PRACTICE PROBLEMS FOR MIDTERM – SOLUTIONS
The test will be given on
Wednesday, April 28
.
It will cover Sections
17-20 and 28-29 (everything covered in this quarter, up to and including
Mean Value Theorem).
In preparing for the test, you can practice solving the problems from the list
below. In addition, take a look at the homework assignments 1-3 (
at least
one problem on the midterm will come from the homework
), at the examples
given in the textbook, and the the examples discussed in class.
1.
Suppose the functions
f
,
g
, and
h
are defined on the interval (
a, b
), and
g
(
x
)
≤
f
(
x
)
≤
h
(
x
) for every
x
∈
(
a, b
). Suppose, furthermore, that
g
and
h
are continuous at
c
∈
(
a, b
), and
g
(
c
) =
h
(
c
). Prove that
f
is continuous
at
c
.
We need to show that, if (
x
n
) is a sequence of elements of (
a, b
), satis-
fying lim
x
n
=
c
, then lim
f
(
x
n
) =
f
(
c
). But
g
(
x
n
)
≤
f
(
x
n
)
≤
h
(
x
n
), and
lim
g
(
x
n
) =
g
(
c
) =
f
(
c
), and lim
h
(
x
n
) =
h
(
c
) =
f
(
c
). By Squeeze Theorem,
lim
f
(
x
n
) =
f
(
c
).
2.
Prove that there exists
x
∈
[0
,
2] s.t.
x
5
−
3
x
−
1 =
√
x
3
+ 1.
Consider
f
(
x
) =
x
5
−
3
x
−
1
−
√
x
3
+ 1.
Then
f
is continuous on [0
,
2].
Indeed,
f
(
x
) =
g
(
x
)
−
h
(
x
), where
g
(
x
) =
x
5
−
3
x
−
1,
h
(
x
) =
radicalbig
k
(
x
),
and
k
(
x
) =
x
3
+ 1.
g
and
k
are polynomials, hence they are continuous.
Moreover,
k
(
x
)
≥
0 for any
x
∈
[0
,
2], hence
h
is also continuous on [0
,
2].
This implies the continuity of
f
=
g
−
h
.
Observe that
f
(0) =
−
2, and
f
(2) = 32
−
6
−
1
−
3 = 22. By Intermediate
value Theorem,
f
(
x
) = 0 for some
x
∈
(0
,
2).
Therefore,
x
5
−
3
x
−
1 =
√
x
3
+ 1.
3.
Let
f
(
x
) = 1
/
√
x
. Prove that
f
is not uniformly continuous on (0
,
1].
Solution 1
. We know that a uniformly continuous function maps a Cauchy
sequence to a Cauchy sequence. This is not the case for
f
. Indeed, let
s
n
=
1
/n
(
n
∈
N
).
The sequence (
s
n
) is convergent, hence Cauchy.
Moreover,
s
n
∈
(0
,
1] for any
n
. However,
f
(
s
n
) =
√
n
, and lim
f
(
s
n
) = +
∞
. Thus,
the sequence (
f
(
s
n
)) is not Cauchy.