# $\frac{1}{f'(x_1)}+\frac{1}{f'(x_2)}=2$

1.Let f be a real-valued differentiable function defined on [0, 1]. If $f(0)=0$ and $f(1)=1$, prove that there exists two numbers $x_1,x_2 \in [0, 1]$ such that $\frac{1}{f'(x_1)}+\frac{1}{f'(x_2)}=2$.

2.Let f be a real-valued differentiable function defined on (0,$\infty$). If $\lim_{x\rightarrow0^+}f(x)=\lim_{x\rightarrow\infty}f(x)=1$, prove that there exists $c\in(1,\infty)$ such that $f'(c)=0$.

1.From the question's statement, I wasn't sure if $x_1$ could equal $x_2$. If it can, is the following proof valid? Clearly, it suffices to show $f'(x)=1$ for some $x \in [0,1]$.

Suppose $f'(x)\neq1$ for all $x \in [0,1]$.

Then either $f'(x)>1$ or $f'(x)<1$ for all relevant x (otherwise, $\exists x_1,x_2$ s.t. $f'(x_1)>1, f'(x_2)<1$ which contradicts the IVT for derivatives).

Assume, wlog, $f'(x)>1$ for all x.

Then $1=f(1)=\int_{0}^{1}f'(x)dx>\int_{0}^{1}dx=1,$ contradiction.

2. Again we proceed by contradiction; suppose $f'(x)\neq0$ for all x. By an identical argument to the one used in 1, either $f'(x)>0$ or $f'(x)<0$ for all x. Wlog, assume $f'(x)>0$. Then f(x) is strictly increasing $\forall x$.

Then, if x>1, $1=\lim_{t\rightarrow0^+}f(t)<f(1)<f(x)$. Hence $\lim_{x\rightarrow\infty}f(x)>1$ (if it exists), contradiction.

First, if my proofs are valid, are there simpler proofs? Second, can someone provide me with a proof of the case where $x_1\neq x_2$ in 1?

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I think in these problems, one could use the mean value theorem and rolle's theorem. – PEV Jan 24 '11 at 1:19

For (1) I think you could use the Mean Value Theorem. Namely, there exists an $x_1 \in (0,1)$ such that $f'(x_1) = \frac{f(1)-f(0)}{1-0} = 1$. Then put $x_2 = x_1$. For (2), you could use the generalization of Rolle's Theorem.

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Would you mind demonstrating? – Benji Jan 24 '11 at 1:22
Wow, that was a bit of an oversight on my part. – Benji Jan 24 '11 at 1:23

Updated for completeness

In fact, there are an uncountable number of solutions to

$\frac{1}{f'(x_1)} + \frac{1}{f'(x_2)} = 2$

with $x_1 \ne x_2$. First, if the graph of $f$ coincides with the main diaognal -- i.e. if $f(x) = x$ for all $x \in [0,1]$ -- then any pair $(x_1,x_2)$ is a solution. So suppose there exists $t$ with $f(t) \ne t$. Then there must exist a point $(u,f(u))$ off the main diagonal with $f'(u) = 1$. To see this, apply the Mean Value Theorem (MVT) to the interval $(l,r)$, where $l$ is the greatest upper bound of those $x \lt t$ with $f(x)=x$, and $r$ is the least upper bound of those points $x \gt t$ with $f(x)=x$.

Now, to fix things, suppose that $f(u) \gt u$. Then we apply the MVT to intervals $[0,u]$ and $[u,1]$, to find points $v \in (0,u)$ and $w \in (u,1)$ with

$f'(v) = f(u)/u > 1$ and $f'(w) = (1 - f(u))/(1-u) < 1$.

Now, given any $s \in (1,f'(v))$ and $t \in (f'(w),1)$, there exist $x_1 \in (v,u)$ and $x_2 \in (u,w)$ such that $f'(x_1) = s$ and $f'(x_2) = t$ (because a derivative, although not necessarily continuous, does obey the Intermediate Value property). There are uncountably many such pairs $(s,t)$ with the property $\frac{1}{s} + \frac{1}{t} = 2$, so there are uncountably many solutions $(x_1, x_2)$.

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