3. If $f$ is a function such that $f(a+b)=\frac{1}{f(a)}+\frac{1}{f(b)}$, find all possible values of $f(2011)$.
3. Consider $f(0)=f(0+0)=\frac{1}{f(0)}+\frac{1}{f(0)}$ which gives $[f(0)]^{2}=2$. Thus, $f(0)= \pm \sqrt{2}$.
Let $x=f(2011)$.
If $f(0)=\sqrt{2}$ then $x=f(2011)=f(2011+0)=\frac{1}{f(2011)}+\frac{1}{f(0)}=\frac{1}{x}+\frac{1}{\sqrt{2}}$. So, $x=\frac{\sqrt{2}+x}{\sqrt{2} x}$ which yields to $\sqrt{2} x^{2}-x-\sqrt{2}=0$.
Solving for $x$ using the quadratic formula yields to,
$$
x=\frac{1 \pm \sqrt{1-4(\sqrt{2})(-\sqrt{2})}}{2 \sqrt{2}}=\frac{1 \pm 3}{2 \sqrt{2}}
$$

Hence, if $f(0)=\sqrt{2}$ then either $f(2011)=\sqrt{2}$ or $f(2011)=-\frac{\sqrt{2}}{2}$.
However, suppose $f(2011)=-\frac{\sqrt{2}}{2}$. Consider $f(0)=f(2011+(-2011))=\frac{1}{f(2011)}+$ $\frac{1}{f(-2011)}$ which implies that $\sqrt{2}=-\sqrt{2}+\frac{1}{f(-2011)}$. Thus, $f(-2011)=\frac{\sqrt{2}}{4}$.
But if we consider $f(-2011)=f(-2011+0)=\frac{1}{f(-2011)}+\frac{1}{f(0)}$, this means that $\frac{\sqrt{2}}{4}=2 \sqrt{2}+\frac{1}{f(0)}$.
Thus, $f(0)=-\frac{2 \sqrt{2}}{7}$ which is a contradiction. Thus for $f(0)=\sqrt{2}, f(2011)=\sqrt{2}$.
If $f(0)=-\sqrt{2}$ then $x=f(2011)=f(2011+0)=\frac{1}{f(2011)}+\frac{1}{f(0)}=\frac{1}{x}-\frac{1}{\sqrt{2}}$. So, $x=\frac{\sqrt{2}-x}{\sqrt{2} x}$ which yields to $\sqrt{2} x^{2}+x-\sqrt{2}=0$.
Solving for $x$ using the quadratic formula yields to,
$$
x=\frac{-1 \pm \sqrt{1-4(\sqrt{2})(-\sqrt{2})}}{2 \sqrt{2}}=\frac{-1 \pm 3}{2 \sqrt{2}}
$$

Hence, if $f(0)=-\sqrt{2}$ then either $f(2011)=-\sqrt{2}$ or $f(2011)=\frac{\sqrt{2}}{2}$.
However, suppose $f(2011)=\frac{\sqrt{2}}{2}$. Consider $f(0)=f(2011+(-2011))=\frac{1}{f(2011)}+$ $\frac{1}{f(-2011)}$ which implies that $-\sqrt{2}=\sqrt{2}+\frac{1}{f(-2011)}$. Thus, $f(-2011)=-\frac{\sqrt{2}}{4}$.
But if we consider $f(-2011)=f(-2011+0)=\frac{1}{f(-2011)}+\frac{1}{f(0)}$, this means that $-\frac{\sqrt{2}}{4}=-2 \sqrt{2}+\frac{1}{f(0)}$.
Thus, $f(0)=\frac{2 \sqrt{2}}{7}$ which is a contradiction. Thus for $f(0)=-\sqrt{2}, f(2011)=-\sqrt{2}$. So the possible values for $f(2011)$ are $\pm \sqrt{2}$.