How would you go about find the values of a,b and c such that they f(0) = f'(0) = f''(0) = 1 ? The function is: 
\begin{align}
  f( x) &= \frac{a+bx}{1+cx} \\\\
\end{align}
The question also tells us that the function and both it's first and second derivatives are equal to the corresponding values of: 
$$\ e^{x}$$
Would finding the values of a,b and c simply be a trial and error process? 

Given $f(0) = f'(0) = f''(0) = 1$ with $$f(x) = \frac{a+bx}{1+cx}$$ quickly leads to $a=1$ and $$f(x) = \frac{1+bx}{1+cx} = \frac{b}{c} - \frac{b-c}{c} \, \frac{1}{1+cx}.$$
Taking derivatives provides:
\begin{align}
f'(x) &= \frac{b-c}{(1+cx)^{2}} \\
f''(x) &= - \frac{2c (b-c)}{(1+cx)^{3}}.
\end{align}
Using the remaining conditions leads to $1 = b-c$ and $1 = -2c(b-c) = -2c$ which is $b = \frac{1}{2}$ and $c = - \frac{1}{2}$. The function then becomes
$$f(x) = \frac{2+x}{2-x}$$  
