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problem A function $f(x)$ is defined for all real values $x$. If $f(a+b)=f(ab)$ for all a and b, and $f(\frac{-1}{2})=\frac{-1}{2}$ , compute the value of $f(1)$.

my steps i know that I can have

$f(\frac{-1}{2}+\frac{-1}{2})=f(-1)=f(\frac{1}{4})$ from there i don't know what to do

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    $\begingroup$ Why is the function odd? It looks even to me: $f(a+b)=f(ab)=f(-a\times-b)=f(-a-b)=f(-(a+b))$ $\endgroup$ – Ian Miller Sep 19 '16 at 23:20
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    $\begingroup$ f(x) = f (0+x)=f (0)=f (0*-1/2)=f (-1/2)=-1/2. So.... $\endgroup$ – fleablood Sep 19 '16 at 23:21
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Taking $a=0$ in the functional equation (which is always a good thing to try when solving functional equations), we get that

$$f(0+b) = f(0\cdot b)$$

$$f(b) = f(0)$$

and thus $f$ is constant, and $f(1) = -\frac{1}{2}$.

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  • $\begingroup$ @JohnRawls We have that the functional equation holds for all $a,b$, so we can choose $a=0$. Then we get that $f(b)=f(0)$ from the equation. Because of that, for any $x$ and $y$, $f(x) = f(0), f(0) = f(y) \implies f(x) = f(y)$. Specifically, $f\left(\frac{-1}{2}\right) = f(1) = -\frac{1}{2}$. I hope that clarifies things. $\endgroup$ – Carl Schildkraut Sep 19 '16 at 23:26

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