Possible explanation why roots of $f(x)$ and $g(x)$ are very similar Made a small observation, and I'm looking for a possible explanation.

Suppose we have $f(x)$ and we define $g(x)$ as: $$g(x):=x-\frac{f(x)}{f'(x)}$$
Why are the roots of $g'(x)$ and $f(x)$ almost identical? 
Almost because although some are numerically equivalent, all the roots are not the same. 
Consider $f(x):=x^2+3 x-1$, then $g'(x)$ is $\frac{2 \left(x^2+3 x-1\right)}{(2 x+3)^2}$, both functions' roots are: $$\left\{x\to \frac{1}{2} \left(-\sqrt{13}-3\right)\right\},\left\{x\to \frac{1}{2} \left(\sqrt{13}-3\right)\right\}$$ Consider $f(x):=x^5-2 x^4+2 x^3-2 x^2-x+\frac{1}{2}$, then the real roots are: $$\{x\to -0.52325\},\{x\to 0.33225\},\{x\to 1.66996\}$$ The roots of $g'(x)$ are: $$\{x\to 0.756324\},\{x\to -0.52325\},\{x\to 0.33225\},\{x\to 1.66996\}$$
 A: You have $g' = \frac{f f''}{f'^2}$, therefore  $g'(x_0) = 0$ if:


*

*$f(x_0) = 0 \ne f'(x_0)$, or

*$f''(x_0) = 0 \ne f'(x_0)$.


In your first example, $f''(x) = 2$ has no zeros, and all zeros of $f$
are simple, therefore $g'$ and $f$ have the same zeros.
In your second example the zero $x \approx 0.756324$ of $g'$ is a zero
of $f''$.
Remark: $g$ is not defined at points where $f'(x)=0$. If $f$ is analytic
(i.e. equal to its Taylor series) then it can be shown that
$g'$ approaches either infinity or a finite non-zero limit at points where
$f'(x_0) = 0$.
A: What you have there is the Newton-Raphson method for approximating a solution to an equation, namely $f(x)=0$. As $a$ gets closer to the root, $g(a)-g(a-k),  k<<a$, too becomes smaller, that is to say, the approximation, while it becomes more accurate, does so at a slower rate as you get closer to the root. At $g(r)$, where $r$ is one root of $f(x)=0$, $g(r)-g(r-k)$ becomes virtually zero, and this is essentially saying $g'(x)=0$. You can see that happening in this graph, notice especially how the $g(x)$ curve flattens as it approaches the root to $f(x)=0$.
