roberts (eer474) – 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
,
−
4]
uniondisplay
[2
,
10)
2.
(
−
10
,
−
4)
uniondisplay
(
−
4
,
10)
correct
3.
(
−
10
,
−
4)
uniondisplay
(
−
4
,
2)
uniondisplay
(2
,
10)
4.
(
−
10
,
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
= 2 in the case of
f
and
x
= 2
,
−
4 in the case of
g
. But the sum
of continuous functions is again continuous,
so
f
+
g
is certainly continuous on
(
−
10
,
−
4)
uniondisplay
(
−
4
,
2)
uniondisplay
(2
,
10)
.
The only question is what happens at
x
0
=
2
,
−
4. 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
= 2,
lim
x
→
2
−
{
f
(
x
) +
g
(
x
)
}
= 2 =
f
(2) +
g
(2)
=
lim
x
→
2+
{
f
(
x
) +
g
(
x
)
}
,
while at
x
0
=
−
4,
lim
x
→ −
4
−
{
f
(
x
) +
g
(
x
)
}
= 4
negationslash
= 6 =
lim
x
→ −
4+
{
f
(
x
) +
g
(
x
)
}
.
Thus,
f
+
g
is continuous at
x
= 2, but not
at
x
=
−
4. Consequently, on (
−
10
,
10) the
sum
f
+
g
is continuous at all
x
in
(
−
10
,
−
4)
uniondisplay
(
−
4
,
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) = 16
2.
f
(
−
4) = 4
3.
f
(
−
4) =
−
16
4.
f
(
−
4) = 8
5.
f
(
−
4) =
−
8
correct
6.
f
(
−
4) =
−
4
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roberts (eer474) – 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
=
−
2
,
6
2.
x
=
−
6
,
6
3.
x
=
−
6
,
−
2
,
6
correct
4.
f
is continuous everywhere
5.
x
=
−
6
,
−
2
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
= 6.
The only other possible candidates are
a
=
−
6 and
x
0
=
−
2:
(ii) At
x
0
=
−
6
lim
x
→−
6
−
f
(
x
) =
−
5
negationslash
=
lim
x
→−
6+
f
(
x
) =
−
2
so the limit does not exist; while
(iii) at
x
0
=
−
2,
f
(
−
2) = 1
negationslash
=
lim
x
→ −
2
f
(
x
) =
−
2
,
so the limit exists but does not have value
f
(
a
). Consequently,
f
fails to be continuous
only at
x
=
−
6
,
−
2
,
6
on (
−
8
,
8).
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 Spring '09
 Gualdani
 Differential Calculus, lim, Continuous function, Quest HW

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