# Intermediate Value Property Justification [closed]

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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Claude Leibovici, JonMark Perry, user91500, M. Vinay
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. – user348202 Jun 23 '16 at 21:42
• Are y'all allowed to use intermediate value theorem? – AJY Jun 23 '16 at 21:55
• 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. – user348202 Jun 23 '16 at 22:18

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.
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 $t_n = \frac{1}{ 2 \pi n + \arcsin (k) } \leq \frac{1}{ 2 \pi (n - 1) } \leq x_2$.