From previous math courses, finding the roots (or zeros) of simple functions, such as linear functions and quadratic functions, are quite simple by using basic formulas and procedures. However, for higher degree polynomials, such as cubic, quartic (4th degree), and quintic (5th degree), such formulas do not exist, or the calculations to find the roots are very laborious.
Newton’s Method is a recursive process for approximating the root of a differentiable function where other methods of finding roots may not exist or are too strenuous to do by using standard math procedures.
Formula: Newton’s Method
Suppose F(x) will be a differentiable function. Select a point x0 close to the root of the function. The Newton Method formula is given as: Where n = 0, 1, 2, 3, 4…… F(x) will be the given function and F’ (x) will be the First Derivative of the function. 
The video below displays Newton’s Method in action:
The most important thing that should be remembered for this method is that this method will be invalid if the first derivative of any function is zero.
Guidelines for Applying Newton’s Method
Important Notes:

Let’s solve some examples to apply this formula.
Example 1:
Find the root of the function F (x) = x^{3} by using Newton’s Method. Complete 4 iterations.
Solution:
Select an x value (x_{1}) that is close to the root you want to approximate.
Let’s start with the value x_{1} = 1/2.
Input x_{1} into the original function F(x) to obtain F(x_{1}).
Our x_{1} is 1, so we will find F(3):
F(1) = (1)^{3} = 1
Therefore, F(x_{1}) = 1.
Find the derivative of F(x). Input x_{1} into F’(x) to obtain F’(x_{1}).
Find the derivative of F (x) = x^{3}
F’(x) = (3)x^{31}=3x^{2}(power rule)
Plugging in x_{1}= 3 to F’(x),
F’(3) = 3(1)^{2}= 3*1 = 3
Therefore, F’(x_{1}) = 3
Substitute the values of x_{1}, F(x_{1}), and F’(x_{1}) into the formula to obtain the new x value x_{n+1} .
Using Newton’s Method:
The next x value to run through the process is x_{2}= 0.666.
We are going to repeat the process. First, find F(0.666)
F(0.666) = (0.666)^{3} = 0.295
Find F’(0.666).
F’(0.666) = 2(0.666)^{2}= 0.887
Applying the formula,
The next x value to run through the process is x_{3}= 0.333.
Once again, find F(0.333):
F(0.333) = (0.333)^{3}= 0.110
Find F’(0.333).
F’(0.333) = 2(0.333)^{2}= 0.221
Applying the formula,
The next x value to run through the process is x_{4}= 0.164
Once again, find F(0.164):
F(0.164) = (0.164)^{3}= 0.004
Find F’(0.164).
F’(0.164) = 2(0.164)^{2}= 0.053
Applying the formula,
Let’s fill out the data table of our calculations.
n  x_{n}  F(x_{n})  F’(x_{n})  
1  1= x_{1}  1  3  0.666 = x_{2} 
2  0.666  0.295  0.887  0.333 = x_{3} 
3  0.333  0.110  0.221  0.164 = x_{4} 
4  0.164  0.004  0.053  0.088 = x_{5} 
This completes the 4th and last iteration. x_{5}= 0.088 and is our approximation of the root of the function. Even though our approximations oscillate from positive to negative, if we continue to do more iterations, the x value gets closer to zero, which is the real root of the function F(x) = x^{3}. The graph of F(x) proves this:
Example 2:
Suppose we have a function F(x) = x^{2}4x . Use Newton’s Method to find the root value of the given function.
Solution:
Select an x value (x_{1}) that is close to the root you want to approximate.
Let’s first look at the graph of F(x):
Let’s say we wanted to approximate what the root value that is highlighted in yellow. Let’s pick a number close to that value, such as 5.
Input x_{1} into the original function F(x) to obtain F(x_{1}).
Our x_{1} is 5, so we will find F(5):
F(1) = (5)^{2}4(5) = 5
Therefore, F(x_{1}) = 5.
Find the derivative of F(x). Input x_{1}into F’(x) to obtain F’(x_{1}).
Find the derivative of F (x) = x^{2}4x
F’(x) = 2x4 (power rule)
Plugging in x_{1}= 5 to F’(x),
F’(5) = 2(5)4 = 6
Therefore, F’(x_{1}) = 6
Substitute the values of x_{1}, F(x_{1}), and F’(x_{1}) into the formula to obtain the new x value x_{n+1} .
