1. ## Mvt

Here is an outline of my question:
-f is continuous on [a,b]
-f is differentiable on (a,b)
-f(a) = a and f(b) = b
-then there are two points c1 and c2 such that 1/f'(c1) + 1/f'(c2) = 2. I know how to do the problem f'(c1) = f'(c2) = (b-a)/(b-a) = 1. I just don't get this. How do we know that the second point c2 exists with this condition? Doesn't the Mean Value Theorem only guarantee one such point? How do we know they are distinct?

2. Just because $\displaystyle \frac{1}{f'(c_1)}+\frac{1}{f'(c_2)}=2$ does not imply that $\displaystyle f'(c_1)=f'(c_2)=1$

Since $\displaystyle f$ is continuous, let $\displaystyle f(c)=\frac{a+b}{2}$ for some $\displaystyle c$ in $\displaystyle (a,b)$.
Apply the MVT to the intervals $\displaystyle [a,c]$ and $\displaystyle [c,b]$. Let $\displaystyle c_1$ and $\displaystyle c_2$ be the corresponding points and you should see that
$\displaystyle \frac{1}{f'(c_1)}+\frac{1}{f'(c_2)}=2$

3. Thank you very much; I didn't even notice that.

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