This preview shows pages 1–4. Sign up to view the full content.
Version 084 – Exam 3 – Radin – (56520)
1
This printout should have 19 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
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.
correct
4.
5.
6.
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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document Version 084 – Exam 3 – Radin – (56520)
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 Fg
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 Fgures 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 Fgures 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 Fgures
only
consists entirely of graphs of antiderivatives
of
f
.
keywords: antiderivative, graphical, graph,
geometric interpretation
002
10.0 points
±ind the value of
f
(0) when
f
′
(
t
) = cos 2
t,
f
p
π
4
P
=
−
1
.
1.
f
(0) =
−
3
2
correct
2.
f
(0) =
−
2
3.
f
(0) =
−
1
4.
f
(0) =
−
1
2
5.
f
(0) = 0
Explanation:
Since
d
dx
sin
mt
=
m
cos
mt,
for all
m
n
= 0, we see that
f
(
t
) =
1
2
sin 2
t
+
C
where the arbitrary constant
C
is determined
by the condition
f
(
π/
4) =
−
1. But
sin 2
t
v
v
v
t
=
π/
4
= sin
π
2
= 1
.
Thus
f
p
π
4
P
=
1
2
+
C
=
−
1
,
and so
f
(
t
) =
1
2
sin 2
t
−
3
2
.
Consequently,
f
(0) =
−
3
2
.
003
10.0 points
±ind
f
(1) when
f
′′
(
x
) = 3
x
and
f
(
−
1) = 4
,
f
′
(
−
1) = 3
.
1.
f
(1) = 8
correct
2.
f
(1) = 10
3.
f
(1) = 6
4.
f
(1) = 9
Version 084 – Exam 3 – Radin – (56520)
3
5.
f
(1) = 7
Explanation:
When
f
′′
(
x
) = 3
x
the most general anti
derivative of
f
′′
is
f
′
(
x
) =
3
2
x
2
+
C
where the arbitrary constant
C
is determined
by the condition
f
′
(
−
1) = 3. For
f
′
(
−
1) = 3
=
⇒
f
′
(
x
) =
3
2
x
2
+
3
2
.
But then
f
(
x
) =
1
2
x
3
+
3
2
x
+
D,
the arbitrary constant
D
being given by
f
(
−
1) = 4
=
⇒ −
2 +
D
= 4
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 06/05/2010 for the course MATH 57420 taught by Professor Radin during the Fall '09 term at University of Texas at Austin.
 Fall '09
 RAdin

Click to edit the document details