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Suppose that $f$ is a continuous function and that $f(-1)=f(1)=0$. Show that there is $c\in(-1,1)$ such that $$f(c)=\frac{c}{1-c^2}$$

I am not sure if this is a new question as I set it this morning, after solved a similar question. I wanted to prove it using the same idea (Intermediate Value Theorem) but it was not that nice...

Can anyone help me on this?

By the way, the 'similar question' I mentioned above is as follow:

Suppose that $f$ is a continuous function and that $f(0)=1$ and $f(1)=2$. Show that there is $c\in(0,1)$ such that $$f(c)=\frac{1}{c}$$ The hint given was to let $g(x)=xf(x)$ and use the Intermediate Value Theorem.

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I guess that $c\in (-1,1)$. –  Sigur Nov 15 '12 at 2:20
    
Have you tried using a similar sort of hint? Like using the intermediate value theorem on some function $g(x)$ which you define (somehow)? –  Jesse Madnick Nov 15 '12 at 2:26
    
@Sigur, thanks for pointing it. I have edited it. –  pipi Nov 15 '12 at 3:33

2 Answers 2

up vote 1 down vote accepted

Hint Let $g(x)=[f(x) (1-x^2)]-x$. Then $g(-1)=1$ and $g(1)=-1$.

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This is nice! It help me to generalize all other possible similar questions! –  pipi Nov 15 '12 at 6:09

If $f(0) = 0$ then we are done with $c=0$. Otherwise $f(0) >0$ or $f(0) <0$. Suppose the former. Define $g(x) = \frac{(1-x^2)f(x)}{x}$. Then $\lim_{x \to 0^+} g(x) = \infty$ so by IVT there is $c$ with $g(c) = 1$. Suppose the latter. Then $\lim_{x \to 0^-} g(x) = \infty$ so by IVT there is $c$ with $g(c) = 1$.

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Thanks for the answer. –  pipi Nov 15 '12 at 6:08

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