The root of the function $f (x) = x^{3} + x - 1$ obtaind after first iteration on application of Newton Raphson scheme using an initial guess of $x_{0} = 1$

0.682

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0.686

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0.750

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1.000

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Solution

The correct option is C $0.750$
Let $f (x) = x^{3} + x - 1$
$f^{'} (x) = 3 x^{2} + 1$
at $x = x_{0} = 1;$
$f (x_{0}) = 1 + 1 - 1 = 1$
$f^{'} (x_{0}) = 3 (1)^{2} + 1 = 4$
$∴ x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}$
$\Rightarrow x_{1} = 1 - \frac{1}{4} = \frac{3}{4} = 0.75$
$∴ x_{1} = 0.750$

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The root of the function f(x)=x3+x−1 obtaind after first iteration on application of Newton Raphson scheme using an initial guess of x0=1

The correct option is C 0.750Let f(x)=x3+x−1 f′(x)=3x2+1 at x=x0=1; f(x0)=1+1−1=1 f′(x0)=3(1)2+1=4 ∴x1=x0−f(x0)f′(x0) ⇒x1=1−14=34=0.75 ∴x1=0.750

The root of the function $f (x) = x^{3} + x - 1$ obtaind after first iteration on application of Newton Raphson scheme using an initial guess of $x_{0} = 1$