syed (sms3768) – Quest HW 3 – seckin – (56425)
1
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21
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001
10.0 points
Functions
f
and
g
are defined on (
−
10
,
10)
by their respective graphs in
2
4
6
8
−
2
−
4
−
6
−
8
4
8
−
4
−
8
f
g
Find all values of
x
where the sum,
f
+
g
, of
f
and
g
is continuous, expressing your answer
in interval notation.
1.
(
−
10
,
2)
uniondisplay
(2
,
10)
correct
2.
(
−
10
,
−
5]
uniondisplay
[2
,
10)
3.
(
−
10
,
−
5)
uniondisplay
(
−
5
,
10)
4.
(
−
10
,
−
5)
uniondisplay
(
−
5
,
2)
uniondisplay
(2
,
10)
5.
(
−
10
,
10)
Explanation:
Since
f
and
g
are piecewise linear, they are
continuous
individually
on (
−
10
,
10) except
at their ‘jumps’,
i.e.
, at
x
=
−
5 in the case of
f
and
x
=
−
5
,
2 in the case of
g
. But the sum
of continuous functions is again continuous, so
f
+
g
is certainly continuous on
(
−
10
,
−
5)
uniondisplay
(
−
5
,
2)
uniondisplay
(2
,
10)
.
The only question is what happens at
x
0
=
−
5
,
2. To do that we have to check if
lim
x
→
x
0
−
{
f
(
x
) +
g
(
x
)
}
=
f
(
x
0
) +
g
(
x
0
)
=
lim
x
→
x
0
+
{
f
(
x
) +
g
(
x
)
}
.
Now at
x
0
=
−
5,
lim
x
→ −
5
−
{
f
(
x
) +
g
(
x
)
}
=
−
4 =
f
(
−
5) +
g
(
−
5)
=
lim
x
→ −
5+
{
f
(
x
) +
g
(
x
)
}
,
while at
x
0
= 2,
lim
x
→
2
−
{
f
(
x
) +
g
(
x
)
}
=
−
8
negationslash
=
−
6 =
lim
x
→
2+
{
f
(
x
) +
g
(
x
)
}
.
Thus,
f
+
g
is continuous at
x
=
−
5, but not
at
x
= 2.
Consequently, on (
−
10
,
10) the
sum
f
+
g
is continuous at all
x
in
(
−
10
,
2)
uniondisplay
(2
,
10)
.
002
10.0 points
If the function
f
is continuous everywhere
and
f
(
x
) =
x
2
−
16
x
−
4
when
x
negationslash
= 4, find the value of
f
(4).
1.
f
(4) =
−
4
2.
f
(4) = 4
3.
f
(4) =
−
8
4.
f
(4) = 8
correct
5.
f
(4) = 16
6.
f
(4) =
−
16
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syed (sms3768) – Quest HW 3 – seckin – (56425)
2
Explanation:
Since
f
is continuous at
x
= 4,
f
(4) =
lim
x
→
4
f
(
x
)
.
But, after factorization,
x
2
−
16
x
−
4
=
(
x
−
4)(
x
+ 4)
x
−
4
=
x
+ 4
,
whenever
x
negationslash
= 4. Thus
f
(
x
) =
x
+ 4
for all
x
negationslash
= 4. Consequently,
f
(4) =
lim
x
→
4
(
x
+ 4) = 8
.
003
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 the values of
x
at which
f
fails to be continuous on (
−
8
,
8).
1.
x
=
−
5
,
−
1
2.
x
=
−
5
,
3
3.
x
=
−
1
,
3
4.
x
=
−
5
,
−
1
,
3
correct
5.
f
is continuous everywhere
Explanation:
The function
f
is continuous at a point
a
in
(
−
8
,
8) when
(i)
f
(
a
) is defined,
(ii) lim
x
→
a
f
(
x
) exists, and
(iii) lim
x
→
a
f
(
x
) =
f
(
a
).
We check where one or more of these condi
tions fails.
(i) This fails at
a
= 3.
The only other possible candidates are
a
=
−
5 and
x
0
=
−
1:
(ii) At
x
0
=
−
5
lim
x
→−
5
−
f
(
x
) =
−
5
negationslash
=
lim
x
→−
5+
f
(
x
) =
−
2
so the limit does not exist; while
(iii) at
x
0
=
−
1,
f
(
−
1) = 1
negationslash
=
lim
x
→ −
1
f
(
x
) =
−
2
,
so the limit exists but does not have value
f
(
a
). Consequently,
f
fails to be continuous
only at
x
=
−
5
,
−
1
,
3
on (
−
8
,
8).
004
10.0 points
Determine which of the following could be
the graph of
f
near the origin when
f
(
x
) =
x
2
−
7
x
+ 10
2
−
x
,
x
negationslash
= 2
,
2
,
x
= 2
.
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 Spring '10
 seckin
 lim, Continuous function, Syed, Quest HW

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