Is it true that $f'(\xi_1)(\xi_1-a)+f'(\xi_2)(\xi_2-b)+f(a)+f(b)=0$ if some conditions are met?

Problem: if function $$f$$ is continuous on the closed interval $$[a,b]$$ and differentiable on the open interval $$(a,b)$$, and $$\int_a^b f(x)dx=0$$ Prove that there exists two distinct real numbers $$\xi_1,\xi_2\in(a,b)$$ such that $$f'(\xi_1)(\xi_1-a)+f'(\xi_2)(\xi_2-b)+f(a)+f(b)=0$$

I suspect I should use some sort of mean value theorem to prove this problem, but I tried all forms of the theorem listed on the Wikipedia page without any success. I am beginning to suspect this problem might be wrong and am looking for counter-example. Any help from you is greatly appreciated!

• Do you have a source for this problem? I'm asking myself why you think the mean value theorem is applicable, for example : is it because you recently learned it, or because the place from where you got the problem suggested it? It may happen that this problem needs a bigger hammer. Commented Sep 1, 2021 at 15:56
• @TeresaLisbon Because the problem is given $\int_a^b f(x)dx=0$ so I conclude that there exists at least one point $x_0\in(a,b)$ such that $f(x_0)=0$, and I naively think it should have something to do with the mean value theorem. I could be wrong. This problem comes from a calculus problem set, in which it does not given any context. So I think all the conditions have been mentioned in the problem statement. Commented Sep 2, 2021 at 1:37

Take $$f(x) = x^{10}-1$$ on the interval $$[0,11^{1/10}]$$, it meets all your conditions;

Given that $$f(0) +f(11^{1/10}) = 9$$ and $$f'(x) = 10x^9$$ we are trying to find $$\alpha,\beta \in (0,11^{1/10})$$ such that

$$10\alpha^{10}+10\beta^9(\beta-11^{1/10}) +9 = 0$$

but notice that $$10\beta^9(\beta-11^{1/10}) +9$$ is always positive (and also $$10\alpha^{10}$$ obviously) so the theorem is not true!

The way I found it :

I started to reason about $$\cos x$$ and $$\arcsin x$$ but then I realized that this "type" of functions attained the same derivative two times and they were also too symmetrical. Therefore I tried $$x^2-1$$ to avoid the problem but it wasn't strong enough (one of the two parts in which I also decomposed the expression resulting from $$x^{10}-1$$ was negative) then I tried $$\alpha x^k + \beta$$ and there were many counter-examples :-)

• Awesome job! How on earth did you come up with $x^{10}-1$? Commented Sep 2, 2021 at 19:09
• thank you @Alann Rosas ! I updated with the way I've found it! Commented Sep 2, 2021 at 19:15