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Find a root of the equation $x^{3}+x-1=0$ correct to three decimal places, by the method of False Position.
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written 3.6 years ago by
prajapatijaimin
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Solution:
$ \begin{aligned} &\longrightarrow \text { Here } f(x)=x^{3}+x-1=0 \\\\ &\text { At } x=0, f(0)=-1(\text {-ve) } \\\\ &x=1, f(1)=1(\text { tve }) \\ \end{aligned} $
$\therefore$, The root lies between 0 and 1.
$ \text { Take } \begin{array}{rl} x_{0}=0 & x_{1}=1 \\\\ f\left(x_{1}\right)=-1 & f\left(x_{1}\right)=1 \\ \end{array} …
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