COLUMBIA UNIVERSITY
Math V1101
Calculus I
Fall 2013
2nd Midterm Exam
10.31.2013
Instructor: S. Ali Altuğ
A
Name & UNI:
Question:
1
2
3
4
5
6
Total
Points:
5
10
6
7
14
0
42
Score:
Instructions:
• Please write your NAME and UNI on EVERY PAGE.
• There are 6 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 question 1) JUSTIFY YOUR
WORK.
• Put your final answer in a box.
• No calculators, cell phones, books, notebooks, notes or cheat sheets are allowed.
10.31.2013
Midterm II
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 f (x) has domain (0, ∞) and does not have a horizontal asymptote then limx→∞ f (x) =
∞ or limx→∞ f (x) = −∞.
(b) (1 point) If f (x) is not differentiable at x = c then f (|x|) is also not differentiable at x = c.
(c) (1 point) If f (x) = (x6 − x4 + 1)5 then f (31) (x) = 0.
(Recall that for any integer r, f (r) (x) denotes the r’th derivative of f (x).)
(d) (1 point) If f (x) has a local minimum or a local maximum at x = c then f 0 (c) = 0.
(e) (1 point) If f (x) is continuous on (a, b) then it has an absolute maximum and an absolute minimum
in (a, b).
Solution:
(a) False. Think of f (x) = sin(x) restricted to (0, ∞).
(b) False. A counter example is given by
(
1
f (x) =
−1
x≥0
x<0
(c) True. The degree of f (x) is 30. Each time we differentiate the degree drops by 1, hence when
we differentiate 31-times we get 0.
(d) False. For example f (x) = −|x| has a maximum at x = 0 but it is not differentiable at x = 0.
(e) False. f (x) =
1
x
is continuous on (0, 1) but it does not have a maximum in that interval.
Page 2
10.31.2013
Midterm II
Name & UNI:
2. In the first four parts find f 0 (x) for the given function f (x), in the last part find the limit (If the limit
does not exist write DN E, if it is infinite write ∞ or −∞ accordingly.).
(a) (2 points) f (x) = 10x sin(x)
(b) (2 points) f (x) =
cos(2x+1)
ln(x)
(c) (2 points) f (x) = earcsin(x)
(d) (2 points) f (x) =
x2 (x+2) tan(x)
(x−5)(x2 +x)(x3 −2x−1)2
(e) (2 points) limx→∞
esin(x)
x+2
Solution:
(a) Recall that
d
x
dx 10
= ln(10)10x . Hence by the product rule we have,
f 0 (x) = ln(10)10x sin(x) + 10x cos(x)
= 10x (ln(10) sin(x) + cos(x))
(b) By the quotient and chan rules,
−2sin(2x + 1) ln(x) + cos(2x+1)
x
(ln(x))2
−2x sin(2x + 1) + cos(2x + 1)
=
x(ln(x))2
f 0 (x) =
(c) By the chain rule,
earcsin(x)
f 0 (x) = √
1 − x2
(d) Taking ln of both sides gives,
ln(f ) = ln(x2 ) + ln(x + 2) + ln(tan(x)) − (ln(x − 5) + ln(x2 + x) + ln((x3 − 2x − 1)2 ))
Differentiating both sides implicity we get,
f0
2
1
sec2 (x)
= +
+
−
f
x x+2
tan(x)
1
2x + 1
2(3x2 − 2)
+ 2
+ 2
x − 5 x + x x − 2x − 1
Therefore,
x2 (x + 2) tan(x)
f (x) =
(x − 5)(x2 + x)(x3 − 2x − 1)2
0
2
1
sec2 (x)
+
+
−
x x+2
tan(x)
(e) Since −1 ≤ sin(x) ≤ 1, e−1 ≤ esin(x) ≤ e. Therefore
e−1
esin(x)
e
≤
≤
x+2
x+2
x+2
Hence by the squeeze theorem we get that the limit is 0.
