Math 4650 Homework #1 Solutions
1. 1.1 #2. Find intervals containing solutions to the following equations.
Solution:
For all of these problems, we need to use the intermediate value theorem.
So we want to find numbers
a
and
b
such that
f
(
a
)
>
0 and
f
(
b
)
<
0; then there is a
solution to
f
(
x
) = 0 for some
x
between
a
and
b
.
(a)
x

3

x
= 0. The function is
f
(
x
) =
x

3

x
. Use trial and error:
f
(0) =

1,
f
(1) = 1

1
3
=
2
3
. Since
f
(0) and
f
(1) have opposite signs, there is a solution in
the interval (0
,
1).
(b) 4
x
2

e
x
= 0. The function is
f
(
x
) = 4
x
2

e
x
.
f
(0) =

1,
f
(1) = 4

e
= 1
.
3
. . .
,
so again there is a solution in the interval (0
,
1).
(c)
x
3

2
x
2

4
x
+3 = 0. The function is
f
(
x
) =
x
3

2
x
2

4
x
+3. We have
f
(0) = 3,
f
(1) =

2, so again there is a solution in (0
,
1).
(d)
x
3
+ 4
.
001
x
2
+ 4
.
002
x
+ 1
.
101 = 0. We have
f
(
x
) =
x
3
+ 4
.
001
x
2
+ 4
.
002
x
+ 1
.
101,
so that
f
(0) = 1
.
101,
f
(1) = 10
.
104,
f
(2)
>
33, etc. We’re going in the wrong
direction. In the other direction, we have
f
(

1) =

1+4
.
001

4
.
002+1
.
101 = 0
.
1,
f
(

2) =

8+16
.
004

8
.
004+1
.
101 = 1
.
101, and
f
(

3) =

27+36
.
009

12
.
006+
1
.
101 =

1
.
896. Since
f
(

2) and
f
(

3) have opposite signs, there is a solution
in the interval (

3
,

2).
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 Spring '10
 To
 Equations, Intermediate Value Theorem, Taylor Series, Continuous function, Natural logarithm

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