# Proving the existence of $ξ$ and $η$ such that $f'(\xi)(\xi-a)+f'(\eta)(\eta-b)+f(a)+f(b)=0$

Let $$f$$ be continuous on $$[a,b]$$, and differentiable on $$(a,b)$$, $$\int_a^b f(x)dx=0$$. Show that there exists two distinct $$\xi,\eta\in(a,b)$$, such that $$f'(\xi)(\xi-a)+f'(\eta)(\eta-b)+f(a)+f(b)=0$$.

My attempt: Let $$F(x)=(x-a)[f(x)+f(a)]-\int_a^x f(t)dt$$, $$G(x)=(x-b)[f(x)+f(b)]-\int_b^x f(t)dt$$. Then $$F(a)=0, F(b)=-G(a), G(b)=0.$$ And what we need is to show some $$\xi\neq \eta$$, such that $$F'(\xi)+G'(\eta)=0$$. Consider $$0, and by the Lagrange intermediate value, it suffices to show for some $$c$$, $$\frac{F(c)}{c-a}+\frac{-G(c)}{b-c}=0$$. But such a $$c$$ is not obvious by continuity.

• Maybe by using a two-steps procedure like in this interesting problem and solution: math.stackexchange.com/q/3715320 but I am not able to find the good auxiliary function... Jun 11 at 10:05

The statement is wrong, a counterexample is $$f(x) = x^2-1/3$$ on $$[a, b]= [-1, 1]$$.

The condition $$\int_{-1}^1 f(x) \, dx = 0$$ is satisfied, but for all $$\xi, \eta \in (-1, 1)$$ is $$f'(\xi)(\xi-a)+f'(\eta)(\eta-b)+f(a)+f(b) \\ = 2\xi(\xi+1) + 2\eta(\eta-1) + \frac 43 \\ = 2 \left(\xi+\frac 12\right)^2 + 2\left(\eta-\frac 12\right)^2 + \frac 13 \ge \frac 13 > 0 \, .$$

Some remarks on how I came up with this example. Of course I tried to prove the statement, and wanted to determine $$\xi$$ and $$\eta$$ for some concrete cases. In order to make things simple, I chose $$[a, b] = [-1, 1]$$.

If $$f$$ is an odd function then $$\eta = -\xi$$ always works: $$f'(\xi)(\xi+1)+f'(\eta)(\eta-1)+f(-1)+f(1) \\ = f'(\xi)(\xi+1)+f'(\xi)(-\xi-1) = 0 \, .$$

The “simplest” even functions on $$[-1, 1]$$ are $$f(x) = x^2+c$$, and with $$c=1/3$$ the condition $$\int_{-1}^1 f(x) \, dx = 0$$ is satisfied. Then I tried to find a solution of the form $$\eta = -\xi$$, but that gave a quadratic equation without real solutions.

This lead to the conjecture that the statement might be wrong, and that $$f(x) = x^2-1/3$$ could be a counterexample.

• I was thinking about this problem since it was asked. Great result! Thanks for sharing. Will give the bounty as soon as the site allows. Jun 12 at 18:49
• @VIVID: So was I :) – Thanks! Jun 12 at 18:51
• Shout out for sharing your working process to find the example. Very insightfull. Great answer. Jun 12 at 19:36
• (+1) Well done! Jul 13 at 16:08