Page 3
1
2x + 1
2(3x2 − 2)
+ 2
+ 2
x − 5 x + x x − 2x − 1
10.31.2013
Midterm II
Name & UNI:
3. Let C be the curve implicitly given by
(x + y)y = (y − x)x
(a) (1 point) Show that the point (0, 1) is on the curve C.
(b) (3 points) Find y 0 .
(c) (2 points) Find the equation of the tangent line to C at the point (0, 1).
Solution:
(a)
1 = (0 + 1)1 = (1 − 0)0 = 1
(b) Taking ln of both sides gives,
y ln(x + y) = x ln(y − x)
Differentiating implicitly, we get
x(y 0 − 1)
y(1 + y 0 )
= ln(y − x) +
x+y
y−x
y 0 ln(x + y) +
Solving for y 0 then gives,
y0 =
ln(y − x) −
ln(x + y) +
y
x+y
y
x+y
−
−
x
y−x
x
y−x
(c) The slope, m, of the tangent line to C at (0, 1) is the value of the derivative, y 0 , at the point
(0, 1). Using part (b) we find the slope to be −1. Therefore the equation of the tangent
line is
y = −x + 1
Page 4
10.31.2013
Midterm II
Name & UNI:
4. The top of a 5m long ladder slides down a vertical wall at a rate of 0.15 m/s.
(a) (4 points) At what rate does the bottom of the ladder slide away from the wall when it is 3m from
the wall?
(b) (3 points) At what rate does the angle between the floor and the ladder decrease when the ladder
is 3m from the wall?
Solution:
(a) Let y be the vertical distance and x the horizontal (follow the figure). Then by the pythagorean
theorem we have,
25 = x2 + y 2
Differentiating this with respect to t we then get,
0 = 2x
dx
dy
+ 2y
dt
dt
Hence,
−y dy
dx
=
dt
x dt
When x = 3 b(y the Pythagorean theorem again) y = 4. Noting that y is decreasing, hence
dy
dt = −0.15, we get,
dx 4
= × 0.15 = 0.2m/s
dt x=3
3
(b) We know that x = 5 sin(θ). Differentiating with respect to t then gives,
dθ
dx
= −5 cos(θ)
dt
dt
By part a) we know that
dx dt x=3
= 0.2. Hence,
dθ −0.2
=
dt x=3
5 sin(θ)
Finally by the Pythagorean theorem we know that when x = 3, y = 4, and hence sin(θ) = 54 .
Therefore,
dθ −0.2
=
= −0.05
dt x=3
4
Page 5
10.31.2013
Midterm II
5. Let f (x) = 1 +
1
x
+
Name & UNI:
1
x2 .
(a) (1 point) Find the domain of f (x).
(b) (1 point) Find horizontal asymptote(s) of f (x) (if there is any).
(c) (2 points) Find vertical asymptote(s) of f (x) (if there is any).
(d) (1 point) Find the critical number(s) of f (x) (if there is any).
Don’t forget that a critical number by definition has to be in the domain of the function!
(e) (2 points) Determine the interval(s) where f (x) is increasing and/or decreasing.
(f) (2 points) Find the absolute minimum and maximum values of f (x) in the interval [−3, −1].
(g) (2 points) Determine the interval(s) where f (x) is concave up and/or concave down.
(h) (1 point) Find the inflection point(s) of f (x) (if there are any).
(i) (2 points) Sketch the graph of f (x).
Solution:
(a) Since the function is rational, it’s domain is everywhere but where the denominator is 0, therefore
the domain is R\{0}.
(b)
1
1
1+ + 2
x→∞
x x
lim f (x) = lim
x→∞
=1
lim f (x) = lim
x→−∞
x→−∞
1+
1
1
+
x x2
=1
The only horizontal asymptote for f (x) is y = 1.
(c) The function cannot be continuous at a vertical asymptote and since f (x) is rational it is continuous on everywhere on it’s domain. Therefore it is enough to chack the limit as x approaches
0 from either side.
