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solve equation x4-x-9=0 by newtonraphson
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set \(f(x)=x^4-x-9\) \[f'(x)=4x^3-1\] because \(f(1)=-9\) and \(f(2)=5\), there exists at least one root \(1<x_r<2\). choose an initial value \(1<x_1<2\) then use the newton's method \[\Large x_{n+1}=x_n+\frac{f(x_n)}{f'(x_n)}\]
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