points the function cannot be continuous. Since f is a product of a rational function with
an algebraic function it is continuous on it’s domain. Therefore we need to look at points that
are not√in √
the domain. Furthermore since f is not defined in any neighborhood of any point
a ∈ (− 5, 5), the limit, limx→a± f (x), does not exists and it is not √
∞. Therefore all we need
to check is the limit of the function as x approaches the points ±3, ± 5.
• 3.
√
Since as x → 3 the function x x2 − 5 → 6 > 0 we get
lim f (x) = ∞
x→3+
Therefore x = 3 is a vertical asymptote.
• −3.
√
Since as x → −3 the function x x2 − 5 → −6 < 0 we get
lim f (x) = −∞
x→3+
Therefore x = −3 is a vertical asymptote.
√
• 5.
√
√
√ +
Since as x → 5 the function x2x−9 → −45 and x2 − 5 → 0 we get
lim
√ +
x→ 5
Therefore x =
√
• − 5.
√
5 is not a vertical asymptote.
√ −
Since as x → − 5 the function
x
x2 −9
→
√
− 5
−4
lim
√ +
x→ 5
Therefore x =
√
f (x) = 0
and
√
x2 − 5 → 0 we get
f (x) = 0
5 is not a vertical asymptote.
To sum up, the vertical asymptotes of f are at x = ±3.
(c) To find the horizontal asymptotes we need to check the limit of f as x → ±∞.
Page 3
10.01.2013
Midterm I
Name & UNI:
• x → ∞.
√
x x2 − 5
lim f (x) = lim
x→∞
x→∞ x2 − 9
p
x x2 (1 − 5/x2 )
= lim
x→∞
x2 (1 − 9/x2 )
p
x|x| 1 − 5/x2
= lim
x→∞ x2 (1 − 9/x2 )
p
1 − 5/x2
= lim
x→∞ 1 − 9/x2
=1
• x → −∞.
√
x x2 − 5
lim f (x) = lim
x→−∞
x→−∞ x2 − 9
p
x x2 (1 − 5/x2 )
= lim
x→−∞
x2 (1 − 9/x2 )
p
x|x| 1 − 5/x2
= lim
x→−∞ x2 (1 − 9/x2 )
p
− 1 − 5/x2
= lim
x→−∞
1 − 9/x2
= −1
Therefore the horizontal asymptotes of f are the lines y = ±1.
Page 4
10.01.2013
Midterm I
Name & UNI:
3. Calculate the following limits. (If the limit does not exist then write DNE, if it is infinite then write ∞
or −∞ accordingly.) (Show your work.)
(a) (3 points) limx→0
sin(3x)
sin(6x)
Hint: Try to rewrite the expression in terms of
1
(b) (3 points) limx→1+ ln(x) sin x−1
√
(c) (3 points) limx→∞ x2 + 5x − 1 − x + 3
sin(3x)
3x
and
sin(6x)
6x .
Solution:
(a)
lim
x→0
sin(3x)
1
sin(3x)/3x
= lim
sin(6x)
2 x→0 sin(6x)/6x
1
=
2
(b) Since −1 ≤ sin(1/(x − 1)) ≤ 1 and ln(x) > 0 for x > 1 (note that we are only approaching
1 on the right) we have
1
− ln(x) ≤ ln(x) sin
≤ ln(x)
x−1
for x > 1. Since we have
x→1+ − ln(x) = lim→1+ ln(x) = 0, by the squeeze theorem we
lim
1
get limx→1+ ln(x) sin x−1 = 0.
(c)
lim
x→∞
p
√
√
( x2 + 5x − 1 − (x − 3))( x2 + 5x − 1 + (x − 3))
√
x→∞
x2 + 5x − 1 + (x − 3)
2
x + 5x − 1 − (x − 3)2
= lim √
x→∞
x2 + 5x − 1 + (x − 3)
11x − 8
= lim √
2
x→∞
x + 5x − 1 + (x − 3)
x(11 − 8/x)
= lim p
2
x→∞
x (1 + 5/x − 1/x2 ) + x(1 − 3/x))
x(11 − 8/x)
p
= lim
x→∞ |x| (1 + 5/x − 1/x2 ) + x(1 − 3/x))
x(11 − 8/x)
p
= lim
x→∞ x( (1 + 5/x − 1/x2 ) + (1 − 3/x))
11
=√
1+1
11
=
2
x2 + 5x − 1 − x + 3 = lim
Page 5
10.01.2013
Midterm I
Name & UNI:
4. Find the following limits. (You don’t have to show your work for this question. The correct
answer gets full credit, however in order to claim partial credit you need to show some
work. If the limit does not exist then write DNE, if it is infinite then write ∞ or −∞ accordingly)
(a) (2 points) limx→0 ln(cos(4πex ))
(b) (2 points) limx→0+ [[x]]
x
Recall that [[x]] is the greatest integer that is less than or equal to x.
