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Find false position method 2x^3-2x-5, a=1, b=4

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`






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