MATHEMATICS 113/114
Midterm Examination, Version 2
Fall 2011
Date: Wednesday, October 26, 2011
Time: 50 minutes
LAST NAME:
FIRST NAME:
(Please, print!)
Please, check your section/instructor!
Section
Instructor X
Math 113, E1 E. Osmanagic
Math 114, D1
X. Chen
Instructions
1. Books, notes or calculators are not permitted.
2. Show all your work.
3. Make sure your examination paper has 5 questions.
ID#:
Name:
Instructions
1. Books, notes or calculators are not permitted.
2. Show all your work.
3. Make sure your examination paper has 5 questions.
Question Value Your Mark
1
20
2
10
3
10
4
10
5
10
Total
60
ID#:
Name:
1. (20 points)
(a) Find the domain of f if f (x) =
√
x2 − 4 +
√
8 − x Solution
Df = {x |x2 − 4 ≥ 0 and 8 − x ≥ 0}.
Since
x2 − 4 ≥ 0 ⇐⇒ x ∈ (−∞, −2] ∪ [2, ∞)
and
8 − x ≥ 0 ⇐⇒ x ∈ (−∞, 8],
we have
Df = ((−∞, −2] ∪ [2, ∞)) ∩ (−∞, 8] = (−∞, −2] ∪ [2, 8].
(b) Find horizontal and vertical asymptotes of f (x) =
x−1
. Justify your answer.
2x + 1
Solution:
1−
x−1
= lim
x−→−∞ 2x + 1
x−→−∞ 2 +
lim
1−
x−1
= lim
x−→∞ 2x + 1
x−→∞ 2 +
lim
1
x
1
x
1
x
1
x
=
=
1
2
1
2
Thus, y = 1/2 is a horizontal asymptote as x −→ −∞ and as x −→ ∞.
Note that x = − 12 is the ”candidate” for the vertical asymptote. Since,
lim
−
x−→− 21
x−1
x−1
= ∞, , lim +
= −∞
1
2x + 1
x−→− 2 2x + 1
the line x = − 21 is the vertical asymptote.
ID#:
Name:
(c) Evaluate
√
lim
x−→0
1+x−1
.
2 sin x
Solution:
√
lim
x−→0
1
1+x−1
= lim
2 sin x
2 x−→0
=
√
√
1
1+x−1
1+x−1
1+x+1
√
= lim
=
·√
sin x
2 x−→0 sin x( 1 + x + 1)
1+x+1
1
1
1
1
=
lim sin x · lim √
2 x−→0 x x−→0 1 + x + 1
4
(d) Let g(x) = e3x+1 f (x3 ). Find g 0 (2) if f (8) = 1, f 0 (8) = 2 .
Solution: By the Chain Rule:
g 0 (x) = e3x+1 3f (x3 ) + e3x+1 f 0 (x3 )3x2 = e3x+1 (3f (x3 ) + 3x2 f 0 (x3 )).
Thus,
g 0 (2) = e7 (3f (8) + 12f 0 (8)) = 27e7 .
ID#:
Name:
2. (10 points) Consider the function y = f (x) =
√3 .
x−2
(a) Use the definition of the derivative to find f 0 (x) and state the domain for f and f 0 .
NOTE: No marks will be given if the definition is not used.
(b) Find an equation of the normal line to the graph of f at the point (a, f (a)) where a = 4.
Solution:
(a)
√ 3
x+h−2
−
√3
x−2
√
√
3 √x−2−3 √x+h−2
x+h−2 x−2
f (x + h) − f (x)
f 0 (x) = lim
= lim
= lim
=
h−→0
h−→0
h−→0
h
h
h
√
√
√
√
√
√
x−2− x+h−2
x−2− x+h−2 x−2+ x+h−2
√
√
√
= 3 lim √
·√
=
= 3 lim √
h−→0 h x + h − 2 x − 2
h−→0 h x + h − 2 x − 2
x−2+ x+h−2
x − 2 − (x + h − 2)
√
√
√
√
=
h−→0 h x + h − 2 x − 2( x − 2 +
x + h − 2)
= 3 lim
−h
√
√
√
√
=
h−→0 h x + h − 2 x − 2( x − 2 +
x + h − 2)
= 3 lim
= 3 lim √
h−→0
−1
√
√
√
=
x + h − 2 x − 2( x − 2 + x + h − 2)
=√
−3
√
√
√
=
x − 2 x − 2( x − 2 + x − 2)
−3
−3
√
√
= √
= p
.
