Let $f:[1,12]\to \mathbb{R}$ be continious and $f(1)=f(6)$. Prove, there exists such $x\in [1,6]$ that $ f(x)=f(2x)$

assume $g:[1,6] \to \mathbb{R}$ , $ g(x):=f(x)-f(2x)$

Now consider where $g(x)=0$ So I look for such $a$ and $b$ that $g(a)\cdot g(b)\le0$

I need $(f(a)-f(2a))(f(b)-f(2b))\le 0$

I suppose there exists $a\in [1,6]$ such that $f(a)\ge f(2a)$ and $f(b)\le f(2b)$ So then I have sollution from Darboux's theorem.

But why exactly $a$ and $b$ have to exists?

  • $\begingroup$ So solution is wrong. Can anyone give me hint? $\endgroup$ – UfmdFkiF Mar 5 '17 at 19:20

Doesn't seem to be right. Define a continuous strictly positive function $f$ first on $[1,2]$ by setting $f(1)=4$, $f(3/2)=1$, $f(2)=8$ (and arbitrarily elsewhere in that interval). Extend now by setting $f(2x)=2f(x)$ to a continuous strictly positive function on $[1,12]$ and it seems we have a counter example, with $f(1)=f(6)=2^2f(3/2)=4$ and $f(2x)-f(x)=f(x)>0$.


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