Version 140 – EXAM 1 – Fouli – (58320)
1
This print-out should have 18 questions.
Multiple-choice questions may continue on
the next column or page – fnd all choices
be±ore answering.
001
10.0 points
Below is the graph o± a ±unction
f
.
2
4
6
−
2
−
4
−
6
2
4
−
2
−
4
Use this graph to determine all o± the values
o±
x
on (
−
7
,
7) at which
f
is discontinuous.
1.
no values o±
x
2.
none o± the other answers
3.
x
=
−
5
4.
x
= 5
5.
x
= 5
,
−
5
correct
Explanation:
Since
f
(
x
) is defned everywhere on (
−
7
,
7),
the ±unction
f
will be discontinuous at a point
x
0
in (
−
7
,
7) i± and only i±
lim
x
→
x
0
f
(
x
)
n
=
f
(
x
0
)
or i±
lim
x
→
x
0
−
f
(
x
)
n
=
lim
x
→
x
0
+
f
(
x
)
.
As the graph shows, the only possible candi-
dates ±or
x
0
are
x
0
=
−
5 and
x
0
= 5. But at
x
0
=
−
5,
f
(
−
5) = 0
n
=
lim
x
→−
5
f
(
x
) =
−
4
,
while at
x
0
= 5,
lim
x
→
5
−
f
(
x
) = 4
n
=
lim
x
→
5+
f
(
x
) = 0
.
Consequently, on (
−
7
,
7) the ±unction
f
is
discontinuous only at
x
= 5
,
−
5
.
002
10.0 points
A ±unction
f
is defned by
f
(
x
) =
6
−
x,
x
≤ −
3,
x
2
,
−
3
< x <
3,
6 +
x,
x
≥
3.
Determine where
f
is continuous, expressing
your answer in interval notation.
1.
(
−∞
,
−
3)
∪
(
−
3
,
3)
∪
(3
,
∞
)
2.
(
−∞
,
−
3)
∪
(
−
3
,
∞
)
3.
(
−∞
,
−
3)
∪
(3
,
∞
)
4.
(
−∞
,
3)
∪
(3
,
∞
)
5.
(
−∞
,
∞
)
correct
Explanation:
The ±unction is piecewise continuous, so we
have to check the le±t and right hand limits at
the points where the defnition o±
f
changes,
i.e.
, at
x
=
−
3 and
x
= 3. Now at
x
=
−
3
lim
x
→−
3
−
f
(
x
) =
lim
x
→−
3
−
6
−
x
= 9
,
lim
x
→−
3+
f
(
x
) =
lim
x
→−
3+
x
2
= 9
,
hence, the ±unction is continuous at the point
x
=
−
3. On the other hand, at
x
= 3
lim
x
→
3
−
f
(
x
) = lim
x
→
3
−
x
2
= (3)
2
= 9
,
lim
x
→
3+
f
(
x
) = lim
x
→
3+
6 +
x
= 9
.