Question

find the positive value of cosx=xe^x using bisection method​

Answer 1

Answer:

f(x)=cosx-x.e^x

f(π/4)=cos(π/4)-(π/4).e^(π/4)

f(π/4)=1.7225

f(π/2)=cos(π/2)-(π/2).e^(π/2)

f(π/2)=0-(π/2).e^(π/2)=7.854

f(π/3)=cos(π/3)-(π/3)*e^(π/3)

f(π/3)=(1/2)-(π/3)e^(π/3)

f(π/3)=-2.484

So the root lies in the interval (π/4,π/3)=(a,b)

As f(π/4) is very near to zero

b'=[(π/3)+(π/4)]/2

b'=7π/24

So new interval is x belongs to (π/4,7π/24)

f(7π/24)=cos(7π/24)-(7π/24)*e^(7π/24)

f(7π/24)=-1.68199

The root lies in the interval x''=[(π/4)+(7π/24)]/2

x''=13π/48

So roots lies in (π/4,13π/48)

f(13π/48)=-1.9371

so new interval is (π/4,

x'''=[(π/4)+(13π/48)]/2

x'''=25π/96

f(25π/96)=-2.83525

x'''"=[(π/4)+(25π/96)]/2

x'''''=(39π)/(192)

So x is in (π/4, 39π/192)

f(39π/192)=cos(39π/192)-(39π/192)*e^(39π/192)

f(39π/192)=-0.787

so x'”'”=((π/4)+(39π/192))/2

x''''''=87π/384

f(87π/384)=-1.3638

On following the iteration the solution is nearly

x=87π/384