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I am working on showing the Intermediate Value Property holds for a certain function. I noticed the one page on here talks about this idea, but I can not follow with what they are trying to say. [The problem is posted below - Ed.] . Any guidance would be appreciated. I just am very unsure what the property says.

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closed as off-topic by user223391, Claude Leibovici, JonMark Perry, user91500, M. Vinay Jul 1 '16 at 13:57

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  • $\begingroup$ I answered part a. I know it is not a continuous function. I am lost past that. I was wondering if I select a interval for the property or if there is a specific one I need to select. $\endgroup$ – user348202 Jun 23 '16 at 21:42
  • $\begingroup$ Are y'all allowed to use intermediate value theorem? $\endgroup$ – AJY Jun 23 '16 at 21:55
  • $\begingroup$ Is it enough to say an interval could be [-10,0]. Then I select x1 to be -9 and x2 to be -1. f(x1)=-.412 and f(x2)=-.841. I look and select k to be between those and find a c value which f(c)=k. $\endgroup$ – user348202 Jun 23 '16 at 22:18
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You should prove it for any interval $[a,b]$.

For any interval not containing $0$, $f(x)$ is continous on that interval and thus satisfies IVP. So you only have to consider the case where $a\leq 0 \leq b$.

Hint: $f(x)$ oscillates $\textbf{a lot}$ as you approach zero from either side.

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So if you believe the intermediate value theorem, then you know that if $x_1, x_2 > 0$ or $x_1, x_2 < 0$, the property follows. So suppose $x_1 \leq 0 < x_2$. It's enough to show that there exists $c \in (0, x_1)$. To do this, consider the sequence $$(t_n)_{n \in \mathbb{n}} = \left( \frac{1}{ 2 \pi n + \arcsin (k) } \right)_{n \in \mathbb{N}} ,$$ along which $f(t_n) = k$. Let $n = \lceil 2 \pi n / x_2 \rceil + 1$. Then $t_n = \frac{1}{ 2 \pi n + \arcsin (k) } \leq \frac{1}{ 2 \pi (n - 1) } \leq x_2$.

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