Find the solution of $e^{x}+2x=0.96$ We have the following function $$f(x)=e^{x}+2x.$$
Using the theorem on the derivative of the inverse function and the differentail of the function find the solution of $f(x)=0.96$.
I started by taking $x=f^{-1}(0.96)$.
Now, from teh differential we can write that $f^{-1}(x+\Delta x) \approx f^{-1}(x)+[f^{-1}(x)]'\Delta x$. I do not know if it is correct and what I should do now. 
I would be grateful for any hits.
 A: If you are approximating something with differentials, start with an $x$ for which you know the value $f(x)$ is near $0.96$.
Hint: Start with $f^{-1}(1) = 0$.
A: This has been written for your curiosity.
In fact, the equation
$$y=e^{a x}+b x$$ shows explicit solution(s) in terms of Lambert function. They write
$$x=\frac{y}{b}-\frac{1}{a}W\left(\frac{a }{b}e^{\frac{a y}{b}}\right)$$ So, for your specifdic case $(a=1,b=2,y=\frac{24}{25})$
$$x=\frac{12}{25}-W\left(\frac{e^{12/25}}{2}\right)$$ Since the argument is "small", you could use, as an approximation built around $t=0$,
$$W(t)=t \,\frac{1+\frac{228 }{85}t+\frac{451 }{340}t^2 } {1+\frac{313 }{85}t+\frac{1193 }{340}t^2+\frac{133 }{204}t^3 }$$ This would give, as an approximate solution $x\approx -0.0130442$ while the exact value would be $x\approx -0.0133630 $.
For the specific case you address, plotting the function, you notice that the solution is close to $x=0$. So, make a series expansion
$$e^x+2x=1+3 x+\frac{1}{2}x^2+O\left(x^3\right)$$ So, using the first term
$$\frac{24}{25}=1+3x\implies x=-\frac 1{75} \approx -0.0133333 $$ Using the first and second term
$$\frac{24}{25}=1+3x+\frac{1}{2}x^2\implies x=\frac{\sqrt{223}}{5}-3\approx -0.0133631$$
Sooner or later, you will alos learn that, better then with Taylor series, we can approximate functions using Padé approximants. If we use the $[1,n]$ of them which will write as
$$f(x)\simeq \frac {f(a)+c_{(n)} (x-a)}{1+\sum_{k=1}^n d_k (x-a)^k}$$ then, starting with a rational value of $a$, we can get rational approximations of the solution
$$x_{(n)}=a-\frac{f(a)}{c_{(n)}}$$
$$\left(
\begin{array}{ccc}
n & x_{(n)} & x_{(n)} \approx \\
 0 &  -\frac{1}{75} &  -0.01333333333   \\
 1 &  -\frac{6}{449} &    -0.01336302895 \\
 2 &  -\frac{1347}{100801} &    -0.01336296267
\end{array}
\right)$$
