$f:[a,b] \to \mathbb R$ is a continuous function and $0 < a < b$ and $f$ is differentiable in $(a,b)$ and $\dfrac{f(a)}{a} = \dfrac{f(b)}{b}$.
Prove that there exists $x \in (a,b)$ so that $xf'(x) = f(x)$.
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$\begingroup$ It should also be assumed that $f$ is continuous at the endpoints. $\endgroup$– Michael HardyJun 16, 2013 at 0:09
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1 Answer
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Hint: Apply the mean value theorem (or Rolle's theorem) to $g(x) = \dfrac{f(x)}{x}$.
answered Jun 16, 2013 at 0:06
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$\begingroup$ OK, so $\frac{\frac{f(b)}{b} - \frac{f(a)}{a}}{b-a}$ = $\frac{f'(x)}{x}$ And I have got 0 in the numerator. Don't know what to do next. $\endgroup$– keriJun 16, 2013 at 0:22
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1$\begingroup$ @keri Differentiation should be applied to all of $\dfrac{f(x)}{x}$. In other words, take $\left(\dfrac{f(x)}{x}\right)' = 0$. Apply the quotient rule and simplify to get the result. $\endgroup$ Jun 16, 2013 at 0:26
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$\begingroup$ Aghh so true! My mistake. $\frac{\frac{f(b)}{b} - \frac{f(a)}{a}}{b-a}$ = ($\frac{f(x)}{x}$)' = $\frac{f'(x)-f(x)}{x^2}$ soooo f(x) = xf'(x) Thank you so much! Now, it is clear! :) $\endgroup$– keriJun 16, 2013 at 0:41