MATH 128A, SUMMER 2010, HOMEWORK 1 SOLUTION
BENJAMIN JOHNSON
Homework 1: Due Monday, June 28
1.1; 1b, 4b, 6, 9abcd, 21, 25abc
1.2; 1b, 3c, 5a, 10ab, 15ab
section 1.1
1. Show that the following equations have at least one solution in the given intervals.
b. (
x

2)
2

ln
x
= 0, [1
,
2] and [
e,
4]
solution:
Let
f
(
x
) = (
x

2)
2

ln
x
.
We have
f
(1) = 1 and
f
(2) =

ln 2.
Zero
is between 1 and

ln 2, so by the intermediate value theorem there is a real number
c
∈
[1
,
2] with
f
(
c
) = 0.
For the interval [
e,
4], we have
f
(
e
) = (
e

2)
2

1 and
f
(4) = 2
2

ln 4. Since
e

2
<
1,
we have
f
(
e
)
<
0 and since ln 4
<
4 we have
f
(4)
>
0. Again the desired result follows
from the intermediate value theorem.
4. Find max
a
≤
x
≤
b

f
(
x
)

for the following functions and intervals.
b.
f
(
x
) = (4
x

3)
/
(
x
2

2
x
), [0
.
5
,
1]
solution:
The denominator is nonzero on [0
.
5
,
1], but the numerator changes sign in this
interval. Because of the absolute value we need to check the point at which the numerator
changes sign as a possible extremum, along with the endpoints of the interval, and any
points in the interval where the derivative might be zero.
For the derivative, we have
f
0
(
x
) =
4(
x
2

2
x
)

(4
x

3)(2
x

2)
(
x
2

2
x
)
2
=
4
x
2

8
x

8
x
2
+ 8
x
+ 6
x

6
(
x
2

2
x
)
2
=

4
x
2
+ 6
x

6
(
x
2

2
x
)
2
We see
f
0
is always negative on [0
.
5
,
1]. The numerator of
f
is zero at
x
= 3
/
4 but zero is
the absolute minimum for the function because it is always nonnegative; so the max must
be at an extreme point. We have
f
(0
.
5) =

1

.
75
= 4
/
3, and
f
(1) =

1
/

1

= 1; so the max
of
f
on this interval is
f
(0
.
5) = 4
/
3.
6. Suppose
f
∈
C
[
a, b
] and
f
0
(
x
) exists on (
a, b
). Show that if
f
0
(
x
)
6
= 0 for all
x
∈
(
a, b
), then
there can exist at most one number
p
in [
a, b
] with
f
(
p
) = 0.
solution:
We argue by contradiction. Suppose to the contrary that there are two points
p, q
∈
[
a, b
] with
f
(
p
) =
f
(
q
) = 0. Since
f
is differentiable on (
a, b
), Rolle’s theorem implies
that there exists a real number
c
in (
a, b
) with
f
0
(
c
) = 0. But this contradicts our hypothesis
Date
: June 28, 2010.
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 Fall '10
 To
 Equations, Continuous function, benjamin johnson

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