Home > Numerical methods calculators > False Position method calculator

 Solve any problem (step by step solutions) Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution will be displayed step by step (In 2 parts)
Solution
 Problem: regula falsi method 2x^3-2x-5, a=1, b=4 [ Calculator, Method and examples ]Solution:Your problem -> regula falsi method 2x^3-2x-5, a=1, b=4Here 2x^3-2x-5=0Let f(x) = 2x^3-2x-51^(st) iteration :Here f(1) = -5 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1 and x_1 = 4x_2 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_2 = 1 - (-5) * (4 - 1)/(115 - (-5))x_2 = 1.125f(x_2)=f(1.125)=2*(1.125)^3-2*(1.125)-5=-4.4023 < 02^(nd) iteration :Here f(1.125) = -4.4023 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1.125 and x_1 = 4x_3 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_3 = 1.125 - (-4.4023) * (4 - 1.125)/(115 - (-4.4023))x_3 = 1.231f(x_3)=f(1.231)=2*(1.231)^3-2*(1.231)-5=-3.7312 < 03^(rd) iteration :Here f(1.231) = -3.7312 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1.231 and x_1 = 4x_4 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_4 = 1.231 - (-3.7312) * (4 - 1.231)/(115 - (-3.7312))x_4 = 1.318f(x_4)=f(1.318)=2*(1.318)^3-2*(1.318)-5=-3.0568 < 04^(th) iteration :Here f(1.318) = -3.0568 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1.318 and x_1 = 4x_5 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_5 = 1.318 - (-3.0568) * (4 - 1.318)/(115 - (-3.0568))x_5 = 1.3875f(x_5)=f(1.3875)=2*(1.3875)^3-2*(1.3875)-5=-2.4331 < 05^(th) iteration :Here f(1.3875) = -2.4331 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1.3875 and x_1 = 4x_6 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_6 = 1.3875 - (-2.4331) * (4 - 1.3875)/(115 - (-2.4331))x_6 = 1.4416f(x_6)=f(1.4416)=2*(1.4416)^3-2*(1.4416)-5=-1.8914 < 06^(th) iteration :Here f(1.4416) = -1.8914 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1.4416 and x_1 = 4x_7 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_7 = 1.4416 - (-1.8914) * (4 - 1.4416)/(115 - (-1.8914))x_7 = 1.483f(x_7)=f(1.483)=2*(1.483)^3-2*(1.483)-5=-1.4431 < 07^(th) iteration :Here f(1.483) = -1.4431 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1.483 and x_1 = 4x_8 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_8 = 1.483 - (-1.4431) * (4 - 1.483)/(115 - (-1.4431))x_8 = 1.5142f(x_8)=f(1.5142)=2*(1.5142)^3-2*(1.5142)-5=-1.0851 < 08^(th) iteration :Here f(1.5142) = -1.0851 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1.5142 and x_1 = 4x_9 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_9 = 1.5142 - (-1.0851) * (4 - 1.5142)/(115 - (-1.0851))x_9 = 1.5374f(x_9)=f(1.5374)=2*(1.5374)^3-2*(1.5374)-5=-0.807 < 09^(th) iteration :Here f(1.5374) = -0.807 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1.5374 and x_1 = 4x_10 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_10 = 1.5374 - (-0.807) * (4 - 1.5374)/(115 - (-0.807))x_10 = 1.5546f(x_10)=f(1.5546)=2*(1.5546)^3-2*(1.5546)-5=-0.5952 < 010^(th) iteration :Here f(1.5546) = -0.5952 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1.5546 and x_1 = 4x_11 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_11 = 1.5546 - (-0.5952) * (4 - 1.5546)/(115 - (-0.5952))x_11 = 1.5672f(x_11)=f(1.5672)=2*(1.5672)^3-2*(1.5672)-5=-0.4363 < 011^(th) iteration :Here f(1.5672) = -0.4363 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1.5672 and x_1 = 4x_12 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_12 = 1.5672 - (-0.4363) * (4 - 1.5672)/(115 - (-0.4363))x_12 = 1.5764f(x_12)=f(1.5764)=2*(1.5764)^3-2*(1.5764)-5=-0.3184 < 012^(th) iteration :Here f(1.5764) = -0.3184 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1.5764 and x_1 = 4x_13 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_13 = 1.5764 - (-0.3184) * (4 - 1.5764)/(115 - (-0.3184))x_13 = 1.5831f(x_13)=f(1.5831)=2*(1.5831)^3-2*(1.5831)-5=-0.2316 < 013^(th) iteration :Here f(1.5831) = -0.2316 < 0 and f(4) = 115 > 0:. Now, Root lies between x_0 = 1.5831 and x_1 = 4x_14 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))

Solution provided by AtoZmath.com
Any wrong solution, solution improvement, feedback then Submit Here
Want to know about AtoZmath.com and me