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Unformatted text preview: 114 1 Solutions . The initial conditions
and
imply that
. So
on
. By continuity it follows that
and
and these constitute the initial
values for the equation
on the interval
. The
general solution on this interval is
and at
we
get
and
.
Solving for and gives
and
so that
.
Putting the two pieces together gives
and if 33. 1.
for all
, while
.
2. It is enough to observe that
does not exist. But letting
gives
for all positive integers , while letting
gives
so there is one sequence
with
while another
sequence
with
so
cannot be continuous at 0.
3. To be piecewise continuous,
would have to have a limit at approaches 0 from above, and this is not true as shown in part 2. Section 5.2
if
if
if 1. Thus, the graph is if
if
if
if 2.
graph is Thus, the ...
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
 Fall '08
 BELL,D

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