Solutions to selected exercises
•
Use the Bisection method to find solutions accurate to within 10
−
2
for
x
3
−
7
x
2
+ 14
x
−
6 = 0 on [0,1].
Solution:
Let
f
(
x
) =
x
3
−
7
x
2
+ 14
x
−
6 = 0. Note that
f
(0) =
−
6
<
0
and
f
(1) = 2
>
0, therefore, based on the Intermediate Value Theorem, since
f
is continuous, there is
p
∈
(0
,
1) such that
f
(
p
) = 0.
Let
a
0
= 0,
b
0
= 1, with
f
(
a
0
)
<
0,
f
(
b
0
)
>
0.
Let
p
0
=
a
0
+
b
0
−
a
0
2
= 0
.
5, and we have
f
(
p
0
) =
−
0
.
6250
<
0 (the same
sign as
f
(
a
0
), therefore
a
1
=
p
0
= 0
.
5,
b
1
=
b
0
= 1 and repeat:
p
1
= 0
.
75, ...
This yields the following results for
p
n
and
f
(
p
n
):
n
p
n
f
(
p
n
)
0
0.5
0.6250000
1
0.75000000
+0.9843750
2
0.62500000
+0.2597656
3
0.56250000
0.1618652
4
0.59375000
+0.0540466
5
0.57812500
0.0526237
6
0.58593750
+0.0010313
•
Use the theorem from the course to find a bound for the number of
iterations needed to achieve an approximation with accuracy 10
−
3
to the
solution of
x
3
−
x
−
1 = 0 lying in the interval [1
,
4].
Solution:
Let’s first verify that
f
has a zero in the interval [1
,
4]:
f
(1) =
−
2
<
0,
f
(4) = 64
>
0, therefore, since
f
is continuous, by the Intermediate
Value Theorem,
f
has a zero in [1
,
4].
By the theorem from the course, we impose:

p
n
−
p
 ≤
b
−
a
2
n
=
3
2
n
≤
10
−
3
,
then
3
·
10
3
≤
2
n
⇒
n
≥
log
10
(3
·
10
3
)
log
10
(2)
≈
11
.
55
•
Use algebraic manipulation to show that each of the following functions
has a fixed point at
p
precisely when
f
(
p
) = 0, where
f
(
x
) =
x
4
+2
x
2
−
x
−
3.
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 Spring '08
 staff
 Intermediate Value Theorem, 3 1 G

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