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: false position method 2x^3-2x-5, a=1, b=4 [ Calculator, Method and examples ]

Solution:
Your problem `->` false position method 2x^3-2x-5, a=1, b=4


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

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


`1^(st)` iteration :

Here `f(1) = -5 < 0` and `f(4) = 115 > 0`

`:.` Now, Root lies between `x_0 = 1` and `x_1 = 4`

`x_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 = 1.125`

`f(x_2)=f(1.125)=2*(1.125)^3-2*(1.125)-5=-4.4023 < 0`


`2^(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 = 4`

`x_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 = 1.231`

`f(x_3)=f(1.231)=2*(1.231)^3-2*(1.231)-5=-3.7312 < 0`


`3^(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 = 4`

`x_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 = 1.318`

`f(x_4)=f(1.318)=2*(1.318)^3-2*(1.318)-5=-3.0568 < 0`


`4^(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 = 4`

`x_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 = 1.3875`

`f(x_5)=f(1.3875)=2*(1.3875)^3-2*(1.3875)-5=-2.4331 < 0`


`5^(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 = 4`

`x_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 = 1.4416`

`f(x_6)=f(1.4416)=2*(1.4416)^3-2*(1.4416)-5=-1.8914 < 0`


`6^(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 = 4`

`x_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 = 1.483`

`f(x_7)=f(1.483)=2*(1.483)^3-2*(1.483)-5=-1.4431 < 0`


`7^(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 = 4`

`x_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 = 1.5142`

`f(x_8)=f(1.5142)=2*(1.5142)^3-2*(1.5142)-5=-1.0851 < 0`


`8^(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 = 4`

`x_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 = 1.5374`

`f(x_9)=f(1.5374)=2*(1.5374)^3-2*(1.5374)-5=-0.807 < 0`


`9^(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 = 4`

`x_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 = 1.5546`

`f(x_10)=f(1.5546)=2*(1.5546)^3-2*(1.5546)-5=-0.5952 < 0`


`10^(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 = 4`

`x_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 = 1.5672`

`f(x_11)=f(1.5672)=2*(1.5672)^3-2*(1.5672)-5=-0.4363 < 0`


`11^(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 = 4`

`x_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 = 1.5764`

`f(x_12)=f(1.5764)=2*(1.5764)^3-2*(1.5764)-5=-0.3184 < 0`


`12^(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 = 4`

`x_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 = 1.5831`

`f(x_13)=f(1.5831)=2*(1.5831)^3-2*(1.5831)-5=-0.2316 < 0`


`13^(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 = 4`

`x_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
  
 

 
Copyright © 2019. All rights reserved. Terms, Privacy





We use cookies to improve your experience on our site and to show you relevant advertising. By browsing this website, you agree to our use of cookies. Learn more