dy/ dx = f’(x) = 3x – 3
The NewtonRaphson formula becomes:
x n+1 = x n  x n³  3x + 1
3x n²  3
I am now going to use this formula to search for the root in the interval [2 , 1]
Let x 1 = 2
So x 2 = 2 – (2³  3(2) +1)
3(2)²  3
= 2 – ( 1/ 9)
= 1.888888889 (1 8/ 9)
Method 2: NewtonRaphson method
Fig1. Shows the function f(x) = 1/3x 3 – 5x + 1.4
Differentiating the function we get: f(x)` = x 2 – 5
So now, using the NewtonRaphson iteration equation we can find the roots
x n+1 = x n – f(x n)/f`(x n)
 Root a:
Therefore,
x a = 3.72456040
to 5 d.p. x a = 3.72455 0.00005
 Root b:
Therefore,
x b = 0.28148691 (8 d.p.)
 Root c:
Therefore,
x c = 4.00604730 (8 d.p.)
NEWTONRAPHSON METHOD DOESN’T ALWAYS WORK!
This method will not work if:
 If the initial value is not close to the root, or is near a turning point, the iteration may diverge or converge to another root!
 This method can break down when the equation is discontinuous.
I will demonstrate this using the equation g(x) = x 3 – 5x 2 + 8x 4.1
In this case, if we use the value x=2 as the starting point we get a “DIVERGENT” pattern. So the iteration cannot go to the upper or lower root.
At x=2 the trial will fail because the derivative there is 0; and in the equation x n+1 = x n – g(x n)/g`(x n), g`(x n) can’t equal 0 as it will lead to an undefined quantity for x n+1.
The Change of Sign method locates the root of an equation by where it crosses the xaxis. The point at which the curve crosses xaxis is the root. When a function changes sign in a certain interval, we can see that the root, or the place where the curve crosses the axis is within that interval. Using a Decimal Search method, it is possible to find intervals to varying degrees of accuracy to help pinpoint the position of a root.
I am going to use the equation:
y = 5x3+4x27x4
The graph of the equation is shown below:
By doing an initial search using the equation, I have found three main intervals where there is a change of sign.
The change of signs are in the intervals [2, 1], [1,0] and [1,2].
To demonstrate the method, I am going to find the root in the interval [1,2].
The above table shows that there is a change of sign between 1.1 and 1.2. Therefore, we now know that the root lies in the interval [1.1,1.2]
We can extend the search further to find the interval of the change of sign with greater accuracy.
The table shows a change of sign between 1.11 and 1.12. The root is in the interval [1.11,1.12]
This table now shows that the root is in the interval [1.11,1.111].
Error Bounds
X=
Failure of the Change of Sign Method
Although the Change of Sign method has been proved to work above, there are many examples of cases or equations which would be wrongly represented by this technique. Examples of these are shown below:

Repeated Roots  in an initial search using the Change of Sign method, the table of values for an equation such as y = (x1.26)2 (x+1.4) would overlook the second root. As you can see from the graph below, there are two roots; one in the interval [2,1] and one in the interval [1,0]. However, the table only shows one change of sign in the interval [2,1].
2. The Change of Sign Method would also fail with an equation where all of the roots fall within the same interval, such as in the equation y = x31.7x2+0.84x0.108. The table below shows only one change of sign in the interval [0,1]. This would indicate that there is only one root, rather than three, in this interval.