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Here is a quote from my textbook:

"If $f(x)$ has a root at $x=\alpha$ with multiplicity $m>1$, then $f'(x)$ has a root at $x=\alpha$ with multiplicity $m-1$. Then, the function \begin{align*} g(x)=f(x)/f'(x) \end{align*} has a root at $x=\alpha$ with multiplicity 1 (a simple root)."

But how is $\alpha$ a root for $g$, since $g(\alpha)=0/0$? I'm quite confused, any help appreciated.

mXdX
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    Take the quotient before evaluating; e.g., $f(x)=x^2$ has a double root at $x=0$, and $g(x)=x^2/2x=x/2$ has a root at $x=0$ – J. W. Tanner Feb 16 '20 at 15:42
  • How can I do that in general? I want to show that the function $g(x)$ has a simple root at $\alpha$. I don't know where to begin. – mXdX Feb 16 '20 at 15:49
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    where you could begin: $f(x)=(x-\alpha)^mq(x)$ – J. W. Tanner Feb 16 '20 at 15:51
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    $g$ as written there isn't defined at $\alpha=0$, but the singularity is removable, because $f(x)=(x-\alpha)^m r_1(x)$ where $r_1(\alpha) \neq 0$ and so $f'(x)=(x-\alpha)^{m-1} r_2(x)$ where $r_2(\alpha) \neq 0$. – Ian Feb 16 '20 at 15:52

1 Answers1

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Write $f(x)=(x-a)^m h(x)$, with $h(a)\ne0$. Then $$ g(x) = \frac{f(x)}{f'(x)} = \frac{(x-a)^m h(x)}{m(x-a)^{m-1}h(x)+(x-a)^m h'(x)} = \frac{(x-a) h(x)}{mh(x)+(x-a)h'(x)} = (x-a)j(x) $$ where $$ j(x)=\frac{h(x)}{mh(x)+(x-a)h'(x)} $$ Since $j(a)=\dfrac{h(a)}{m h(a)}=\dfrac{1}{m}\ne0$, we get that $a$ is a simple root of $g$.

lhf
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