Consider the function $f$ defined on $\mathbb { R }$ by

$$f ( x ) = x ^ { 3 } \mathrm { e } ^ { x }$$

It is admitted that the function $f$ is differentiable on $\mathbb { R }$ and we denote $f ^ { \prime }$ its derivative function.

\begin{enumerate}
  \item The sequence $(u _ { n })$ is defined by $u _ { 0 } = - 1$ and, for all natural integer $n$, $u _ { n + 1 } = f \left( u _ { n } \right)$.\\
a. Calculate $u _ { 1 }$ then $u _ { 2 }$.

Exact values will be given, then approximate values to $10 ^ { - 3 }$.\\
b. Consider the function fonc, written in Python language below.

Recall that in Python language, ``i in range (n)'' means that i varies from 0 to n -1.

\begin{verbatim}
def fonc (n):
    u =- 1
    for i in range(n) :
        u=u**3*exp(u)
    return u
\end{verbatim}

Determine, without justification, the value returned by fonc (2) rounded to $10 ^ { - 3 }$.
  \item a. Prove that, for all real $x$, we have $f ^ { \prime } ( x ) = x ^ { 2 } \mathrm { e } ^ { x } ( x + 3 )$.\\
b. Justify that the variation table of $f$ on $\mathbb { R }$ is the one represented below:

\begin{center}
\begin{tabular}{ | l | l c l | }
\hline
$x$ & $- \infty$ & - 3 & $+ \infty$ \\
\hline
 & 0 &  & $+ \infty$ \\
$f$ &  &  & $+ 27 \mathrm { e } ^ { - 3 }$ \\
\end{tabular}
\end{center}

c. Prove, by induction, that for all natural integer $n$, we have:

$$- 1 \leqslant u _ { n } \leqslant u _ { n + 1 } \leqslant 0$$

d. Deduce that the sequence $(u _ { n })$ is convergent.\\
e. We denote $\ell$ the limit of the sequence $(u _ { n })$.

Recall that $\ell$ is a solution of the equation $f ( x ) = x$.\\
Determine $\ell$. (For this, it will be admitted that the equation $x ^ { 2 } \mathrm { e } ^ { x } - 1 = 0$ has only one solution in $\mathbb { R }$ and that this solution is strictly greater than $\frac { 1 } { 2 }$).
\end{enumerate}