Version 140 – EXAM 1 – Radin – (58305)
1
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printout
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have
18
questions.
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before answering.
001
10.0 points
Below is the graph of a function
f
.
8
7
6
5
4
3
2
1
0
1
2
3
4
5
6
7
2
4
6
−
2
−
4
−
6
2
4
−
2
−
4
Use this graph to determine all of the values
of
x
on (
−
7
,
7) at which
f
is discontinuous.
1.
none of the other answers
2.
x
= 3
3.
no values of
x
4.
x
= 3
,
−
3
correct
5.
x
=
−
3
Explanation:
Since
f
(
x
) is defined everywhere on (
−
7
,
7),
the function
f
will be discontinuous at a point
x
0
in (
−
7
,
7) if and only if
lim
x
→
x
0
f
(
x
)
negationslash
=
f
(
x
0
)
or if
lim
x
→
x
0
−
f
(
x
)
negationslash
=
lim
x
→
x
0
+
f
(
x
)
.
As the graph shows, the only possible candi
dates for
x
0
are
x
0
=
−
3 and
x
0
= 3. But at
x
0
=
−
3,
f
(
−
3) = 0
negationslash
=
lim
x
→ −
3
f
(
x
) = 4
,
while at
x
0
= 3,
lim
x
→
3
−
f
(
x
) =
−
4
negationslash
=
lim
x
→
3+
f
(
x
) = 0
.
Consequently, on (
−
7
,
7) the function
f
is
discontinuous only at
x
= 3
,
−
3
.
002
10.0 points
A function
f
is defined by
f
(
x
) =
2
−
x,
x
≤ −
2,
x
2
,
−
2
< x <
4,
12 +
x,
x
≥
4.
Determine where
f
is continuous, expressing
your answer in interval notation.
1.
(
−∞
,
−
2)
∪
(4
,
∞
)
2.
(
−∞
,
−
2)
∪
(
−
2
,
4)
∪
(4
,
∞
)
3.
(
−∞
,
∞
)
correct
4.
(
−∞
,
−
2)
∪
(
−
2
,
∞
)
5.
(
−∞
,
4)
∪
(4
,
∞
)
Explanation:
The function is piecewise continuous, so we
have to check the left and right hand limits at
the points where the definition of
f
changes,
i.e.
, at
x
=
−
2 and
x
= 4. Now at
x
=
−
2
lim
x
→−
2
−
f
(
x
) =
lim
x
→−
2
−
2
−
x
= 4
,
lim
x
→−
2+
f
(
x
) =
lim
x
→−
2+
x
2
= 4
,
hence, the function is continuous at the point
x
=
−
2. On the other hand, at
x
= 4
lim
x
→
4
−
f
(
x
) =
lim
x
→
4
−
x
2
= (4)
2
= 16
,
lim
x
→
4+
f
(
x
) =
lim
x
→
4+
12 +
x
= 16
.
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Version 140 – EXAM 1 – Radin – (58305)
2
Thus, the function is also continuous at the
point
x
= 4.
003
10.0 points
Let
f
be a continuous function on [
−
3
,
1]
such that
f
(
−
3) =
−
1
,
f
(1) = 7
.
Which of the following is a consequence of the
Intermediate Value Theorem without further
restrictions on
f
?
1.
f
′
(
c
) = 1 for some
c
in (
−
3
,
1)
2.
f
(0) = 0
3.
f
(
c
) = 1 for some
c
in (
−
3
,
1)
correct
4.
f
(
c
) = 0 for some
c
in (
−
3
,
0)
5.
−
1
≤
f
(
x
)
≤
7 for all
x
in (
−
3
,
1)
Explanation:
Because
f
is continuous on [
−
3
,
1] the In
termediate Value Theorem ensures that for
each
M,
−
1
< M <
7
,
there exists at least
one choice of
c
in (
−
3
,
1) for which
f
(
c
) =
M
.
In particular,
f
(
c
) = 1 for some
c
in (
−
3
,
1)
.
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 Fall '09
 Gualdani
 Sin, Limit, Continuous function, lim g

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