Solution to AMM problem 11892

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)}
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)}.


Author: cskyon

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