4. Determine the highest real root of
f (x) = 2
– 11.7
+ 17.7x – 5
a) Graphically
b) Fixed point iteration method (three iteration, 
= 3).Compute the approximate percent relative errors for your solutions.
[Answer]
First iteration
= 3
= 
When i = 0 then = 3
=
= 
= 3.181
The approximate error is
= 
x 100%
= 
x 100%
= 5.68%
Second iteration
= 3.181
=
When i = 1 then = 3.181
=
= 
= 3.333
The approximate error is
= x 100%
= 
x 100%
= 4.59%
The rest of iterations are shown as below. The steps are same as first iteration and second iteration.
i |
|
|
0 | 3 | |
1 | 3.181 | 5.684 |
2 | 3.334 | 4.594 |
3 | 3.443 | 3.154 |
The root is 3.443 which has 3.154% error.
c) Newton-Raphson method (three iterations, = 3).Compute the approximate percent relative errors for your solutions.
[Answer]
The Newton Raphson formula is shown as below,
Then the f(x) is shown below,
f (x) = 2 – 11.7 + 17.7x – 5
After differentiate,
f’ (x) = 6 – 23.4
+ 17.7
Then, let initial guesses as 3
First iteration
= 3, i=0
You have to substitute = 3 into two equations
f (x) = 2 – 11.7 + 17.7x – 5
f (3) = 2
– 11.7
+ 17.7(3) – 5
f (3) = -3.2
f’ (x) = 6 – 23.4 + 17.7
f’ (3) = 6 – 23.4
+ 17.7
f’ (3) = 1.5
Then substitute = 3, f (3) = -3.2, and f’ (3) = 1.5 into equation below
The approximate error is
= x 100%
= 
x 100%
= 41.55%
Second iteration
= 5.133, i=1
You have to substitute = 5.133 into two equations
f (x) = 2 – 11.7 + 17.7x – 5
f (5.133) = 2
– 11.7
+ 17.7(5.133) – 5
f (5.133) = 48.09
f’ (x) = 6 – 23.4 + 17.7
f’ (5.133) = 6 – 23.4
+ 17.7
f’ (5.133) = 55.687
Then substitute = 5.133, f (5.133) = 48.09, and f’ (5.133) = 55.687 into equation below
The approximate error is
= x 100%
= 
x 100%
= 20.21%
The rest of iterations are shown in table
i |
| f (x) | f’ (x) |
|
0 | 3 | -3.2 | 1.5 | |
1 | 5.133 | 48.09 | 55.687 | 41.56 |
2 | 4.269 | 12.957 | 27.173 | 20.23 |
3 | 3.793 | 2.948 | 15.263 | 12.57 |
The root is 3.793 which 12.57%
d) Secant method (three iterations, 
= 3, = 4 ).Compute the approximate percent relative errors for your solutions.
[Answer]
The formula of secant method is shown as
First iteration
i=0
= 3
= 2 – 11.7 + 17.7x – 5
= 2(
– 11.7 + 17.7(3) – 5
= -3.2
= 4
= 2 – 11.7 + 17.7x – 5
= 2
– 11.7
+ 17.7(4) – 5
= 6.6
Substitute = 3, = 4, = -3.2, = 6.6 into
The approximate error is
= x 100%
= 
x 100%
= 20.23%
The first iteration is shown in table
i | |
|
|
-1 | 3 | -3.2 | |
0 | 4 | 6.6 | |
1 | 3.3265 | -1.969 | 20.25 |
The second iteration is shown as
i | |
|
|
0 | 4 | 6.6 | |
1 | 3.3265 | -1.969 | |
2 | 3.4813 | -0.796 | 4.445 |
The third iteration is shown as
i | |
|
|
1 | 3.3265 | -1.969 | |
2 | 3.4813 | -0.796 | |
3 | 3.586 | 0.248 | 2.927 |
The root is 3.586 which has 2.927% error.