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23 Maths -- Numerical Computations

Using bisection method, find the root of the equation f(x)=x2 -6logx-3=0(2<=x<=3)'correct to 4 decimal places with an accuracy of 10 -2 .

Using bisection method, find the root of the equation f(x)=x2 -6logx-3=0(2<=x<=3)'correct to 4 decimal places with an accuracy of 10 -2 .

2 Answers

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Here,  f(x)=x2 -6 logx-3=0

f(2)=4-6 log2-3=-0.806

f(3)=9-6 log3-3=3.1373

f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.

Hence, the root lies between 2 and 3

c0 =(2+3)/2=2.5

f(2.5)=6.25-6 log 2.5-3=0.8623

Now

n
 a(-ve)
 b(+ve)
 cn
 f(cn)
0
 2
 3
 2.5
 0.8623
1
 2
 2.5
 2.25
 -0.050595
2
 2.25
 2.5
 2.375
 0.38664
3
 2.25
 2.375
 2.3125
 0.1631658
4
 2.25
 2.3125
 2.28125
 0.05506
5
 2.25
 2.28125
 2.265625
 0.001925

From the table,

f(2.265625)=0.001928<10-2 

Therefore, the root of the given equation correct upto four paces of decimal=2.2656

Here,  f(x)=x2 -6 logx-3=0

f(2)=4-6 log2-3=-0.806

f(3)=9-6 log3-3=3.1373

f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.

Hence, the root lies between 2 and 3

c0 =(2+3)/2=2.5

f(2.5)=6.25-6 log 2.5-3=0.8623

Now

n
 a(-ve)
 b(+ve)
 cn
 f(cn)
0
 2
 3
 2.5
 0.8623
1
 2
 2.5
 2.25
 -0.050595
2
 2.25
 2.5
 2.375
 0.38664
3
 2.25
 2.375
 2.3125
 0.1631658
4
 2.25
 2.3125
 2.28125
 0.05506
5
 2.25
 2.28125
 2.265625
 0.001925

From the table,

f(2.265625)=0.001928<10-2 

Therefore, the root of the given equation correct upto four paces of decimal=2.2656

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