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? 
 A: 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
$$f(x) = \frac{2+x}{2-x}$$  
A: Well firstly since $f(0)= 1 \implies a/1 = 1 \implies a=1$
Now $f'(x)= \dfrac{b(1+cx)+c(a+bx)}{(1+cx)^2} $
$f'(0) =  b+bc +ca +bc = c(2b+a)+b = 1 \implies 2cb+1+b=b(2c+1)+1=1 \implies  b(2c+1)=0$ therefore either $b=0$ or $c=-1/2$
You do the same thing for $f''(x)$  
And you will find the solutions of $a,b$ and $c$ easily
A: Just multiply the denominator
$$
\frac{a+bx}{1+cx}=1+x+\tfrac12x^2+O(x^3)\\\iff\\ 
a+bx =(1+cx)(1+x+\tfrac12x^2+O(x^3))=1+(1+c)x+(\tfrac12+c)x^2+O(x^3)
$$
so comparing coefficients you get linear equations 
$$
a=1,\;b=1+c,\;0=\tfrac12+c\iff a=1,\;c=-\tfrac12,\;b=\tfrac12
$$
