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Nguyen (ln4328) – HW #5 – treisman – (54540)
1
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001
10.0 points
Suppose
f
is a continuous Function on
[
−
2
,
2] such that
f
(
−
2) = 1
,
f
(2) =
−
1
.
Which oF the properties below Follow with
out Further restriction on
f
by applying the
Intermediate Value Theorem?
A.
f
(
c
) = 0 For some
c
in (
−
1
,
1);
B.
f
(
x
) + 1
≥
0 on (
−
2
,
2);
C.
f
2
(
c
) =
1
4
For some
c
in (
−
2
,
2).
1.
B only
2.
all oF them
3.
C only
correct
4.
none oF them
5.
B and C only
6.
A and B only
7.
A only
8.
A and C only
Explanation:
The Intermediate Value Theorem (IVT)
says:
If
f
is continuous on an interval
[
a, b
]
, then
for each
N
between
f
(
a
)
and
f
(
b
)
there exists
c
in
(
a, b
)
at which
f
(
c
) =
N
.
Consequently,
A. ±alse:
f
(
x
) =
b
1
,
−
2
≤
x <
1
3
−
2
x,
1
≤
x
≤
2
.
B. ±alse: (choose any
f
so that
f
(0) =
−
2).
C. True: (by IVT,
f
(
c
) =
1
2
For some
c
in
(
−
2
,
2)).
002
10.0 points
IF a construction worker drops a wrench
From the top oF a building 450 Feet high the
wrench will be
s
(
t
) =
−
16
t
2
+ 450
Feet above the ground aFter
t
seconds.
Determine the velocity oF the wrench 4 sec
onds aFter it was was dropped.
Correct answer:
−
128 Ft
/
sec.
Explanation:
The velocity oF the wrench is given by
v
(
t
) = lim
h
→
0
s
(
t
+
h
)
−
s
(
t
)
h
,
so its velocity aFter
a
seconds is
v
(
a
) = lim
h
→
0
s
(
a
+
h
)
−
s
(
a
)
h
=
(
−
16(
a
+
h
)
2
+ 450)
−
(
−
16
a
2
+ 450)
h
=
−
16
h
(2
a
+
h
)
h
=
−
32
a .
Thus
v
(
a
) =
−
32
a
Ft/sec
,
the negative sign indicating that the wrench
is Falling to the ground. At
a
= 4, thereFore,
velocity =
−
128 Ft
/
sec Ft/sec
.
003
10.0 points
±ind an equation For the tangent line at the
point
P
(1
, f
(1)) on the graph oF
f
when
f
is
defned by
f
(
x
) =
x
2
−
3
x
+ 4
.
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View Full DocumentNguyen (ln4328) – HW #5 – treisman – (54540)
2
1.
y
−
x
+ 5 = 0
2.
y
+
x
+ 5 = 0
3.
y
+
x
−
3 = 0
correct
4.
y
−
x
−
3 = 0
5.
y
+
x
−
5 = 0
6.
y
−
x
+ 3 = 0
Explanation:
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 Fall '06
 McAdam
 Calculus

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