Can someone please help me in proving this? Let $k_{2}>k_{1}>0$, prove that for any $x>0$, $f(x)$ is a monotonically increasing function.  
$$
f(x)=\frac{1-e^{-k_{1} x}}{1-e^{-k_{2} x}}.
$$
We can have $f'(x)=\frac{(k_{2}-k_{1})e^{-(k_{1}+k_{2})x}+k_{1}e^{-k_{1}x}-k_{2}e^{-k_{2}x}}{(1-e^{-k_{2}x})^2}$. Then what?

Try $$\frac{f'(x)}{f(x)}=\frac{k_1e^{-k_1x}}{1-e^{-k_1x}}-\frac{k_2e^{-k_2x}}{1-e^{-k_2x}}$$
Now consider $g(k)=\frac{ke^{-kx}}{1-e^{-kx}}$ and show that it is monotonic as a function of $k$, for any particular value of $x>0$.
