Somehow I can't figure out how to prove this seemingly obvious result (assuming we're only dealing with $\mathbb{R}$):
if $\forall a\neq 0, f(x)=af(ax)$, then $\exists k, f(x)=\frac{k}{x}$
Prove if $\forall a\neq 0, f(x)=af(ax)$, then $\exists k, f(x)=\frac{k}{x}$
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calculus
algebra-precalculus
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1Since the identity is true for all $a, x \neq 0$, simply relate $a$ and $x$ so that $ax = 1$. Then $f(x) = \frac{f(1)}{x}$. – 2017-01-17
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0Let $f (1)=k$ for sx\ne 0$, $f (x)=x*f (1)=x*k $ so doesn't appear to be true. – 2017-01-17