MHB Functional Equation: Solving for f(2012)

[juantheron]

If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $

 

[Sudharaka]

If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $

Hi jacks, :)

Since, \(f(x+y) = f(xy)\) we have,

\[f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{2}+0\right)=f\left(-\frac{1}{2}\times 0\right)=f(0)\]

Since, \(f\left(-\dfrac{1}{2}\right) = -\dfrac{1}{2}\)

\[f(0)=-\frac{1}{2}\]

Now,

\[f(2012)=f(2012+0)=f(2012\times 0)=f(0)=-\frac{1}{2}\]

Kind Regards,
Sudharaka.

 

[CaptainBlack]

If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $

Observe:

\[f(x+y)=f(xy) \Rightarrow f(x)=f(0)\]

Hence \( f(x) \) is a constant function, so \(f(-1/2)=-1/2 \Rightarrow f(2012)=-1/2\) \)

CB