# There exist $x_1, x_2, x_3$ such that $\frac{1}{f'(x_1)} + \frac{1}{f'(x_2)} + \frac{1}{f'(x_3)} = 3$

Let $f$ be a real-valued function defined in $[a, b] \subset \mathbb{R}$, with $f(a) = a, f(b) = b$. Suppose that $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$. Show that there exist three distinct points $x_1, x_2, x_3$ such that

$$\frac{1}{f'(x_1)} + \frac{1}{f'(x_2)} + \frac{1}{f'(x_3)} = 3$$

My hunch is to use the mean-value theorem or Rolle's theorem somehow. But these theorems only guarantee the existence of a certain point, and not a triple of points, so I am stuck.

• Closely related to math.stackexchange.com/questions/18674/… – Niklas Jul 6 '14 at 17:32
• Here is what I have tried. Maybe you can make it work out. Mean value theorem gives you a $x_1\in (a,b)$ such that $f'(x_1)=1$. Now for other two points: wlog little to left of $x_1$ slope is little less than one and a little to the right the slope is a little more than one. – john w. Jul 6 '14 at 17:33
• @gnometorule constant function does not fit your hypothesis about $a$ and $b$. Ian's comment answers your question. – john w. Jul 6 '14 at 17:41
• @johnw. I think his point was that $f'$, not $f$, could be constant, but in that case the problem is trivial. – Ian Jul 6 '14 at 17:42

## 2 Answers

Let $z_1$ and $z_2$ with $z_1<z_2$ be such that

$$f(z_1)=\frac{2a+b}{3}$$ and $$f(z_2)=\frac{a+2b}{3}$$

Then it is easy to see that the slopes formed by the segements $$(a,a), (z_1,f(z_1)), (z_2,f(z_2)), (b,b)$$ sum in their reciprocals to $3$.

Namely,

$$\left(\frac{\frac{b-a}{3}}{z_1-a}\right)^{-1}+ \left(\frac{\frac{b-a}{3}}{z_2-z_1}\right)^{-1}+ \left(\frac{\frac{b-a}{3}}{b-z_2}\right)^{-1}=3$$ This the gives the points $x_1 \in (a,z_1)$ $x_2 \in (z_1,z_2)$ $x_3 \in (z_2,b)$ by the mean value theorem.

$$f'(x_1)=\frac{f(a)-f(b)}{a-b}=1$$

If that is true then we can simplyfy the task to:

$$\frac{1}{f'(x_2)}+\frac{1}{f'(x_3)}=2$$

For which there already is an answer here