13.2 (a)
The function can be plotted
120
80
40
0
40
2
1
0
1
2
(b)
The function can be differentiated twice to give
2
4
24
45
)
(
"
x
x
x
f
−
−
=
Thus, the second derivative will always be negative and hence the function is concave for all
values of
x
.
(c)
Differentiating the function and setting the result equal to zero results in the following
roots problem to locate the maximum
12
8
9
0
)
(
'
3
5
+
−
−
=
=
x
x
x
f
A plot of this function can be developed
400
200
0
200
400
2
1
0
1
2
A technique such as bisection can be employed to determine the root. Here are the first few
iterations:
iteration
x
l
x
u
x
r
f
(
x
l
)
f
(
x
r
)
f
(
x
l
)
×
f
(
x
r
)
ε
a
1
0.00000
2.00000
1.00000
12
5
60.0000
2
0.00000
1.00000
0.50000
12
10.71875
128.6250
100.00%
3
0.50000
1.00000
0.75000
10.71875
6.489258
69.5567
33.33%
4
0.75000
1.00000
0.87500
6.489258
2.024445
13.1371
14.29%
5
0.87500
1.00000
0.93750
2.024445
1.10956
2.2463
6.67%
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View Full DocumentThe approach can be continued to yield a result of
x
= 0.91692.
13.3
First, the golden ratio can be used to create the interior points,
2361
.
1
)
0
2
(
2
1
5
=
−
−
=
d
2361
.
1
2361
.
1
0
1
=
+
=
x
7639
.
0
2361
.
1
2
2
=
−
=
x
The function can be evaluated at the interior points
1879
.
8
)
7639
.
0
(
)
(
2
=
=
f
x
f
8142
.
4
)
2361
.
1
(
)
(
1
=
=
f
x
f
Because
f
(
x
2
) >
f
(
x
1
), the maximum is in the interval defined by
x
l
,
x
2
, and
x
1
.where
x
2
is the
optimum. The error at this point can be computed as
%
100
%
100
7639
.
0
0
2
)
61803
.
0
1
(
=
×
−
−
=
a
ε
For the second iteration,
x
l
= 0 and
x
u
= 1.2361. The former
x
2
value becomes the new
x
1
, that
is,
x
1
= 0.7639 and
f
(
x
1
) = 8.1879. The new values of
d
and
x
2
can be computed as
7639
.
0
)
0
2361
.
1
(
2
1
5
=
−
−
=
d
4721
.
0
7639
.
0
2361
.
1
2
=
−
=
x
The function evaluation at
f
(
x
2
) = 5.5496. Since this value is less than the function value at
x
1
,
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 Spring '10
 ROCKE
 Derivative, unimodal function

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