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

Solution:
Your problem -> regula falsi 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 x_0 = 1 and x_1 = 2

x_2 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))

x_2 = 1 - (-5) * (2 - 1)/(7 - (-5))

x_2 = 1.4167

f(x_2)=f(1.4167)=2*(1.4167)^3-2*(1.4167)-5=-2.147 < 0

2^(nd) iteration :

Here f(1.4167) = -2.147 < 0 and f(2) = 7 > 0

:. Now, Root lies between x_0 = 1.4167 and x_1 = 2

x_3 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))

x_3 = 1.4167 - (-2.147) * (2 - 1.4167)/(7 - (-2.147))

x_3 = 1.5536

f(x_3)=f(1.5536)=2*(1.5536)^3-2*(1.5536)-5=-0.6076 < 0

3^(rd) iteration :

Here f(1.5536) = -0.6076 < 0 and f(2) = 7 > 0

:. Now, Root lies between x_0 = 1.5536 and x_1 = 2

x_4 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))

x_4 = 1.5536 - (-0.6076) * (2 - 1.5536)/(7 - (-0.6076))

x_4 = 1.5892

f(x_4)=f(1.5892)=2*(1.5892)^3-2*(1.5892)-5=-0.1506 < 0

4^(th) iteration :

Here f(1.5892) = -0.1506 < 0 and f(2) = 7 > 0

:. Now, Root lies between x_0 = 1.5892 and x_1 = 2

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