Prove this : When the graphs of two differentiable functions have the minimum distance then the secants at those points are parallel .
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
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).