Question : Use method of false position to find root of x^3 − 5x + 2 = 0 lying in the interval (0,1) correct to three decimal places.
Solution :
Iteration 1:
f(a)= f(0)= 2, f(b)= f(1)= -2
By the Method of false position
\[c =\frav{ af(b)- bf(a)}{(f(b)-f(a) }\]
c=0.5
Now,
f(c) = f(0.5) = -0.375.
Since, f(a)f(c) < 0, the root lies in the interval (0, 0.5)
Iteration 2:
a= 0.000, f(a)= 2.00
b= 0.500,f(b)= -0.375
c= 0.250,f(c)= 0.766
Since, f(b)f(c) < 0, the root lies in the interval (0.5, 0.25)
Iteration 3:
a= 0.250, f(a)= 0.766
b= 0.500,f(b)= -0.375
c= 0.375,f(c)= 0.178
Since, f(b)f(c) < 0, the root lies in the interval (0.5, 0.375)
Iteration 4:
a= 0.375, f(a)= 0.178
b= 0.500,f(b)= -0.375
c= 0.438,f(c)= -0.104
Since, f(a)f(c) < 0, the root lies in the interval (0.375, 0.438)
Iteration 5:
a= 0.375, f(a)= 0.178
b= 0.438,f(b)= -0.104
c= 0.406,f(c)= 0.0358
Since, f(b)f(c) < 0, the root lies in the interval ( 0.438,0.406)
Iteration 6:
a= 0.406 f(a)= 0.0358
b= 0.438 ,f(b)= -0.104
c= 0.422 ,f(c)= -0.0343
Since, f(a)f(c) < 0, the root lies in the interval ( 0.406,0.422)
iteration 7:
a= 0.406, f(a)= 0.0358
b= 0.422,f(b)= -0.0343
c= 0.414 , f(c)= 0.000678
Since, f(b)f(c) < 0, the root lies in the interval ( 0.406,0.414)
iteration 8:
a= 0.414, f(a)= 0.000678
b= 0.422,f(b)= -0.03.43
c= 0.418,f(c)= -0.0168
Since, f(a)f(c) < 0, the root lies in the interval ( 0.414,0.418)
iteration= 9
a= 0.414,f(a)= 0.000678
b= 0.418 ,f(b)= -0168e
c= 0.416 ,f(c)= -0.00808
Since, f(a)f(c) < 0, the root lies in the interval ( 0.414,0.416)
iteration= 10
a= 0.414,f(a)= 0.000678
b= 0.416 ,f(b)= -0.00808
c= 0.415 ,f(c)= -0.00370
Since, f(a)f(c) < 0, the root lies in the interval ( 0.414,0.415)
iteration= 11
a= 0.414, f(a)= 0.000678
b= 0.415 ,f(b)= -0.00370
c= 0.415,f(c)= -0.00151
Since, f(a)f(c) < 0, the root lies in the interval ( 0.414,0.415)
iteration= 12
a= 0.414, f(a)= 0.000678
b= 0.415 ,f(b)= -0.00151
c= 0.414 ,f(c)= -0.000417
Since, f(a)f(c) < 0, the root lies in the interval ( 0.414,0.414)
Therefore
Root of the given equation lying in the interval (0,1) after 12 iterations of False position method is given as 0.414 which is correct to three decimal places.
and graph of the function is
