Bisection method Example-2 f(x)=2x^3-2x-5

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Home > Numerical methods calculators > Bisection method example

1. Bisection method example ( Enter your problem )

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1. Algorithm & Example-1 `f(x)=x^3-x-1`
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3. Example-3 `x=sqrt(12)`
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2. Example-2 `f(x)=2x^3-2x-5`

Find a root of an equation `f(x)=2x^3-2x-5` using Bisection method

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

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

Here

`x`	0	1	2
`f(x)`	-5	-5	7

`1^(st)` iteration :

Here `f(1) = -5 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `1` and `2`

`x_0=(1+2)/2=1.5`

`f(x_0)=f(1.5)=2*1.5^(3)-2*1.5-5=-1.25 < 0`

`2^(nd)` iteration :

Here `f(1.5) = -1.25 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `1.5` and `2`

`x_1=(1.5+2)/2=1.75`

`f(x_1)=f(1.75)=2*1.75^(3)-2*1.75-5=2.21875 > 0`

`3^(rd)` iteration :

Here `f(1.5) = -1.25 < 0` and `f(1.75) = 2.21875 > 0`

`:.` Now, Root lies between `1.5` and `1.75`

`x_2=(1.5+1.75)/2=1.625`

`f(x_2)=f(1.625)=2*1.625^(3)-2*1.625-5=0.33203 > 0`

`4^(th)` iteration :

Here `f(1.5) = -1.25 < 0` and `f(1.625) = 0.33203 > 0`

`:.` Now, Root lies between `1.5` and `1.625`

`x_3=(1.5+1.625)/2=1.5625`

`f(x_3)=f(1.5625)=2*1.5625^(3)-2*1.5625-5=-0.49561 < 0`

`5^(th)` iteration :

Here `f(1.5625) = -0.49561 < 0` and `f(1.625) = 0.33203 > 0`

`:.` Now, Root lies between `1.5625` and `1.625`

`x_4=(1.5625+1.625)/2=1.59375`

`f(x_4)=f(1.59375)=2*1.59375^(3)-2*1.59375-5=-0.09113 < 0`

`6^(th)` iteration :

Here `f(1.59375) = -0.09113 < 0` and `f(1.625) = 0.33203 > 0`

`:.` Now, Root lies between `1.59375` and `1.625`

`x_5=(1.59375+1.625)/2=1.60938`

`f(x_5)=f(1.60938)=2*1.60938^(3)-2*1.60938-5=0.1181 > 0`

`7^(th)` iteration :

Here `f(1.59375) = -0.09113 < 0` and `f(1.60938) = 0.1181 > 0`

`:.` Now, Root lies between `1.59375` and `1.60938`

`x_6=(1.59375+1.60938)/2=1.60156`

`f(x_6)=f(1.60156)=2*1.60156^(3)-2*1.60156-5=0.0129 > 0`

`8^(th)` iteration :

Here `f(1.59375) = -0.09113 < 0` and `f(1.60156) = 0.0129 > 0`

`:.` Now, Root lies between `1.59375` and `1.60156`

`x_7=(1.59375+1.60156)/2=1.59766`

`f(x_7)=f(1.59766)=2*1.59766^(3)-2*1.59766-5=-0.03926 < 0`

`9^(th)` iteration :

Here `f(1.59766) = -0.03926 < 0` and `f(1.60156) = 0.0129 > 0`

`:.` Now, Root lies between `1.59766` and `1.60156`

`x_8=(1.59766+1.60156)/2=1.59961`

`f(x_8)=f(1.59961)=2*1.59961^(3)-2*1.59961-5=-0.01322 < 0`

`10^(th)` iteration :

Here `f(1.59961) = -0.01322 < 0` and `f(1.60156) = 0.0129 > 0`

`:.` Now, Root lies between `1.59961` and `1.60156`

`x_9=(1.59961+1.60156)/2=1.60059`

`f(x_9)=f(1.60059)=2*1.60059^(3)-2*1.60059-5=-0.00017 < 0`

Approximate root of the equation `2x^3-2x-5=0` using Bisection method is `1.60059`

`n`	`a`	`f(a)`	`b`	`f(b)`	`c=(a + b)/2`	`f(c)`	Update
1	1	-5	2	7	1.5	-1.25	`a = c`
2	1.5	-1.25	2	7	1.75	2.21875	`b = c`
3	1.5	-1.25	1.75	2.21875	1.625	0.33203	`b = c`
4	1.5	-1.25	1.625	0.33203	1.5625	-0.49561	`a = c`
5	1.5625	-0.49561	1.625	0.33203	1.59375	-0.09113	`a = c`
6	1.59375	-0.09113	1.625	0.33203	1.60938	0.1181	`b = c`
7	1.59375	-0.09113	1.60938	0.1181	1.60156	0.0129	`b = c`
8	1.59375	-0.09113	1.60156	0.0129	1.59766	-0.03926	`a = c`
9	1.59766	-0.03926	1.60156	0.0129	1.59961	-0.01322	`a = c`
10	1.59961	-0.01322	1.60156	0.0129	1.60059	-0.00017	`a = c`

This material is intended as a summary. Use your textbook for detail explanation.
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