Newton's Method Calculator

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**Step 1 :**

**Step 2 :**

**Step 3 :**

**Step 4 :**

Newton's method is the method used to get the root of an equation of the form:

**Newtons Method Calculator** calculates the root for any given function for the desired number of decimal places.

Default function with initial point is given in the calculator below. Roots are calculated by differentiating the function and substituting the initial value in the formula given in the steps. You will see the values of the function when you click on "Solve".

ax^{2} + bx + c = 0

It gives the root value to any desired level of accuracy.Default function with initial point is given in the calculator below. Roots are calculated by differentiating the function and substituting the initial value in the formula given in the steps. You will see the values of the function when you click on "Solve".

Read the given problem. Note down the given equation, initial point x_{o} and no of decimal places n.

Check where exactly the root lies by find the values of f(x). Start from initial point, give different values of f(x), if the values of given function f(x) are such that one is positive and other is negative. It means that the root lies between those two numbers.

Differentiate f(x) to get f'(x).

Find the value of f(x) and f'(x) for x_{0}.

Substitute x_{0} in the formula:

x_{n+1} = x_{n} - $\frac{f(x)}{f'(x)}$

We get the value of x_{1}.

Again check where the root lies by substituting the value of x_{1} in x_{2} and also substitute the value of x_{1} get the value of x_{2}. Carry on the same process till the last decimal place. List out all the values for the given equation as x_{0}, x_{1}, x_{2}......x_{n}, which gives you the answer.

Find the roots between 0 and 1 of the equation x

^{3}− 5x + 2 = 0 correct to three decimal places.**Step 1 :**The given function is: x

^{3}- 5x + 2 = 0,

Initial point x

_{o}= 0,

no of decimal places n = 3.

**Step 2 :**f(0) = 0

^{3}- 5(0) + 2 = 2

f(1) = 1

^{3}- 5(1) + 2 = -2

The Root lies between 0 and 1.

**Step 3 :**For x

_{o}= 1, f(1) = 1^{3}- 5 (1) + 2 = -2

f'(1) = 3(1)

^{2}- 5 = -2

x

_{1}= x_{o}- $\frac{f(x)}{f'(x)}$

= 1 - $\frac{-2}{-2}$

= 0.

**Step 4 :**for x1 = 0, f(0) = 0

^{3}- 5(0) + 2 = 2

f'(0) = 0-5 = - 5

x

_{2}= x_{1}- $\frac{f(x)}{f'(x)}$

= 0 - $\frac{2}{-5}$

= 0 - $\frac{2}{-5}$

= 0.4.

f(0.4) = (0.4)

^{3}- 5(0.4) + 2

= 0.064

f'(0.4) = 3(0.4)

^{2}- 5

= - 4.52

x

_{3}= x_{2}- $\frac{0.064}{- 4.52}$

= 0.4 - 0.014159.

= 0.414159.

**Answer :**The root is 0.414159 correct to three decimal numbers.

The root of x

^{3}−6x+4 = 0 arex

_{o}= 0x

_{1}= 0.4x

_{2 }= 0.414159.