Answer to Question #177898 in Linear Algebra for william

Question

Find the root of the equation x^3 - x - 11 = 0 using bisection method.

up to six iterations.

Answer 1

$x^3 - x - 11 = 0$

$\text{let } f(x) = x^3 - x - 11$

$x= 0 ;f(x)= 0-0-11<0$

$x=1;f(x)=1-1-11<0$

$x=2;f(x)= 8-2-11<0$

$x=3;f(x)=27-3-11>0$

$x\isin[2;3]\exists x$ :f(x) =0

$\text{1 th iteration}$

$[2;3] ;c=\frac{3+2}{2}=2.5$

$f(2.5)=2.5^3-2.5-11=2.125>0$

$\text{2 th iteration}$

$[2;2.5];c=\frac{2+2.5}{2}=2.25$

$f(2.25)=2.25^3-2.25-11=-1.859375<0$

$\text{3 th iteration}$

$[2.25;2.5];c=\frac{2.25+2.5}{2}=2.375$

$f(2.375)=2.375^3-2.375-11=0.021484375>0$

$\text{4 th iteration}$

$[2.25;2.375];c=\frac{2.25+2.375}{2}=2.3125$

$f(2.3125)=2.3125^3-2.3125-11=-0.946044921875<0$

$\text{5 th iteration}$

$[2.3125;2.375];c=\frac{2.3125+2.375}{2}=2.34375$

$f(2.34375)=2.34375^3-2.34375-11=\newline =-0.469146728515625<0$

$\text{6 th iteration}$

$[2.34375;2.375];c=\frac{2.34375+2.375}{2}=2.359375$

Answer: 2.359375 the root of the equation after 6 iterations

on the interval [2;3]