When the derivative approaches $-\infty$

I was recently working on a problem where I needed to show that a function could not be exclusively positive over an interval. Despite all my efforts, the solution seemed to elude me, until I realized that, possibly the derivative could help me if I could show it to be negative on some interval (you will understand once you have read my conjecture). Notwithstanding, it turned out that the derivative, evaluated at the point I considered, approaches $-\infty$, which is not quite what I expected. The conjecture reads:

Conjecture. You are given a continuous function $f:\mathbb R_{\geqslant 0}\to\mathbb R$ which is differentiable everywhere, such that $f(0)=0$ and $$\lim_{x\to 0^+}f'(x)=-\infty$$ Prove that there exists a real $\varepsilon>0$ such that $f$ is strictly decreasing on the interval $[0, \varepsilon)$, and, hence, there exists some interval $(0, \varepsilon)$ where $f$ only yields negative values.

Intuitively, it seems to me that, if $\lim_{x\to 0}f'(x)$ would have been some negative real value, this statement should be true (even though I don't know how to prove it), but $-\infty$ makes me think of a "vertical tangent", rather than a "decreasing tangent", so I am not sure, if the statement will hold.

Any ideas, hints, or proofs?

Thanks in advance.

Solution 1:

HINT: from $$\lim_{x \to 0^+} f'(x)= - \infty$$ you can deduce that there exists a right neighbourhood $(0, \varepsilon)$ such that for all $x \in (0, \varepsilon)$ you have $f'(x) <-1$.

This means that the function is strictly decreasing. Since $f(0)=0$, all the values in that interval will be negative.