B U Department of Mathematics
Math 101 Calculus I
Fall 2004 Final
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1.)
Let
f
(
x
) =
√

x
if
x <
0
3

x
if 0
≤
x <
3
(
x

3)
2
if
x >
3
Determine the point(s) at which
f
(
x
) is discontinuous. Explain in detail.
Solution:
Check the points
x
= 0 and
x
= 3 because these are candidate points of discontinuity.
x
= 0
:
lim
x
→
0

f
(
x
) = lim
x
→
0

√

x
= 0
lim
x
→
0
+
f
(
x
) = lim
x
→
0
+
(3

x
) = 3
⇒
lim
x
→
0
f
(
x
) doesn’t exist and therefore
f
is discontinuous at
x
= 0.
x
= 3
:
lim
x
→
3

f
(
x
) = lim
x
→
3

(3

x
) = 0
lim
x
→
3
+
f
(
x
) = lim
x
→
3
+
(
x

3)
2
= 0
⇒
lim
x
→
3
f
(
x
) = 0 but
f
(3) is undefined and therefore
f
is discontinuous at
x
= 3, too
because for continuity we must have lim
x
→
a
f
(
x
) =
f
(
a
)
.
2.)
For what values of
r
does the function
y
=
e
rx
satisfy the equation
y
00
+ 5
y
0

6
y
= 0?
Solution:
y
=
e
rx
⇒
y
0
=
re
rx
and
y
00
=
r
2
e
rx
Now substitute in the equation above
r
2
e
rx
+ 5
re
rx

6
e
rx
= 0
⇒
e
rx
(
r
2
+ 5
r

6) = 0
e
rx
cannot be zero and hence
r
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 Calculus, Convergence, lim

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