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Chapter 7
which implies that
f
(
t
) is continuous from the right at
t
=1
.Thus
f
(
t
) is continuous at
t
=1
and, therefore, is continuous at any
t
in [0
,
10].
23.
All the functions involved in the deFnition of
f
(
t
), that is,
g
1
(
t
)
≡
1,
g
2
(
t
)=
t
−
1, and
g
3
(
t
)=
t
2
−
4, are continuous on (
−∞
,
∞
). So,
f
(
t
), being a restriction of these functions, on
[0
,
1), (1
,
3), and (3
,
10], respectively, is continuous on these three intervals. At points
t
=1
and 3,
f
(
t
) is not deFned and so is not continuous. But onesided limits
lim
t
→
1
−
f
(
t
) = lim
t
→
1
−
g
1
(
t
)=
g
1
(1) = 1
,
lim
t
→
1
+
f
(
t
) = lim
t
→
1
+
g
2
(
t
)=
g
2
(1) = 0
,
lim
t
→
3
−
f
(
t
) = lim
t
→
3
−
g
2
(
t
)=
g
2
(3) = 2
,
lim
t
→
3
+
f
(
t
) = lim
t
→
3
+
g
3
(
t
)=
g
3
(3) = 5
,
exist and pairwise di±erent. Therefore,
f
(
t
) has jump discontinuities at
t
=1and
t
=3
,and
hence piecewise continuous on [0
,
10].
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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