Answer
$\displaystyle \qquad f(x)=\frac{4}{1+3(3^{-x})} $
Work Step by Step
Goal: $\displaystyle \quad f(x)=\frac{N}{1+Ab^{-x}}$,
where $N$ = limiting value = $4$ (given)
$f(0)=1\Rightarrow\left\{\begin{array}{ll}
1 & =\dfrac{4}{1+Ab^{0}}\\
1+A & =4\\
A & =4-1=3
\end{array}\right.\qquad\Rightarrow\qquad A=3$
$f(x)=\displaystyle \frac{4}{1+3b^{-x}}$
$f(1)=2\Rightarrow\left\{\begin{array}{ll}
2 & =\dfrac{4}{1+3b^{-1}}\\
1+3b^{-1} & =4/2\\
3b^{-1} & =2-1=1\\
b^{-1} & =1/3\\
b&=3
\end{array}\right.$
Thus, $\displaystyle \qquad f(x)=\frac{4}{1+3(3^{-x})} $