Using Newton’s Method:
The next x value to run through the process is x_{2}= 4.16667.
We are going to repeat the process. First, find F(4.16667)
F(4.16667) = (4.1667)^{2}4(4.16667) = 0.6947
Find F’(4.16667).
F’(4.16667) = 2(4.16667)4= 4.3333
Applying the formula,
The next x value to run through the process is x_{3}= 4.0064.
Once again, find F(4.0064):
F(4.0064) = (4.0064)^{2}4(4.0064)= 0.0256
Find F’(4.0064).
F’(4.0064) = 2(4.0064)4= 4.0128
Applying the formula,
Let’s fill out the data table of our calculations.
n  xn  F(x_{n})  F’(x_{n})  
1  5 = x_{1}  5  6  4.16667= x_{2} 
2  4.16667  0.6947  4.3333  4.0064 = x_{3} 
3  4.0064  0.0256  4.0128  4.0000 = x_{4} 
From three iterations, x_{4}= 4.0000, which is our approximation of the root. If we continue to do iterations, it is clear that the root of the function near x = 5 is x = 4. Let’s graph the function F(x) and zoom in to prove the root is x = 4:
Example 3:
Suppose we are given the function, F (x) = x^{3}3x^{2}+1. Using the initial value x_{1} = 0.6, use the Newton Method to find a root of the given function.
Solution:
Select an x value (x_{1}) that is close to the root you want to approximate.
Let’s start with the value x_{1} = 0.6.
Input x_{1} into the original function F(x) to obtain F(x_{1}).
Our x_{1} is 1, so we will find F(0.6):
F(0.6) = (0.6)^{3}3(0.6)^{2}+1= 0.136
Therefore, F(x_{1}) = 0.136.
Find the derivative of F(x). Input x_{1} into F’(x) to obtain F’(x_{1}).
Find the derivative of F (x) = x^{3}3x^{2}+1
F’(x) = (3)x^{31}(2)3^{21}+0=3x^{2}6x(power rule)
Plugging in x_{1}= 0.6 to F’(x),
F’(0.6) = 3(0.6)^{2}6(0.6)= 2.52
Therefore, F’(x_{1}) = 2.52
Substitute the values of x_{1}, F(x_{1}), and F’(x_{1}) into the formula to obtain the new x value x_{n+1} .
Using Newton’s Method:
The next x value to run through the process is x_{2}= 0.6539.
We are going to repeat the process. First, find F(0.6539)
F(0.6539) = (0.6539)^{3}3(0.6539)^{2}+1= 0.0031
Find F’(0.6539).
F’(0.6539) = 3(0.6539)^{2}6(0.6539)= 2.6406
Applying the formula,
The next x value to run through the process is x_{2}= 0.6527.
We are going to repeat the process. First, find F(0.6527)
F(0.6527) = (0.6527)^{3}3(0.6527)^{2}+1= 0
Find F’(0.6527).
F’(0.6539) = 3(0.6527)^{2}6(0.6527)= 2.6381
Applying the formula,
Let’s fill out the data table with our calculations:
n  x_{n}  F(x_{n})  F’(x_{n})  
1  0.6 = x_{1}  0.136  2.52  0.6539= x_{2} 
2  0.6539  0.0031  2.6406  0.6527 = x_{3} 
3  0.6527  0  2.6381  0.6527 = x_{4} 
From three iterations, x_{4}= 0.6527, which is our approximation of the root. If we continue to do iterations, it is clear that the root of the function near x = 0.6 is x = 0.6527 or another value extremely close to this. Let’s graph the function F(x) and zoom in on the root in this region:
As we see, the root is x = 0.6527.
Summary of Section
Newton’s Method is an algorithm used to create increasingly accurate approximations of roots of a function. Essentially, the steps are as follows:
And so on and so forth. This repeated interaction gives us the formula for Newton’s Method: Repeat the iteration until the desired root is found to the desired accuracy. 
References:
http://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx
https://www.desmos.com/calculator/kgwfrkiyh8
http://scidiv.bellevuecollege.edu/dh/Calculus_all/CC_2_7_NewtonsMethod
http://fourier.eng.hmc.edu/e176/lectures/ch2/node5.html
https://upload.wikimedia.org/wikipedia/commons/e/e0/NewtonIteration_Ani.gif
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