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Problem: bisection method 2x^3-2x-5 [ Calculator, Method and examples ]

Solution:
Your problem -> bisection method 2x^3-2x-5

Here 2x^3-2x-5=0

Let f(x) = 2x^3-2x-5

Here
 x f(x) 0 1 2 -5 -5 7

1^(st) iteration :

Here f(1) = -5 < 0 and f(2) = 7 > 0

:. Now, Root lies between 1 and 2

x_0=(1+2)/2=1.5

f(x_0)=f(1.5)=2*(1.5)^3-2*(1.5)-5=-1.25 < 0

2^(nd) iteration :

Here f(1.5) = -1.25 < 0 and f(2) = 7 > 0

:. Now, Root lies between 1.5 and 2

x_1=(1.5+2)/2=1.75

f(x_1)=f(1.75)=2*(1.75)^3-2*(1.75)-5=2.2188 > 0

3^(rd) iteration :

Here f(1.5) = -1.25 < 0 and f(1.75) = 2.2188 > 0

:. Now, Root lies between 1.5 and 1.75

x_2=(1.5+1.75)/2=1.625

f(x_2)=f(1.625)=2*(1.625)^3-2*(1.625)-5=0.332 > 0

4^(th) iteration :

Here f(1.5) = -1.25 < 0 and f(1.625) = 0.332 > 0

:. Now, Root lies between 1.5 and 1.625

x_3=(1.5+1.625)/2=1.5625

f(x_3)=f(1.5625)=2*(1.5625)^3-2*(1.5625)-5=-0.4956 < 0

5^(th) iteration :

Here f(1.5625) = -0.4956 < 0 and f(1.625) = 0.332 > 0

:. Now, Root lies between 1.5625 and 1.625

x_4=(1.5625+1.625)/2=1.5938

f(x_4)=f(1.5938)=2*(1.5938)^3-2*(1.5938)-5=-0.0911 < 0

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