Let $f:(0,\infty)\to\mathbb R$ be a differentiable function for which $f'\left(x\right)\cdot f'\left(\frac{1}{x}\right)<0$ for every $0<x\ne1$. Then $f'(1)=0$
I have this question, and I was wondering how to switch the fact the f is differentiable for every positive x into differentiable in a chosen interval that will help me with my proof.. I tried to make quiet a general interval in $[\frac{1}{x},x]$ but I'm afraid that I can't acctualy do that, and that it's not accurate..
d. True
We know $f$ is differential in $(0,\infty)$, therefore $f$ is also differentiable in the interval $[\frac{1}{x},x]$ for any $x ∈ R$ such that $0<\frac{1}{x}<1<x$.
Since $f'\left(x\right)\cdot f'\left(\frac{1}{x}\right)<0$, by Darboux's theorem there exist a point $c ∈ R$ such that
$\frac{1}{x} < c < x$ and $f'(c)=0$.
Assume toward contradiction that c ≠ 1, then we have $f'\left(c\right)\cdot f'\left(\frac{1}{c}\right)<0$.
But since $f'(c)=0$, we have $0 \cdot f'\left(\frac{1}{c}\right)<0 \ \ \ ➜ \ \ \ 0 < 0$. Contradiction!
Therefore $c = 1$ and $f'(1) = 0$.
As @Proxava said, your proof is OK.
You can also show that $f$ is strictly monotone on $(0,1] \ \& \ [1,+\infty)$, in opposite directions.