(c) (2 points) limx→−1−
(d) (2 points) limx→−∞
x+1
|1+x|
√
x2 + 2x − 1 − x + 1
Solution:
(a) The function is continuous on it’s domain since it is a composition of logarithmic, trigonometric
and exponential functions, and 0 is in the domain of the function. Therefore
lim ln(cos(4πex )) = ln(cos(4πe0 )) = ln(cos(4π)) = ln(1) = 0
x→0
(b) For x ∈ (0, 1/2), [[x]] = 0. Therefore limx→0+
[[x]]
x
= 0.
−
(c) Since x → −1 , x is less than −1. This in particular implies that 1 + x < 0 and therefore
x+1
x+1
|1 + x| = −(1 + x). Hence limx→−1− |1+x|
= limx→−1− −(1+x)
= −1.
(d)
p
p
x2 + 2x − 1 − x + 1 = lim
x2 (1 + 2/x − 1/x2 ) − x(1 − 1/x)
x→−∞
x→−∞
p
= lim |x| 1 + 2/x − 1/x2 − x(1 − 1/x)
lim
x→−∞
Since x → −∞, x < 0. Therefore |x| = −x. Hence,
p
p
lim |x| 1 + 2/x − 1/x2 − x(1 − 1/x) = lim −x 1 + 2/x − 1/x2 − x(1 − 1/x)
x→−∞
x→−∞
p
= lim −x( 1 + 2/x − 1/x2 + 1 − 1/x)
x→−∞
= lim −2x
x→−∞
=∞
Page 6
10.01.2013
Midterm I
Name & UNI:
5. (a) (2 points) Find the value of c which makes the following function continuous everywhere.
(
f (x) =
cx2 + 2x + 2
x3 − cx − 3
if x < 2
if x ≥ 2
(b) (2 points) Show that the function f (x), with the value of c that you have found in part (a), has a
root.
Solution:
(a) Away from x = 2 the function is given by polynomials and therefore is continuous. At x = 2,
we are given
f (2) = 5 − 2c
On the other hand we have
lim f (x) = lim− x3 − xc − 3 = 5 − 2c
x→2−
x→2
and
lim f (x) = lim cx2 + 2x + 2 = 4c + 6
x→2+
x→2+
In order for the function to be continuous at 2 we need to have all these to be the same. i.e.
5 − 2c = 4c + 6, therefore c = −1
6 .
(b) By part (a) the function is continuous everywhere. f (0) = −2, f (3) = e4 (recall that c = 0), in
16
particular f (−1) = −1
6 and f (2) = 3 so by the intermediate value theorem there is a c ∈ (−1, 2)
such that f (c) = 0. (The interval (−1, 2) is just suggestive, you can take any interval (a, b) where
f (a)f (b) < 0.)
Page 7
10.01.2013
Midterm I
Name & UNI:
6. The volume V (r) of a sphere of radius r is given by the formula
V (r) =
4 3
πr
3
(a) (3 points) Calculate the difference quotient (i.e. the derivative),
V (r + h) − V (r)
h→0
h
lim
(b) (0 points) (Yes, it is worth 0 points! So, if you like, you don’t have to even read this part.) The
formula you found in part (a) is the surface area of a sphere of radius r. You may want to think
(when you are done with the exam!) how this happened.
Solution:
(a)
4
π(r + h)3 − 34 πr3
V (r + h) − V (r)
= lim 3
h→0
h→0
h
h
4
3
2
π(r
+
3r
h + 3rh2 + h3 ) − 34 πr3
= lim 3
h→0
h
4
2
2
π(3r h + 3rh + h3 )
= lim 3
h→0
h
4
2
πh(3r
+
3rh + h2 )
= lim 3
h→0
h
4
2
= lim π(3r + 3rh + h2 )
h→0 3
= 4πr2
lim
Page 8
10.01.2013
Midterm I
Name & UNI:
7. (4 points (bonus)) A fixed point of a function f is a point x ∈ Dom(f ) such that f (x) = x. Let a, b ∈ R
such that a < b, and let f be a continuous function from [a, b] to [a, b]. Show that it has a fixed point.
The idea is to consider the function F (x) = f (x)−x. If F (c) = 0 this means that f (c)−c = 0 and hence c
is a fixed point for f . Since f is a function from [a, b] to [a, b] we know that a ≤ f (x) ≤ b. Consider F (a),
by the previous argument a ≤ f (a), if f (a) = a, then a is a fixed point of f and we are done. Assume
not, then f (a) > a, i.e. F (a) = f (a) − a > 0. Now consider the point b. Again by the above argument
f (b) ≤ b. If f (b) = b then b is a fixed point and we are done. Assume not, then F (b) = f (b) − b < 0. In
this case since F (a) > 0 and F (b) < 0, and F (x) = f (x) − x being the sum of two continuous functions
is continuous on [a, b], by the intermediate value theorem there exists a c ∈ (a, b) such that F (c) = 0.
For this c we have 0 = F (c) = f (c) − c, and hence c is a fixed point of f .
Page 9