Solve the equation $x = e^{- x}$ using Newton-Raphson method. Starting with an initial guess values $x_{0} = 0$ , the value of x after first iteration is .

-0.050

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-1

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0.50

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Solution

The correct option is C 0.50
$f (x) = x - e^{- x}$
Newton Raphson method
First iteration, $x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}$
Here, $f^{'} (x) = 1 + e^{- x}$
$x_{1} = 0 - \frac{f (0)}{f^{'} (0)}$
= $\frac{- (0 - e^{- 0})}{1 + e^{- 0}}$
= $\frac{1}{2} = 0.50$

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Solve the equation x=e−x using Newton-Raphson method. Starting with an initial guess values x0=0, the value of x after first iteration is .

The correct option is C 0.50f(x)=x−e−x Newton Raphson method First iteration, x1=x0−f(x0)f′(x0) Here, f′(x)=1+e−x x1=0−f(0)f′(0) =−(0−e−0)1+e−0 =12=0.50

Solve the equation $x = e^{- x}$ using Newton-Raphson method. Starting with an initial guess values $x_{0} = 0$ , the value of x after first iteration is .

The correct option is C 0.50
$f (x) = x - e^{- x}$
Newton Raphson method
First iteration, $x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}$
Here, $f^{'} (x) = 1 + e^{- x}$
$x_{1} = 0 - \frac{f (0)}{f^{'} (0)}$
= $\frac{- (0 - e^{- 0})}{1 + e^{- 0}}$
= $\frac{1}{2} = 0.50$