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Prove this : When the graphs of two differentiable functions have the minimum distance then the secants at those points are parallel .

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Do you mean that the tangent lines are parallel? – Brian M. Scott Sep 19 '12 at 7:32
Take $f(x)=0$, $g(x)=x$, then the minimum distance between the two graphs is 0, at the point $x=0$, but at this point, neither the secant (I'm guessing orthogonal) lines nor the tangent lines are parallel. – Najib Idrissi Sep 19 '12 at 7:47
Assuming the two curves don't intersect, here's a hint: If the lines are parallel, then what does it tell you about their slopes? – celtschk Sep 19 '12 at 8:10

Your conjecture is false, as @nik shows in a comment. However, consider the function $$ \delta \colon x \mapsto (f(x)-g(x))^2, $$ which represents the square of the vertical distance between the graphs of $f$ and $g$. Now, $$ \delta'(x)=2 (f(x)-g(x))(f'(x)-g'(x)). $$ If $x_0$ is a minimum of $\delta$, then either $f(x_0)=g(x_0)$, or $f'(x_0)=g'(x_0)$. This suggests that you should modify your conjecture as follows:


Assume $f$ and $g$ are differentiable on $(a,b)$. If the graphs of $f$ and $g$ do not cross and if the infimum of their distance is attained at some $x_0$, then the graphs of $f$ and $g$ are parallel at $x_0$ (in the sense that the tangent lines to the graphs at $x_0$ are parallel).

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Thank you all for for your immediate response and excuse me for my bad English I wanted to ask a question like the conjecture above . Note that the ordinates of two points are not equal Thanks again – Geokal Sep 19 '12 at 8:29
Be careful. Consider $f(x)=-x$ and $g(x)=(x-1)^2$ on (say) $[0,2]$. If you define the distance as $\inf\{|f(x)-g(y)| \mid x,y \in [0,2]\}$, then the distance between $f$ and $g$ is zero: $f(0)=0=g(1)$. You should decide what is the distance between the graphs of two functions. – Siminore Sep 19 '12 at 8:58
@Geokal: Maybe you should add a more complete explanation what distance you mean. Since the first (and — at least to me, and obviously also to Siminore — most obvious) interpretation doesn't apply, here's my next-best guess: Do you mean the minimal distance of the curves in the plane, that is $\min_{x_1,x_2}\sqrt{(x_1-x_2)^2+(f(x_1)-f(x_2))^2}$? In that case, as long as the curves have well defined tangents in the points realizing the minimum, the answer is even simpler: The connecting line between the points must be orthogonal on both tangents, or else you could make the distance smaller … – celtschk Sep 19 '12 at 12:25
by moving the point in the direction with the smaller angle to the connection line. Two tangents which are both orthogonal to the same line are of course parallel to each other. Of course this again only applies if the curves don't intersect. – celtschk Sep 19 '12 at 12:26
@celtschk : Now it's all clear . I was confused applying Fermat's theorem . Thank you – Geokal Sep 19 '12 at 15:37

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