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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
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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,
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 Fall '08
 schultz
 Differential Calculus

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