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(Math021)[2009](f)midterm~hmhcheung^_10400.pdf
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HKUST
MATH021 Concise Calculus
Midterm Examination Name: 13th Oct 2009 Student I.D.: Tutorial Section:
Directions:
.
DO NOT open the exam until instructed to do so.

.
All mobile phones and pagers should be switched o. during the examina-tion.

.
You must show the steps in order to receive full credits.

.
Electronic calculators are not allowed in the exam.


Question No. Points Out of
1 1
2 2
3 2
4 2
5 6
6 6
7 6


The following formula are given for reference:
sin(x + y) = sin x cos y + cos x sin y sin(x . y) = sin x cos y . cos x sin y cos(x + y) = cos x cos y . sin x sin y cos(x . y) = cos x cos y + sin x sin y
1. Let
2

f (x) = ln x.
What is the domain of f ?

solution:
Since x2 > 0 unless x = 0, the domain of f is the set of all nonzero
numbers.

2. Let
2

f (x)= x for all x. Evaluate the derivative of f , by making use of the de.nition of derivatives.
solution:
For every number x,
f (x)
f (x+h).f (x)

= limh.0 h
2
(x+h)2 .x

= limh.0 h
22
x +2xh+h2 .x

= limh.0 h = limh.0 2x + h =2x.
3. Let
1

f (x) = for all x.
1+ ex

It is known that f is a one-to-one function. Compute the inverse of f .
solution:
If y = f (x),

1
y =
1+ex
1 x
= 1+ e
y 1 x
. 1= e
y

ln( 1 . 1) = x.
y

Therefore, the inverse of f is a function g where
1

g(y) = ln( . 1).
y

4.
Give an example of a continuous function whose domain is the whole real line and whose graph has exactly one horizontal asymptote (Give the ex-plicit de.nition of your function rather than just sketching its graph).

solution:
Let f (x) = 0for all x be a constant function. Since limx.+. f (x)=
limx... f (x) = 0. The graph of f has exactly one horizontal asymptote.


5.
Evaluate each of the following limits, if it exists.


(a)
x
lim
x.+. 1+ x
(b) sin(sin x)lim
x.+.
x
(c) sin(sin x)lim
x.0 x

solution:
(a)
. 1
x
x

lim = lim
x.+. 1+ x x.+. 1 +1
x

but
11
lim . =0and lim =0,
x.+. x.+.
xx

thus,
x 0

lim ==0.
x.+. 1+ x 1+0
(b)
Since .1 . sin sin x . 1 for all x
we have
1 sin sin x 1
.. . for all x> 0,
xxx
and
1 .1

lim =0= lim .
x.+. x.+.
xx

By sandwich theorem,
sin sin x
lim =0.

x.+.
x

(c)
Since
lim x.0 sin x x = 1 and lim x.0 sin sin x sin x = lim y.0 sin y y = 1,
we have
sin sin x sin sin x sin x
lim = lim = 1.
x.0 x x.0 sin x x

6. Let
x
e

f (x)= .
1 . tan x

(a)
State the quotient rule without proof.

(b)
Evaluate the derivative of f .

(c)
Find the point on the graph of f whose x-coordinate is 0.

(d)
Evaluate the slope of tangent to the graph of f at the point in part c).


Solution:
(a)
u(x)

Quotient Rule: For function g(x) = where v(x) = 0 the derivative of
v(x)

g(x) is

uv . uv

g (x)= ,
2
v

or equivalently
dv(x) d u(x) . u(x) d v(x)
dx