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Let $f:[0,1]\to\mathbb{R}$ be a differentiable function, such that $f(0)=0$ and $f(1)=1$. Prove that there exist different $x_0, x_1\in (0,1)$, such that $$\frac{f'(x_0)}{x_0}+\frac{f'(x_1)}{x_1^2}=5$$

I have thought of possibly using Cauchy's theorem, so that we would only need to prove that there exist $k, l\in (0,1)$ such that $f(k)=k^2$ and $f(l)=l^3$, but I don't know how to prove these. Any hint?

Edit 1: Apparently my thought here is wrong, so any ideas?

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  • $\begingroup$ How finding such $k,l$ would help you? $f(k) = k^2$ for some $k$ doesn't neccesarily mean that $f'(k) = 2k$ $\endgroup$ – Stefan4024 Apr 24 '16 at 18:55
  • $\begingroup$ Interesting question, the only thing you for sure is that $f'(c)=1$ for some $c\in (0,1)$, but I'm not sure how to go from there. $\endgroup$ – Mathematician 42 Apr 24 '16 at 18:57
  • $\begingroup$ @Stefan4024 Yes, you are right... But then the exercise becomes impossible!!! $\endgroup$ – Jason Apr 24 '16 at 18:59
  • $\begingroup$ The latter statement is not provable. Think of $f(x)=(x)$, then finding $k$ and $l$ such that $k=k^2$ and $l=l^3$ is not possible on $(0,1)$, and this will be the case for many more $f$'s $\endgroup$ – konewka Apr 24 '16 at 19:04
  • $\begingroup$ No correct solution here too, right? $\endgroup$ – Jason May 13 '16 at 9:53
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Consider the function $g(x) = \frac{1}{3}x^3$. Then as it's differentiable on $(0,1)$ and continuous on $[0,1]$, by Cauchy Mean Value Theorem we have $\exists c \in (0,1)$, s.t.:

$$\frac{f(1) - f(0)}{g(1) - g(0)} = \frac{f'(c)}{g'(c)} \implies \frac{f'(c)}{c^2} = 3$$

Similarly now taking $h(x) = \frac{1}{2}x^2$, by Cauchy Mean Value Theorem $\exists d \in (0,1)$, s.t.:

$$\frac{f(1) - f(0)}{h(1) - h(0)} = \frac{f'(d)}{h'(d)} \implies \frac{f'(d)}{d} = 2$$

Hence the proof.

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  • $\begingroup$ How can you show that $c$ and $d$ are different? $\endgroup$ – Paolo Apr 24 '16 at 19:11
  • $\begingroup$ @Paolo I'm trying to prove but $c=d=\frac 23$ is the only "bad" point. Maybe proving that there exists yet another point such that it satisfies one of the equations. $\endgroup$ – Stefan4024 Apr 24 '16 at 19:29
  • $\begingroup$ I'm trying to prove that, too. But I don't know how, at the moment... $\endgroup$ – Paolo Apr 24 '16 at 19:30
  • $\begingroup$ No answer yet ? $\endgroup$ – Jason Jun 20 '17 at 22:26

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