Question on finding Inverse Functions It is given that the following function is one-to-one
 $$f(x)=e^x+x$$ Find $$(f^{-1})(e+1)$$
I tried to begin by finding $$f^{-1}(x)$$ but could not know where/how to begin. 
 A: Method 1 - By Inspection
As pointed out in the comments, since $f$ is a monotonically increasing function for all $x$ its inverse exists for all $x$. Observing that $f(1) = e + 1$, one has $f^{-1} (e + 1) = 1$.
Method 2 - The Lambert W function way
An inverse for the function $f$ can be explicitly found in terms of the Lambert W function.
If we write $y = e^x + x$, to find its inverse, interchanging the roles for $x$ and $y$ and solve for $y$. Doing so we have
\begin{align*}
x &= e^y + y\\
x - y &= e^y\\
(x - y)e^{-y} &= 1\\
(x - y) e^{x - y} &= e^x
\end{align*}
As this last equation is now exactly in the form for the defining equation for the Lambert W function, namely
$$\text{W}(x) e^{\text{W}(x)} = x,$$
we have
$$x - y = \text{W}_0 (e^x),$$
or
$$f^{-1} (x) = x + \text{W}_0 (e^x).$$
Note the principal branch of the Lambert W function is selected since its argument is positive for all real $x$. 
Now at the point $x = e+1$ we have
$$f^{-1} (e + 1) = e + 1 - \text{W}_0 (e^{e + 1}).$$
The term containing the Lambert W function can be simplified by making use of the following simplification rule
$$x = \text{W}_0 (x e^x), \qquad x \geqslant -1.$$
Thus
$$\text{W}_0 (e^{e+ 1}) = \text{W}_0 (e \cdot e^e) = e,$$
giving
$$f^{-1} (e + 1) = e + 1 - e = 1,$$
as expected.
A: Since your $f$ is one-to-one there is an inverse function $f^{-1}$.  They are related by $f(x) = y \Leftrightarrow f^{-1}(y) = x$.
$f^{-1}(e+1) = x$ is the same as which $x$ gives $f(x) = e+1$, so solve for $x$:
$$
e^x + x = e + 1.
$$
A: Note you don't need to find the inverse function $f^{-1}(x)$ rather the inverse of one particular image point i.e. $e+1$
Note the similarity between 
$$
e^{x}+x
$$
and
$$
e+1
$$
One might then recognise that if $x=1$ then $e^{x}+x = e + 1$, and that is the only value of $x$ generating $e+1$, by the 1-1 property of $f$.
This means the pre-image (inverse) of $e+1$ is $1$ i.e. 
$$f^{-1}(e+1)=1$$
