Problem:
Consider the continuous function $f$ which is $k$ times differentiable: $f(\alpha )=f'(\alpha )=\cdots=f^{(k-1)}(\alpha )=0$ and $f^{(k)}(\alpha )\neq 0$. Assume that $\alpha$ is a root to $f$ with multiplicity $k$, i.e: $f(x)=(x-\alpha )^k g(x)$ where $g(\alpha)\neq0$.
I need to prove that $g'(\alpha)\neq0$. Any ideas?
I tried to differentiate the expression $f(x)=(x-\alpha )^k g(x)$ one time, and then find the expression of $g'(x)$ in terms of $f(x)$ and $f'(x)$, and take the limit as $x$ tends to $\alpha$, but it doesn't work. Any help is appreciated.