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Is there a nonlinear function $f:\mathbb{R}\to\mathbb{R}$, differentiable on $\mathbb{R}$, such that any tangent line is tangent to the graph of $f$ at two distinct points (at least)?

goon65
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I will prove there is no $y=y(x)$ that solves $f'(x)=f'(y)$ while also solving $f(y)-f(x)=(y-x)f'(x) $ for all $x$, unless $y=x$ or $f$ is linear, with the assumptions that $y\in C^1$ and $f\in C^2$. Assume it is true and take the derivative of the second equation to get

$y'(x)f'(y)-f'(x)=y'(x)f'(x)-f'(x)+(y-x)f''(x)$.

Letting $f'(x)=f'(y)$ hold, this expression simplifies to read

$0=(y-x)f''(x)$.

Therefore, either $y=x$, or $f''(x)=0$, and $f$ is a linear function.

alphanzo
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