2 x−2 x−2 x−2
2 (x − 2)3
The domain for f and f 0 is the same set, namely (2, ∞).
√
(b) First√note that f (4) = 3 2 2 . Thus, we need the equation of the normal line at the point
(4, 3 2 2 ). If mT denotes the slope of the tangent line then the slope of the normal line
is − m1T . On the other hand, mT = f 0 (4). Since by part (a) ( or by the Chain Rule,
3
1
differentiating f (x) = 3(x − 2)− 2 ), the derivative f 0 (x) of f is f√0 (x) = − 23 (x − 2)− 2 we
get mT = f 0 (4) = − 4√3 2 . Hence, the slope of the normal line is 4 3 2 . Finally, the equation
of the equation of the normal line is y −
√
3 2
2
=
√
4 2
(x
3
− 4).
ID#:
Name:
3. (10 points)
(a) Find all possible values of a and b so that the

 x + 2b,
bx − a,
f (x) =

4x − 4,
function
x<1
1≤x≤ 2
x>2
is continuous at every x. Justify your answer.
Solution: First not that f is continuous on (−∞, 1) ∪ (1, 2) ∪ (2, ∞) for all values of a
and b as a polynomial degree one. There are two points left, x = 1 and x = 2.
For f to be continuous at x = 1 , the following equations must hold:
f (1) = lim− f (x) = lim+ f (x).
x−→1
x−→1
f (1) = b − a
lim f (x) = lim− (x + 2b) = 1 + 2b.
x−→1−
x−→1
lim f (x) = lim+ (bx − a) = b − a
x−→1+
x−→1
Thus
b − a = 1 + 2b.
For f to be continuous at x = 2 , the following equations must hold:
f (2) = lim− f (x) = lim+ f (x).
x−→2
x−→2
f (2) = 2b − a
lim f (x) = lim− (bx − a) = 2b − a.
x−→2−
x−→2
lim f (x) = lim+ (4x − 4) = 4
x−→2+
x−→2
Thus
2b − a = 4.
Finally, f is continuous on the set of real numbers if b = 1 and a = −2.
(b) Differentiate:
f (x) = cos(sin(cos(x3 ))).
Solution: By the Chain Rule,
f 0 (x) = (− sin(sin(cos(x3 )))(cos(cos(x3 )))(− sin(x3 ))3x2 .
.
ID#:
Name:
4. (10 points) Evaluate the limit or explain why the limit does not exist:
(a)
sin(2x)
.
x−→0 tan(3x)
lim
Solution:
sin(2x)
sin(2x)
sin(2x)
lim
= lim sin(3x) = lim
lim cos(3x) = lim
x−→0 sin(3x) x−→0
x−→0 tan(3x)
x−→0
x−→0
cos(3x)
(b)
lim (x −
x−→∞
√
sin(2x)
2x
sin(3x)
3x
2
2
=
3
3
x2 + x)
Solution:
lim (x −
x−→∞
√
x2
+ x) = lim (x −
x−→∞
√
x2
−x
q
x−→∞
x + |x| 1 +
= lim
√
x2 − (x2 + x)
x + x2 + x
√
√
= lim
=
+ x) ·
x + x2 + x x−→∞ x + x2 + x
−x
q
x−→∞
x(1 + 1 +
= lim
1
x
1
x
1
=− .
2
ID#:
Name:
5. (10 points)
A function y = y(x) is implicitly defined by the equation:
2ey cos x − 2 = sin(xy).
(a) Find
dy
.
dx
(b) Find the slope of the tangent line to the graph of y = y(x) at the point (2π, 0).
Solution:
(a) Differentiating both sides of the equation with respect to the variable x taking into
account that y = y(x) we get:
2(ey
or
dy
dy
cos x + ey (− sin x)) = cos(xy)(y + x ),
dx
dx
dy y
(2e cos x − x cos(xy)) = y cos(xy) + 2ey sin x.
dx
Thus,
dy
y cos(xy) + 2ey sin x
= y
,
dx
2e cos x − x cos(xy)
2ey cos x − x cos(xy) 6= 0.
(b) The slope of the tangent line is:
dy
0
/(2π,0) =
= 0.
dx
2 − 2π