Of course, we can easily solve this one by inspection.

But I've included it because it illustrates a case when Newton's Method is not very efficient.

Newton's Method works well when the slope fo the line is reasonably steep, but in cases where it's quite flat near the root (similar to Example 2 above), it does not converge very quickly.

In fact, even after 12 steps of the method, we are still not very close to the root.

Let's start at `x=-4`.

The derivative:

`dy/dx = 2x`

So

`x_1=x_0-(f(x_0))/(f’(x_0))`

` = -4-(f(-4))/(f’(-4))`

` = -4-16/(-8)`

` = -2`

Step 2:

`x_2=x_1-(f(x_1))/(f’(x_1))`

` = -2-(f(-2))/(f’(-2))`

` = -2-4/(-4)`

` = -1`

Continuing on we get:

x3 = -0.5

x4 = -0.25

x5 = -0.125

x6 = -0.0625

x7 = -0.03125

x8 = -0.015625

x9 = -0.0078125

x10 = -0.00390625

x11 = -0.001953125

x12 = -0.0009765625

Generally, however, Newton's Method is a simple and neat way to find roots of equations. It is commonly used (in conjunction with other efficiency algorithms) by computers and calculators.

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