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Actually let $g(x)=0$ when $f(x)=0$ otherwise $g(x)=f(x)/f'(x)$. Seems clear to me that if $x_0$ is an $n$-order root of $f(x)$ where $n$ is a positive integer, and $f(x)$ can be expressed as a Taylor series about $x_0$, then $x_0$ is a first-order root of $g(x)$. Does this also hold under les strict conditions? Perhaps when the first $n+1$ derivatives are continuous at $x_0$.

f(x) need not be a polynomial. I defined g(x) to ensure it is zero when f(x)=f'(x)=0. Notice we could have f(x)=sin(x) in which case g(x)=tan(x).

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By root of $g(x)$ do you mean a zero of the rational function $\frac{f(x)}{f'(x)}$? $g(x)$ is not necessarily a polynomial.. –  VSJ Feb 10 '12 at 22:29
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Something like it can be derived by using the Lagrange form of the remainder in Taylor's Theorem. I have not checked what is the minimal continuity assumption. It looks as if $n+2$ times continuous differentiability should work easily, but one may very well be able to do a little better. –  André Nicolas Feb 10 '12 at 23:45

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