Given equation f(x) = x3 – 3x – 5
Differentiate with respect to x, we get f ‘(x) = 3x2 – 3
Now, first find the range, where the real roots lie in i.e. f(2) = –3 and f(3) = 13
Since f(2) is a negative value and f(3) is a positive value. Therefore, our one real root of the equation lies between x = 2 and 3.
Now, by Newton-Raphson formula
Now, substitute the value of f(x) and f'(x) in this formula
Choose the initial approximate value of the root x0 = 2 and putting n = 0, we have the first approximate value
Now, for the second approximation, replacing n = 1 in equation (1) we get
Now, for the third approximation, replacing n = 2 in equation (1) we get
Now, for the Fourth approximation, replacing n = 3 in equation (1) we get
Here x3 and x4 obtained the same value; therefore one root of the given equation = x3 – 3x – 5 using the Newton Raphson Method is 2.2790.
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