The Newton's method is preferably used to find the root for the form of the equation: ax2 + bx + c = 0.
This method uses the iterative method and finds the approximate root of any given function.
The Newton's method formula is given by
The Newton's method Calculator calculates the root of any given function for desired level of accuracy if the function and its initial point are entered in the blocks provided. It gives the approximation up to five decimal places.

## Newton's Method Problems

Steps for Solving Newton's Method:
1. Read the given problem and observe the given function and list out the given initial point and no of decimal places to be calculated.
2. Calculate the function f(x) for all values starting from initial point till you get value such that for one function should have the value with opposite sign with respect to another. It means that root lies between those two numbers.
3. Use the formula : xn+1 = xn - $\frac{f(x)}{f'(x)}$. Differentiate f(x) and f'(x) and find the values of xn+1 for desired number of approximations.

Below you could see some solved example problems

### Solved Example

Question: Find the approximation of the function x3 - 8x + 6 correct to four decimal places.
Solution:

The given function is f(x) = x3 - 8x + 6
f'(x) = 3x2 - 8
Take 0 as initial point f(0) = 03 - 8(0) + 6 = 6
f(1) = (1)3 - 8(1) + 6  = -1
The root lies between 0 and 1
f(1) = (1)3 - 8(1) + 6 = -1
f'(1) = 3(1)2 - 8 = -5

The first approximation is given by
x1 = x0 - $\frac{f(x)}{f'(x)}$
= 1 - $\frac{-1}{-5}$
= 1 - 0.2 = 0.8

The second approximation is given by
x2 = x1 - $\frac{f(x)}{f'(x)}$
= 0.8 - $\frac{0.8^{3} - 8 (0.8) + 6}{3(0.8)^{2} - 8}$
= 0.818

The third approximation is
x3 = x2 - $\frac{f(x)}{f'(x)}$
= 0.818 - $\frac{(0.818)^{3} - 8(0.818) + 6}{3(0.818)^{2} - 8}$
= 0.8185

The fourth approximation is
x4 = x3 - $\frac{f(x)}{f'(x)}$
= 0.8185 - $\frac{(0.8185)^{3} - 8(0.8185) + 6}{3(0.8185)^{2} - 8}$
= 0.818558.

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