Critical point problem so I was doing my calculus homework and ran into this tricky question.
Determine the total number of critical points of the function $f(x)=(x+e^x)^k$, where $k>0$ is an integer
So I got the derivative is $f'(x)=k(x+e^x)^{k-1} \times (1+e^x)$, but I couldn't find the point, however when I look at the graph there is a point around $0.567$ or W($1$), so I am a bit lost.
Thanks! 
 A: $$f'(x)=k(x+e^x)^{k-1} \times (1+e^x)=0 $$
has only one solution which is where $x+e^x=0$ and that is the point that you want to approximate. 
The answer should be negative so $x=0.567$ is problematic. 
A: Solution
Notice $$f'(x)=k(x+e^x)^{k-1}(1+e^x).$$
Let $f'(x)=0$. Then we obtain $$x+e^x=0$$
This equation has no closed-form solution. But by graphing $y=-x$ and $y=e^x$, you may intuitively find there exists only one root over $(-1,0)$.

Moreover, we may prove this fact. Since $f''(x)=1+e^x>0.$ Then $f'(x)$ is rigorously increasing over $(-\infty,+\infty)$. Thus, there exists at most one solution for $f'(x)=0$. Notice that $f'(-1)=-1+e^{-1}<0$ and $f'(0)=0+e^0>0$. Thus, by intermediate value theorem, we may claim the fact.
A: At critical points $f'(x)=0$ so
$0=k(x+e^x)^{k-1} \times (1+e^x)$
k and $(1+e^x)$ are positive so $0=x+e^x$
$x_n=-e^{x_{n+1}}$ so $x \approx -0.56714329$
A: Given
\begin{align} 
f'(x)&=k(x+\exp(x))^{k-1} \times (1+\exp(x))
\end{align}  
the function $f$ have no critical points for $k=1$
and one critical point for $k>1$ at
\begin{align}
x&=-\operatorname{W}(1)
\approx -0.567143290409784
.
\end{align}
Edit
We need to solve $f'(x)=0$, so since it is given that $k>0$
and we know that $\exp(x)>0$ for all $x\in \mathbb{R}$,
we can ignore the factor $(1+\exp(x))$, 
and all we left to consider is when
\begin{align}
(x+\exp(x))^{k-1}&=0
.
\end{align}
for $k=1$ we have
\begin{align}
f(x)&=x+\exp(x)
\\
f'(x)&=1+\exp(x)>0
\quad \forall x\in\mathbb{R}
.
\end{align}
For $k>1$:
\begin{align}
x+\exp(x)&=0
,\\
-x&=\exp(x)
,\\
-x\exp(-x)&=1
\tag{1}\label{1}
.
\end{align}
At this point we have an equation of the form
$u\exp(u)=v$, and a long time ago
it was suggested to write an exact
solution of this equation as 
$u=\operatorname{W}(v)$,
where $\operatorname{W}$
is the
Lambert W function,
which for equations of the form 
$u\exp(u)=v$ 
plays exactly the same role 
as the function $\ln$
plays for equations of the form 
$\exp(u)=v$:
to express the exact solution of 
$\exp(u)=v$, we write $u=\ln(v)$.
So, to continue with \eqref{1},
we have
\begin{align}
\operatorname{W}(-x\exp(-x))&=\operatorname{W}(1)
,\\
-x&=\operatorname{W}(1)
,\\
x&=-\operatorname{W}(1)
\tag{2}\label{2}
.
\end{align}
Now, we need to recall that 
when dealing with $\operatorname{W}$
(very much like with quadratic equation, for example)
we can face with from 0 to 
at most two real solutions,
which have special names $\operatorname{W_0}$
and $\operatorname{W_{-1}}$.
The number of solutions and their range 
is solely defined by the argument $v$ in $\operatorname{W}(v)$:
\begin{align}
v&<-\tfrac1{\mathrm{e}}\quad\text{no real solutions}
;\\
-\tfrac1{\mathrm{e}}<
v&<0\quad\text{two real solutions:}
\quad\operatorname{W_0}(v),
\operatorname{W_{-1}}(v),
\text{ for which holds}\quad
\operatorname{W_{-1}}(v)<-1
<\operatorname{W_0}(v)<0
;\\
v&\ge0
\quad\text{just one real solution}
\quad\operatorname{W_0}(v)\ge0
;\\
v&=
-\tfrac1{\mathrm{e}}\quad\text{the case when}
\quad\operatorname{W_0}(v)=
\operatorname{W_{-1}}(v)=-1
.
\end{align}
Thus, in our case \eqref{2}, the argument of $\operatorname{W}$
is $1$, that means, we have just one real solution,
\begin{align}
x&=-\operatorname{W_0}(1)
.
\end{align}
