# If $f(a+b)=f(ab)$ for all a and b and $f(\frac{-1}{2})=\frac{-1}{2}$ find $f(1)$

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

• Why is the function odd? It looks even to me: $f(a+b)=f(ab)=f(-a\times-b)=f(-a-b)=f(-(a+b))$ – Ian Miller Sep 19 '16 at 23:20
• f(x) = f (0+x)=f (0)=f (0*-1/2)=f (-1/2)=-1/2. So.... – fleablood Sep 19 '16 at 23:21

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 $f(1) = -\frac{1}{2}$.
• @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. – Carl Schildkraut Sep 19 '16 at 23:26