I'm struggling with this question. I tried proving by contradiction but I don't know if I'm doing it right. Any help is highly appreciated!
Let $f:\mathbb{R} \to \mathbb{R}$ be smooth such as $f'(x_{1}) \gt 0$ and $f'(x_{2}) \gt 0$ for $x_{1} \lt x_{2}$ roots of $f$. Show that $f$ has at least one root in $(x_{1},x_{2})$.
My atempt:
By Rolle's Theorem, $\exists c \in (x_{1},x_{2}):f'(c)=0$, therefore, $f(c) \gt 0$.
Suppose by contradiction that $\nexists k \in (x_{1},x_{2}):f(k)=0.$ Therefore, $\forall x \in (c,x_{2}), f'(x) \lt 0$.
Since $f$ is smooth, $f'$ is continuous. So, $\exists \delta \gt 0: f'(y) \gt 0, \forall y \in (x_{2}-\delta,x_{2}+\delta)$. Let $y=x_{2}-\frac{\delta}{2}$. Since $y \in (c,x_{2}) \implies f'(y) \lt 0$. Contradiction.
Your proof is not correct, since there is no way you can deduce that $f(c)>0$.
Since $f(x_1)=0$ and $f'(x_1)>0$, there is some number $y_1$ in $(x_1,x_2)$ such that $f(y_1)>0$. And since $f(x_2)=0$ and $f'(x_2)>0$, there is some number $y_2$ in $(x_1,x_2)$ such that $f(y_2)<0$. But then, by the intermediate value theorem, there is some $y$ berween $y_1$ and $y_2$ such that $f(y)=0$. And, sonce both $y_1$ and $y_2$ belong to $(x_1,x_2)$, so does $y$.