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

Solution:
Your problem `->` bisection 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 `1` and `4`

`x_0=(1+4)/2=2.5`

`f(x_0)=f(2.5)=2(2.5)^(3)-2(2.5)-5=21.25 > 0`


`2^(nd)` iteration :

Here `f(1) = -5 < 0` and `f(2.5) = 21.25 > 0`

`:.` Now, Root lies between `1` and `2.5`

`x_1=(1+2.5)/2=1.75`

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


`3^(rd)` iteration :






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