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I need to prove the following question:

Given strictly monotonic increasing function $f:\mathbb{R}\to\mathbb{R}$ and strictly monotonic decreasing function $g:\mathbb{R}\to\mathbb{R}$ if there is point $x_0$ such that $f(x_0)<g(x_0)$ then there is point $x_1$ such as $f(x_1)>g(x_1)$.

So what i know is that for $x_0<x_1$, $f(x_0)<f(x_1)$ and $g(x_0)>g(x_1)$. Let's say that I'm multiplying the following inequality $f(x_0) < g(x_0)$ by $(-1)$ so the result is $-f(x_0) > -g(x_0)$ so what i need to is to prove that exists such point as $x_1$ for which $f(x_1) = -f(x_0)$ and $g(x_1) = -g(x_0)$. How I can do that? Maybe there is another way to prove it?

Thanks for help.

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    $\begingroup$ Don't try to prove this... $\endgroup$ – Julien Mar 9 '13 at 20:00
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    $\begingroup$ If he could prove, there's no need to be here.... $\endgroup$ – HipsterMathematician Mar 9 '13 at 20:01
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You need a stronger hypothesis: consider the functions $f(x)=-e^{-x}$ and $g(x)=e^{-x}$.

If you assume that either $\lim_{x\to\infty}f(x)=\infty$ or $\lim_{x\to\infty}g(x)=-\infty$, you could look at the function $h(x)=g(x)-f(x)$.

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  • $\begingroup$ Wow, you are right and I was exactly thinking of these two functions ! $\endgroup$ – Damien L Mar 9 '13 at 20:04
  • $\begingroup$ Thank you! I need to learn more about different types of functions... this the second time today that because I didn't considered this kind of functions i wasn't able to solve a problem. $\endgroup$ – EMDB1 Mar 9 '13 at 20:06
  • $\begingroup$ @EMDB1: You’re welcome! I’ve added a little to my answer: an extra hypothesis that makes the conclusion true. $\endgroup$ – Brian M. Scott Mar 9 '13 at 20:09

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