how to use method of false position to find root of x^3 − 5x + 2 = 0

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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


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