I am having these two results : Let $f \in C^1([0,1], \mathbb{R})$ :
(1) If $f(0) = 0$ and $f'(0) >0$ then there is a $\epsilon > 0$ such that $f(x) \geq 0$ on $]0, \epsilon[$
(2) If $f(0) = f(1) = 0$ and $f$ is positive there is $\epsilon > 0$ such that $f'$ is positive on $[0, \epsilon[$ and negative on $]1-\epsilon, 1]$
Note that these are assumptions so I am not sure both these results are true.
I think I might have a solution for $(1)$ using an aysmptics. Since $f$ is $C^1$ we have : $f(h) = f(0) + hf'(0) + o(h)$ in a neighboorhood of $0$. Thus $f(h) = hf'(0) + \epsilon(h)$ where $\epsilon$ is a continuous function which goes to $0$ when $h \to 0$.Yet since $hf'(0) = O(h)$ (since $f'(0) > 0)$ then $hf'(0) \geq \epsilon(h)$ for $h$ small enough. Hence there is an $\epsilon$ which veryfy the conditions of $(1)$ Is it correct ? If it's correct I am really interested in an other way of proving this result.
Moreover I don't at all how to proceed for $(2)$.
Thank you.
I would address (1) by contradiction. Suppose: for every $\epsilon > 0$ small enough, there exists an $h$ such that $0 < h < \epsilon$ and $f(h) < 0$. It follows that there exists a monotonically decreasing sequence $h_{n}$ of strictly positive terms that converges to $0$.
For every $h_{n}$, we have: $$ {f(0 + h_{n}) - f(0) \over h_{n}} = {f(h_{n}) \over h_{n}} < 0. $$ By taking the limit of this as $n \rightarrow \infty$, we obtain $f'(0)$, which therefore must be nonpositive: $$ f'(0) \leq 0. $$ This contradicts the assumption that $f'(0) > 0$.