Version 140 – EXAM 1 – Radin – (58305)
1
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Multiple-choice questions may continue on
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beFore answering.
001
10.0 points
Below is the graph oF a Function
f
.
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 defned 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
)
n
=
f
(
x
0
)
or iF
lim
x
→
x
0
−
f
(
x
)
n
=
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
n
=
lim
x
→−
3
f
(
x
) = 4
,
while at
x
0
= 3,
lim
x
→
3
−
f
(
x
) =
−
4
n
=
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 defned 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 defnition 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
.