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Find all differentiable functions $f \colon (0,\infty) \to \mathbb R$ for which there is a positive real number $k$ such that: $$ f(x) \cdot f'(k/x) = x, \qquad\text{for all }x > 0. $$

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up vote 4 down vote accepted

In equation $$ f(x)f'\left(\frac{k}{x}\right)=x $$ we make change of variables $k/x\to x$ then for all $x>0$ we get $$ f\left(\frac{k}{x}\right)f'(x)=\frac{k}{x} $$ Hence $$ \left(f(x)f\left(\frac{k}{x}\right)\right)'=f'(x)f\left(\frac{k}{x}\right)+f(x)f'\left(\frac{k}{x}\right)\frac{-k}{x^2}=\frac{k}{x}+x\frac{-k}{x^2}=0 $$ The rest is clear.

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hmm. we get f(x) f(k/x) = C f(x) = x/f'(k/x) f'(k/x)/f(k/x) = C/x ln(f(k/x)) = Clnx + D f(k/x) = e^D x^C Is this correct? – John Chang Nov 13 '12 at 7:55
@JohnChang You made two mistakes. The first: you must arrive at $f'(k/x)/f(k/x) = x/C$. The second: even if there were not the first misatke the step from $f'(k/x)/f(k/x) = C/x$ to $\ln(f(k/x)) = Clnx + D$ is incorrect. You need to make change $k/x\to x$ in $f'(k/x)/f(k/x) = C/x$. – Norbert Nov 13 '12 at 8:17

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