using bisection method,find the zero of the function f(x)=e^x - 2x^2 in the interval [2.5, 3] with the tolerance `epsi` = 10^-2. Record the values a`kappa` , b`kappa` , and x`kappa` as well as an estimate for the error at every step. Answer: zero `~~`  _________

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`f(x) = e^x -2*x^2`

We observe that

`f(2.5) = e^2.5 -2*2.5^2 =-0.317<0`

`f(3) =e^3 -2*3^2 =2.085 >0`

Iteration number:

1.

`a_1 =2.5` , `b_1 =3` , `x_1 =2.5+(3-2.5)/2 =2.75`

`f(x_1) =e^2.75-2*2.75^2 =0.517 >0`

error `|x-x_1| <=(b_1-a_1)/2^1 =(3-2.5)/2 =0.25 >0.01`

2.

`a_2 =2.5` , `b_2 =2.75` , `x_2 =2.5 +(2.75-2.5)/2 =2.625`

`f(x_2) =e^2.625 -2*2.625^2 =0.023 >0`

`|x-x_2| <= (b_1-a_1)/2^2 =(3-2.5)/4 =0.125 >0.01`

3.

`a_3 =2.5` ,`b_3 =2.625` , `x_3 =2.5 +(2.625-2.5)/2 =2.5625`

`f(x_3) =e^2.5625 -2*2.5625^2 =-0.164 <0`

`|x-x_3| <= (b_1-a_1)/2^3 =(3-2.5)/8 =0.0625 >0.01`

4.

`a_4 =2.5625` , `b_4 =2.625` ,

`x_4 = 2.5625 +(2.625-2.5625)/2 =2.59375`

`f(x_4) =-0.075 <0`

`|x-x_4| <=(b_1-a_1)/2^4 =(3-2.5)/16 =0.03125 >0.01`

5.

`a_5 =2.59375` , `b_5 =2.625` , `x_5 = 2.609375`

`f(x_5) =-0.027 <0`

`|x -x_5|<= (b_1-a_1)/2^5 =(3-2.5)/32 =0.015625 >0.01`

6.

`a_6 =2.609375` , `b_6 = 2.625` , `x_6 =2.6171875`

`f(x_6) =-0.0022 <0`

`|x-x_6|<=(b_1-a_1)/2^6 =(3-2.5)/64 =0.078125 <0.01`

Answer: zero `~~` 2.6171875

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