Mahon, Kevin – Exam 1 – Due: Sep 29 2005, 11:00 pm – Inst: Edward Odell
1
This
printout
should
have
19
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
AFter
t
seconds the displacement,
s
(
t
), oF a
particle moving rightwards along the
x
axis is
given (in Feet) by
s
(
t
) = 4
t
2

2
t
+ 7
.
Determine the average velocity oF the particle
over the time interval [1
,
2].
1.
average vel. = 10 Ft/sec
correct
2.
average vel. = 12 Ft/sec
3.
average vel. = 14 Ft/sec
4.
average vel. = 11 Ft/sec
5.
average vel. = 13 Ft/sec
Explanation:
The average velocity over a time interval
[
a, b
] is given by
dist travelled
time taken
=
s
(
b
)

s
(
a
)
b

a
.
±or the time interval [1
,
2], thereFore,
average vel. =
s
(2)

s
(1)
2

1
Ft/sec
.
Now
s
(2) = 4
×
4

2
×
2 + 7 = 19 Feet
,
while
s
(1) = 4

2 + 7 = 9 Feet
.
Consequently,
average vel. = 19

9 = 10 Ft/sec
.
keywords: Stewart5e,
002
(part 1 oF 1) 10 points
Below is the graph oF a Function
f
.
2
4
6

2

4

6
2
4
6
8

2

4
Use the graph to determine the leFt hand limit
lim
x
→
2

f
(
x
)
.
1.
the limit does not exist
2.
lim
x
→
2

f
(
x
) = 2
correct
3.
lim
x
→
2

f
(
x
) = 0
4.
lim
x
→
2

f
(
x
) =

4
5.
lim
x
→
2

f
(
x
) = 1
Explanation:
As the graph shows,
lim
x
→
2

f
(
x
) = 2.
keywords: Stewart5e,
003
(part 1 oF 1) 10 points
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2
When
f
is the function deFned by
f
(
x
) =
‰
3
x

7
,
x
≤
3 ,
2
x

3
,
x >
3
,
determine if the limit
lim
x
→
3+
f
(
x
)
exists, and if it does, Fnd its value.
1.
limit = 0
2.
limit does not exist
3.
limit = 1
4.
limit = 4
5.
limit = 3
correct
6.
limit = 2
Explanation:
The right hand limit
lim
x
→
3+
f
(
x
) depends
only on the values of
f
to the right of 3. Thus
lim
x
→
3+
f
(
x
) =
lim
x
→
3+
2
x

3
.
Consequently,
limit = 2
×
3

3 = 3
.
keywords: Stewart5e,
004
(part 1 of 1) 10 points
Determine the limit
lim
x
→
4
6
(
x

4)
2
.
1.
none of these
2.
limit =

3
2
3.
limit =
∞
4.
limit =
3
2
5.
limit =
∞
correct
Explanation:
Since (
x

4)
2
≥
0 for all
x
, we see that
lim
x
→
4
6
(
x

4)
2
=
∞
.
keywords: Stewart5e,
005
(part 1 of 1) 10 points
±ind the value of
lim
x
→
2
‡
1
x

1
2
·‡
3
x

2
·
.
1.
limit
=

3
2
2.
limit
=
3
4
3.
limit
=
3
2
4.
limit does not exist
5.
limit
=

3
4
correct
Explanation:
After the Frst term in the product is
brought to a common denominator, the given
expression becomes
3(2

x
)
2
x
(
x

2)
=

3
2
x
so long as
x
6
= 2. Thus
lim
x
→
2
‡
1
x

1
2
·‡
3
x

2
·
=

lim
x
→
2
3
2
x
.
Consequently,
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 schultz
 Derivative, Differential Calculus, Limit, lim, Continuous function

Click to edit the document details