This is my conjecture:

If a line never intersects a differentiable curve then there exists at least one tangent to the curve whose slope is that of the line

I have tried proving it by maybe Lagrange's Mean Value Theorem but to no avail...


Not true. Consider the curve $y=e^x$. It never intersects the $x$-axis, but there is no point where the tangent has slope zero.

(No, there's nothing special about the slope being zero: Take the curve $y=ax+e^x$ and the line $y=ax$.)


As you have alredy been informed, the statement is false. However, if we are dealing with a loop or, to be more precise, with a differentiable function $c$ whose domain is $S^1$, then it is true. Just take the point $\theta\in S^1$ such that the distance from $c(\theta)$ to the line is the smallest (or the greatest) possible. The compactness of $S^1$ (and the continuity of $c$) assures us that such a $\theta$ exists. Then the line tangent to the curve at $c(\theta)$ will be parallel to the line. Therefore, they will have the same slope.

  • $\begingroup$ +1. Suitably phrased this would be a very nice exercise in Lagrange multipliers. $\endgroup$ – Jose27 Jun 4 '18 at 17:21
  • $\begingroup$ @Jose27 Indeed it would. $\endgroup$ – José Carlos Santos Jun 4 '18 at 17:23

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