I wante to find some $v>0$ verifying this property Let us consider a function $f: (0,+∞)→(0,+∞)$. Assuming that $f$ is continuous differentiable and strictly increasing for all $t>0$. 
I want to find some $v>0$ verifying this property:
$$f(a)<f(v)≤c$$
where $a>0$  and $c>0$ are given real numbrers. I have no idea how to start.

Let $a, c > 0$ such that $f(a) < c$ (otherwise it obviously doesn't hold). Then, because $f$ is continuous at $a$, there exists $\delta > 0$ such that $|v - a| < \delta \implies |f(v) - f(a)| \le c - f(a)$.
For any $v \in \langle a, a + \delta\rangle$ we have $f(a) < f(v)$ so:
$$0 < f(v) - f(a) = |f(v) - f(a)| \le c - f(a)$$
Which implies $f(a) < f(v) \le c$.
