Finding zeros of a function I need to find when $f(x)=0$, where:
\begin{equation}
f(x)=1-\dfrac{k}{x}-\dfrac{k}{3}\dfrac{e^{-ax}}{x}+\dfrac{4k}{3}\dfrac{e^{-bx}}{x}
\end{equation}
Here, $k$, $a$ and $b$ are positive constants. I think this problem analytically is impossible, but I really don't know. I've tried to use Mathematica but it doesn't give any answer and I'm pretty desperate.
Thank you so much.
 A: I prefer to add another answer for better notations.
Let $$b=\lambda\, a \qquad \text{and} \qquad t=2\,\frac{ a\, k \,(4 \lambda -1)-3}{a^2 \,k \,\left(4 \lambda ^2-1\right)}$$
Now
$$x=\sum_{n=1}^\infty \Bigg[\frac a {4 \lambda ^2-1}\Bigg]^{n-1}\, \frac{P_{n}(\lambda)}{c_n}\,\, t^n$$ where the first $c_n$ are
$$\{1,3,36,270,6480,45360,2721600,\cdots \}$$ and the $P_n$ are polynomials of degree $3(n-1)$ in $\lambda$. The very first are
$$\left(
\begin{array}{cc}
n & P_n(\lambda) \\
 1 & 1 \\
 2 & 4 \lambda ^3-1 \\
 3 & 80 \lambda ^6+12 \lambda ^4-64 \lambda ^3+12 \lambda ^2+5 \\
 4 & 1088 \lambda ^9+456 \lambda ^7-1800 \lambda ^6+618 \lambda ^5-222 \lambda
   ^4+450 \lambda ^3-114 \lambda ^2-17 
\end{array}
\right)$$
Using this notations, it will be quite simple to generate as many terms as required.
Edit
After a comment about the case $(k=1,a=5,b=2)$, I realise that the first condition is not $g'(0) <0$ but $g(0) \leq 0$. If we face the case where $g'(0)=0$, the problem is a bit different.
$$g'(0)=1+\frac{a k}{3}-\frac{4 a \lambda  k}{3}\implies k=\frac{3}{a (4 \lambda -1)}$$ In such a case, the series expansion of $g(x)$ around $x=0$ starts with a term in $x^2$.
$$g(x)=\frac{x^2} {4\lambda-1}\,\sum_{n=0}^\infty (-1)^n \frac{a^{n+1} \left(4 \lambda ^{n+2}-1\right)}{(n+2)!}\, x^n$$
The inversion of the sum does not make any problem but leads to other polynomials in $$t=\frac{3 \left(4 \lambda ^2-1\right)}{a \left(4 \lambda ^3-1\right)}$$
