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Question

While finding the roots of f(x)=x24=0 using Newton - Raphson method, initial value of (x, x1 = 1). If the value obtained after first iteration isx2.

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Solution

The Newton-Raphson method uses an iterative process to approach one root of a function. The specific root that the process locates depends on the initial, arbitrarily chosen x-value.
xn+1=xnf(xn)f(xn)
Here, xn is the current known x-value, (f(xn)) represents the value of the function at xn, and f(xn) is the derivative (slope) at xn.
xn+1 represents the next x-value that you are trying to find.

Here, we have f(x)=x22 and x1=1
f(x)=2x f(x1)=f(1)=12=1
f(x1)=f(1)=2
x2=x1f(x1)f(x1)
=1(12)=1+0.5=1.5
Thus, the approximate root is 1.5.


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