0
$\begingroup$

If any two polynomials touch exactly at a point (but do not cross), will they always have the same gradient at that touching point? If not, is it almost always and there are a few subtle conditions? I'm assuming it's true but I can't think how to prove it and couldn't seem to find any answer online.

$\endgroup$
  • 1
    $\begingroup$ Polynomials don't touch. I think what you mean is the graphs of the polynomials touch. So, consider the tangent to each graph at the point where the graphs touch. What can you say about those tangent lines? $\endgroup$ – Gerry Myerson Mar 14 '16 at 11:38
  • $\begingroup$ I do. The tangent lines should be equal so I assume so they do have the same gradient at that point. Right? $\endgroup$ – Tom Mar 14 '16 at 11:39
  • $\begingroup$ I think you need a precise definition of "touch but not cross" here, but it seems (to me) correct for any two differentiable functions (not just polynomials). $\endgroup$ – coffeemath Mar 14 '16 at 11:39
  • $\begingroup$ Ok, I'm not sure how to be more precise than "touch but not cross". I think touching but not crossing would imply that one function is always greater than the other in a region around a point where they are equal. $\endgroup$ – Tom Mar 14 '16 at 11:42
2
$\begingroup$

Let $f(x)$ and $g(x)$ be the functions in question. If the functions "touch but do not cross" at a point $x_0$, then it must be the case that $f(x) \geq g(x)$ in some neighborhood of $x_0$, with equality only holding at $x = x_0$. (Note that this can be done without loss of generality—define $f(x)$ to be the function that is never less than the other function in this neighborhood.) In particular, this means that $f(x) - g(x) \geq 0$ in this neighborhood, with equality (again) only at $x = x_0$. Thus, $x = x_0$ is a local minimum of the function $f(x) - g(x)$.

Now, if $f(x)$ and $g(x)$ are both differentiable functions in a neighborhood of $x_0$, then $f(x) - g(x)$ is also differentiable. Since $f(x) - g(x)$ has a local minimum at $x_0$, this implies that $f'(x_0) - g'(x_0) = 0$, which gives us that $f'(x_0) = g'(x_0)$ as desired.

$\endgroup$
  • $\begingroup$ Ah neat I like this, thank you. $\endgroup$ – Tom Mar 14 '16 at 11:50
  • $\begingroup$ Or $f(x)\le g(x)$. $\endgroup$ – Akiva Weinberger Mar 14 '16 at 11:59
  • $\begingroup$ I think you have to remove "the derivative of". $\endgroup$ – TonyK Mar 14 '16 at 12:24
  • $\begingroup$ @AkivaWeinberger: $f(x)$ can be chosen to be the greater of the two functions without loss of generality; I've edited my answer to reflect this. $\endgroup$ – Michael Seifert Mar 14 '16 at 12:34
  • $\begingroup$ @TonyK: Thanks for the correction. $\endgroup$ – Michael Seifert Mar 14 '16 at 12:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.