Example: Start with interval  a0 , b0  and use the 6 steps of False Position Method to find an interval
that contains a solution of the given equation
f  x   2 x  2x  5  0
Solution:
There are two methods for separation of roots.
1. Graphical Methods
2. Analytical Methods
1. Graphical Method
f  x   2 x  2x  5  0
y1  2 x and y2  5  2 x
There exist root on the interval 1, 2 , since f 1 f  2   0 , f 1  1  0, f  2   1.81  0 .
Prepared by Fatoş Rizaner, 2010
1
2. Analytical Method
The function f  x   2 x  2 x  5  0 and the derivative of the function f   x  
point is x 
1
 2  0 , the critical
x
1
.
4
x
0
1
8
1
4
1
3
2
2
Sign of f  x 
-
-
-
-
+
+
 3
There exist root on the interval 1, 2 or 1,  .
 2
Now, we use False Position Method for the function f  x   2 x  2 x  5  0 with the interval 1, 2 .
We should use
xi  b  f  b 
ba
, i  1, 2,3, 4,...
f b  f  a 
Step1. a  1, b  2, f 1  1, f 2   1.828427
2 1
x1  2  1.828427 
 1.353553
1.828427  1
x
1
1.353553
2
Sign of f  x 
-
+
+
Step2 a  1, b  1.353553, f 1  1, f 1.353553  0.033952
1.353553  1
x2  1.353553   0.033952 
 1.341943
0.033952  1
x
Sign of f  x 
1 1.341943 1.353553
-
+
+
Prepared by Fatoş Rizaner, 2010
2
Step3. a  1, b  1.341943, f 1  1, f 1.341943  0.000731
x3  1.341943   0.000731
1.341943  1
 1.341693
0.000731  1
x
1
Sign of f  x 
-
1.341693 1.341943
+
+
Step4. a  1, b  1.341693, f 1  1, f 1.341943  0.000015
1.341693  1
x4  1.341693   0.000015
 1.3416879
0.000015  1
x
1
Sign of f  x 
-
1.3416879 1.341693
+
+
Step5. a  1, b  1.3416879, f 1  1, f 1.3416879   0.00000845
1.3416879  1
x5  1.3416879   0.000000845 
0.000000845  1
 1.3416876
x
1
1.3416876
1.3416879
Sign of f  x 
-
-
+
Step6. a  1.3416876, b  1.3416879, f 1.3416876  0.0000000138, f 1.3416879  0.00000845
x6  1.3416879   1.38 108 
 1.341687905
1.3416879  1.3416876
8.45 107  1.38 108
x
Sign of
f  x
1.3416876 1.341687905 1.3416879
-
+
+
The results for 6 steps of False Position Methods for the function f  x   2 x  2 x  5  0 is shown
below.
Prepared by Fatoş Rizaner, 2010
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False Position Method
xk  b  f  b 
ba
f b  f  a 
f  xk 
xk 1  xk
1.353553
0.033952
1
1.353553
1.341943
0.000731
0.353553
1
1.341943
1.341693
0.000015
0.341943
3
1
1.341693
1.3416879
0.000000845
0.341693
4
1
1.3416879
1.3416876
-0.0000000138
0.3416879
1.341687905
0.0000008595
0.0000003
n
xk
xk 1
0
1
2
1
1
2
5
1.3416876 1.3416879
Prepared by Fatoş Rizaner, 2010
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