Jun 26, 2012 Thread starter #1 J jacks Well-known member Apr 5, 2012 226 If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $
Jun 27, 2012 #2 Sudharaka Well-known member MHB Math Helper Feb 5, 2012 1,621 jacks said: If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $ Click to expand... 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.
jacks said: If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $ Click to expand... 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.
Jun 27, 2012 #3 C CaptainBlack Well-known member Jan 26, 2012 890 jacks said: If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $ Click to expand... 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
jacks said: If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $ Click to expand... 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