24. [12] Let $f(x)=\frac{1}{1-x}$. Let $f^{k+1}(x)=f\left(f^{k}(x)\right)$, with $f^{1}(x)=f(x)$. What is $f^{2008}(2008)$ ?
Answer: $\frac{-1}{2007}$ Notice that, if $x \neq 0,1$, then $f^{2}(x)=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}$, which means that $f^{3}(x)=$ $\frac{1}{1-\frac{x-1}{x}}=x$. So $f^{n}$ is periodic with period $n=3$, which means that $f^{2007}(x)=x$ so $f^{2008}(2008)=$ $f(2008)=\frac{-1}{2007}$.