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Find a root of $x^{3}-x-1=0$ using Bisection methed correct to three decimal places.
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written 3.6 years ago by
prajapatijaimin
• 3.3k
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Solution:
$\rightarrow$ Here, $f(x)=x^{3}-x-1=0$
at $x=1$, f(i) $=x-x-1=-1 \quad(-v e)$
$ x=2, \quad f(2)=8-2-1=5(+v e) \\ $
$\therefore$ The root lies between 1 and 2
$$ \begin{array}{|c|c|c|c|c|} \hline \text { Sr.No. } & \text { Negative } & \text { positive } & x_{i}=\frac{4+b}{2} & f(x i) \\ & \text …
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