Version 063 – Exam 3 – gualdani – (56410)
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printout
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have
18
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001
10.0 points
If the graph of
f
is
which
one
of
the
following
contains
only
graphs of antiderivatives of
f
?
1.
2.
3.
4.
5.
6.
correct
Explanation:
If
F
1
and
F
2
are antiderivatives of
f
then
F
1
(
x
)
−
F
2
(
x
) = constant
independently of
x
; this means that for any
two antiderivatives of
f
the graph of one
is just a vertical translation of the graph of
the other.
But no horizontal translation of
the graph of an antiderivative of
f
will be
the graph of an antiderivative of
f
, nor can
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Version 063 – Exam 3 – gualdani – (56410)
2
a horizontal and vertical translation be the
graph of an antiderivative.
This rules out
two sets of graphs.
Now in each of the the remaining four fig
ures the dotted and dashed graphs consist of
vertical translations of the graph whose line
style is a continuous line. To decide which of
these figures consists of antiderivatives of
f
,
therefore, we have to look more carefully at
the actual graphs. But calculus ensures that
(i) an antiderivative of
f
will have a local
extremum at the
x
intercepts of
f
.
This eliminates two more figures since they
contains graphs whose local extrema occur at
points other than the
x
intercepts of
f
.
(ii) An antiderivative of
f
is increasing on
interval where the graph of
f
lies above the
x
axis, and decreasing where the graph of
f
lies below the
x
axis.
Consequently, of the two remaining figures
only
consists entirely of graphs of antiderivatives
of
f
.
keywords:
antiderivative, graphical, graph,
geometric interpretation
002
10.0 points
Find
f
(
x
) on
(
−
π
2
,
π
2
)
when
f
′
(
x
) = 3
√
2 sin
x
+ 2 sec
2
x
and
f
(
π
4
)
= 3.
1.
f
(
x
) = 2
−
2 tan
x
+ 3
√
2 sin
x
2.
f
(
x
) = 2 tan
x
−
3
√
2 cos
x
+ 4
correct
3.
f
(
x
) = 8
−
2 tan
x
−
3
√
2 cos
x
4.
f
(
x
) = 2 tan
x
+ 3
√
2 sin
x
+ 4
5.
f
(
x
) = 2 tan
x
+ 3
√
2 cos
x
−
2
Explanation:
The most general antiderivative of
f
′
(
x
) = 3
√
2 sin
x
+ 2 sec
2
x
is
f
(
x
) =
−
3
√
2 cos
x
+ 2 tan
x
+
C
with
C
an arbitrary constant.
But if
f
parenleftBig
π
4
parenrightBig
= 3, then
f
parenleftBig
π
4
parenrightBig
=
−
3 + 2 +
C
= 3
,
so
C
= 4
.
Consequently,
f
(
x
) = 2 tan
x
−
3
√
2 cos
x
+ 4
.
003
10.0 points
A particle moves along a straight line so
that its acceleration at any given time
t
is
a
(
t
) = 2
t
−
10
(in units of feet and seconds).
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 Fall '09
 Gualdani
 Derivative, 0 g, 2 seconds, 1 seconds, 0 g

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