Let $f:[a,b]\to \mathbb{R}$ be a continous function. Prove that $\exists c\in(a,b)$ so that $$f(c)=\frac{1}{a-c}+\frac{1}{b-c}.$$
I tried to take another function like $g(x)=f(x)-f(c)$ and calculated the limits from $x \to a$ and $x\to b$. I don't think this will work. Any other ideas? Also I'm pretty sure this should be solved using Cauchy-Bolzano or Darboux.
Hint. Consider the function $$F(x):=f(x)-\frac{1}{a-x}-\frac{1}{b-x}$$ which is continuous in $(a,b)$ and evaluate the limits $$\lim_{x\to a^+} F(x)\quad\mbox{and}\quad \lim_{x\to b^-} F(x).$$ Then apply the Intermediate Value Theorem (or Bolzano's Theorem).