Is there a differentiable function whose derivative is greater than or equal to 1 everywhere but 0 at 0? Does there exist a differentiable function $f: \mathbb{R} \longrightarrow \mathbb{R}$ such that $f^{'}(0)=0$ and $f^{'}(x) \ge 1$ $\forall x \ne 0?$ Prove either way.
I was thinking of coming up with a counterexample such as $x^3$ but adding an expression to make $f' \ge 1$ but that makes $f'(0) \ne 0.$ What would be a rigorous proof otherwise if no such $f$ exists?

Darboux Theroem  says that the derivative of any differentiable function has IVP. So  no such function exists. Ref: https://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis)
