# Solution to AMM problem 11892

Problem:
Let $f$ be a real-valued continuously differentiable function on $[a, b]$ with positive derivative on $(a, b)$. Prove that, for all pairs $(x_{1}, x_{2})$ with $a\leq x_{1} < x_{2} \leq b$ and $f(x_{1})f(x_{2})>0$, there exists $t\in (x_{1}, x_{2})$ such that
$\displaystyle \frac{x_{1}f(x_{2})-x_{2}f(x_{1})}{f(x_{2})-f(x_{1})}=t-\frac{f(t)}{f'(t)}$
Proof:
Because $f(x_{1})f(x_{2})>0$, $f(x_{1})$ and $f(x_{2})$ are the same sign. Without lost generality, we can assume that $f(x_{1})$ and $f(x_{2})$ are positive (Otherwise, we consider -f). Then for all $x\in [x_{1}, x_{2}]$, by Lagrange’s Theorem we get $f(x)-f(x_{1})=f'(c(x))(x-x_{1}) >0$, hence $f(x)>f(x_{1})>0$.
Now, consider two function: $g(x)=\frac{x}{f(x)}$ and $h(x)=\frac{1}{f(x)}$ are continuously differentiable on $[x_{1}, x_{2}]$, by Cauchy’s Theorem there is $t\in (x_{1}, x_{2})$
$\displaystyle \frac{\frac{x_{2}}{f(x_{2})}-\frac{x_{1}}{f(x_{1})}}{\frac{1}{f(x_{2})}-\frac{1}{f(x_{1})}}=\frac{\frac{f(t)-tf'(t)}{f^{2}(t)}}{-\frac{f'(t)}{f^{2}(t)}} \Longleftrightarrow \frac{x_{1}f(x_{2})-x_{2}f(x_{1})}{f(x_{2})-f(x_{1})}=t-\frac{f(t)}{f'(t)}.$

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## Author: cskyon

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