everyone:
The other day I came up with a question which I do not know how to solve. Consider a differentiable function $f:\mathbb{R}\longrightarrow \mathbb{R}$ and form each $x_0\in \mathbb{R}$ let $\gamma_{x_0}$ be the tangent line of $f$ in $x_0$. By definition, $$x_0\in\ker{(f-\gamma_{x_0})},$$ therefore $|\ker{(f-\gamma_{x_0})}|\geq1$. Now, is there any $f$ satisfying the previous conditions such that for all $x_0\in\mathbb{R}$ is $|\ker{(f-\gamma_{x_0})}|=2$?. I conjecture that there is not, but I have not been able to prove it.
As an example, we can consider this two functions:
- If $f(x)=x^3$, it is easy to check that $|\ker{(f-\gamma_{x_0})}|=2$ for all $x_0\in\mathbb{R}\setminus\{0\}$, so it is almost the desired funciton.
- If $f(x)=e^{-|x|}$, the requirement is satisfied in all the points where the function is differentiable, but it is not differentiable in $\mathbb{R}$.
Any ideas?
