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Let $f,g:[0,1] \to \mathbb{R}$ such that $f'(x)>0,\ g'(x) >0,\ \forall x\in [0,1]$. Moreover, $f(0)= g(0)$ and $f(1)=g(1)$. Prove that exists $x_1, x_2 \in [0,1]$ such that $$f(x_1) = g(x_2), \ \text{and}\ \ f'(x_1) = g'(x_2)$$


I've tried to use Mean Value Theorem with:

a) $h(x)= f(x)-g(x)$

b) $h(x) = f(x) - xg(x)$

c) $h(x) = f(x)g(x)$

d) Using $g(1-x)$ instead of $g(x)$

But nothing seems to help in order to find at least one of $x_1$ or $x_2$. I think that, finding one of them, the other could be easy to get.

Thanks for the help!

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2 Answers 2

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Strategy: there are lots of pairs $x_1,x_2$ such that $f(x_1)=g(x_2)$: just take any $y\in[0,1]$ and force $f(x_1)=y$ and $g(x_2)=y$ — in other words, set $x_1=f^{-1}(y)$ and $x_2=g^{-1}(y)$. So let's use that parametrization to study the other desired equality $f'(x_1)=g'(x_2)$.

Proof: Note that $f(f^{-1}(y)) = y = g(g^{-1}(y))$ for all $y\in[0,1]$. By the chain rule, $$ f'(f^{-1}(y)) \frac d{dy}f^{-1}(y) = g'(g^{-1}(y)) \frac d{dy}g^{-1}(y) \tag{1} $$ for all $y$.

Set $a=f(0)=g(0)$ and $b=f(1)=g(1)$. Note that $f^{-1}(a) = 0 = g^{-1}(a)$ and $f^{-1}(b) = 1 = g^{-1}(b)$. Hence, by the mean value theorem (indeed, Rolle's theorem) applied to $f^{-1}-g^{-1}$ on the interval $[a,b]$, there exists $y_0\in(a,b)$ such that $\frac d{dy}f^{-1}(y_0) = \frac d{dy}g^{-1}(y_0)$.

Therefore, by (1), $f'(f^{-1}(y_0)) = g'(g^{-1}(y_0))$. Now set $x_1 = f^{-1}(y_0)$ and $x_2 = g^{-1}(y_0)$.

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  • $\begingroup$ Everything is so clear, except for $f^{-1}(0)=(0)$. I don't see that $\endgroup$
    – FormerMath
    Oct 12, 2014 at 17:51
  • $\begingroup$ Woops, I misread that part of the problem. Edited to fix. $\endgroup$ Oct 12, 2014 at 21:55
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I suggest that you look at $h(x) = g^{-1}(f(x))$. Suppose that for some number $x_1$, $h'(x_1) = 1$. What would that tell you?

Also: How do you know $g^{-1}$ even exists? You need to answer this before you go using it willy-nilly.

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