1
1
lim f (x) = lim+ 1 + + 2
x x
x→0+
x→0
1
= lim+ 2 x2 + x + 1
x→0 x
1
= lim 2
+
x→0 x
=∞
1
1
lim− f (x) = lim− 1 + + 2
x x
x→0
x→0
1
= lim 2 x2 + x + 1
−
x→0 x
1
= lim− 2
x→0 x
=∞
Therefore x = 0 is the only vertical asymptote of f (x).
Page 6
10.31.2013
Midterm II
Name & UNI:
(d)
−1
2
− 3
x2
x
−(x + 2)
=
x3
f 0 (x) =
f 0 (x) = 0 if and only if x = −2, and it is not defined when x = 0, however 0 is not in the domain
of f (x). Therefore the only critical number of f (x) is x = −2.
(e) To find the intervals of increase and decrease we need to find the sign of the derivative. Since
it is the product of two functions, −(x + 2) × x13 , if we know the signs of these then we know
the sign of the derivative.
We know that −(x + 2) is positive if x < −2 and negative if x > −2. On the other hand
nehgative for x < 0 and positive for x > 0. Therefore,


− x < −2
f 0 (x) = + −2 < x < 0


− x>0
1
x3
is
Therefore f is decreasing on (−∞, −2) ∪ (0, ∞) and increasing on (−2, 0).
(f) We need to check the values of f at the critical points in [−3, −1] and at the end points. The
only critical point in the interval is −2. Then,
1 1
7
+ =
3 9
9
3
1 1
f (−2) = 1 − + =
2 4
4
f (−1) = 1 − 1 + 1 = 1
f (−3) = 1 −
Since 1 >
7
9
> 34 , f (−1) is the absolute max and f (−2) is the absolute min of f in [−3, −1].
(g)
2
6
+ 4
x3
x
2(x + 3)
=
x4
f 00 (x) =
In order to find the intervals of concavity we need the sign of f 00 . Since the denominator is a
square it is always ≥ 0 and hence does not affect the sign. Therefore the sign of f 00 is the sign
of (x + 3), which is positive if x > −3 and negative if x < −3. Therefore f is concave down on
(−∞, −3) and concave up on (−3, ∞).
(h) Since at x = −3 f changes from concave down to up, x = −3 is an inflection point. Since this
is the only point f changes concavity, it is the only inflection point.
Page 7
10.31.2013
Midterm II
Name & UNI:
6. (4 points (bonus)) (Each part 2 points.) Find the following limits. (If the limit does not exist then write
DNE, if it is infinite then write ∞ or −∞ accordingly.)
(In case you know L’Hospital’s rule, solutions that use it it will not get any credit.)
(a) limx→0+
esin(x) −1
x
[[x+1]]
(b) limx→∞ [[x−1]]
Recall that [[x]] is the greatest integer that is less than or equal to x.
Solution:
(x)
. Now at the point x = 0 this
(a) Recall the definition of the derivative, f 0 (x) = limh→0 f (x+h)−f
h
reads as,
f (h) − f (0)
f 0 (0) = lim
h→0
h
Let f (x) = esin(x) and notice that,
esin(x) − 1
esin(0+x) − esin(0)
= lim
x→0
x→0
x
x
= f 0 (0)
= cos(x)esin(x) lim
x=0
=1
(b) Since [[x + 1]] is the greatest integer that is less than or equal to x + 1, for any x we have
x ≤ [[x + 1]] ≤ x + 1
(?)
Similarly
x − 2 ≤ [[x − 1]] ≤ x
Then,
1
1
1
≤
≤
x
[[x − 1]]
x−2
Since x → ∞, x > 0. Therefore from (?) adn (??) we get,
x
[[x + 1]]
x+1
≤
≤
x
[[x − 1]]
x−2
Now since limx→∞
x+1
x−2
= 1, by the squaeeze theorem we get,
lim 1 ≤ lim
x→∞
x→∞
[[x + 1]]
x+1
[[x + 1]]
≤ lim
⇒ 1 ≤ lim
≤1
x→∞
x→∞
[[x − 1]]
x−2
[[x − 1]]
[[x + 1]]
⇒ lim
=1
x→∞ [[x − 1]]
Page 8
(??)