Proving N-derivative test Suppose $f:(a,b) \rightarrow \mathbb{R}$ is differentialbe on $(a,b)$ and $c \in (a,b)$ has $f(c) = f'(c) = \dots = f^{n-1}(c) = 0$ and $f^{(n)}(c) > 0$. If $n$ is even, then $f$ has a local min at $c$.
My attempt: Consider the interval $a < c < \beta < B$, then by Taylor's Theorem, there exists $x_0 \in (c, \beta)$ such that $f(\beta) = \frac{1}{n!}f^{(n)}(x_0)(\beta - c)^n$.
This is the part where I am stuck, I know that normally $f(c) \not = 0$ where you can prove with $f(\beta) - f(c) > 0$ but since $f(c) = 0$ in this case, I have no idea where to proceed.

Assuming the n-th derivative of $f$ is continuous, $f^n(c) >0$ implies there exist $\delta > 0 $ such that $f^n(x) > 0 $ for all $ x \in(c-\delta , c+ \delta)$. Now for all $ \beta \in (c-\delta , c+ \delta) $, by writing taylor expansion around $ x = c$ we get :
$$ f(\beta) = \frac{1}{n!}f^{(n)}(x_0)(\beta - c)^n $$ for some $ x_0 \in (c-\delta , c+ \delta) $ this shows  $ f(\beta) \geq 0 = f(c) $ . 
