In this exercise, the plane is equipped with an orthonormal coordinate system.

The curve with equation is represented below:
$$y = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } - 2 \right) .$$
This curve is called a ``catenary''.

We are interested here in ``catenary arcs'' delimited by two points of this curve that are symmetric with respect to the $y$-axis. Such an arc is represented on the graph below in solid line. We define the ``width'' and ``height'' of the catenary arc delimited by the points $M$ and $M^{\prime}$ as indicated on the graph.

The purpose of the exercise is to study the possible positions on the curve of the point $M$ with strictly positive abscissa so that the width of the catenary arc is equal to its height.

\begin{enumerate}
  \item Justify that the problem studied reduces to finding the strictly positive solutions of the equation
$$( E ) : \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } - 4 x - 2 = 0$$

  \item Let $f$ be the function defined on the interval $[ 0 ; + \infty [$ by:
$$f ( x ) = \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } - 4 x - 2 .$$
a. Verify that for all $x > 0 , f ( x ) = x \left( \frac { \mathrm { e } ^ { x } } { x } - 4 \right) + \mathrm { e } ^ { - x } - 2$.\\
b. Determine $\lim _ { x \rightarrow + \infty } f ( x )$.

  \item a. Let $f ^ { \prime }$ denote the derivative function of $f$. Calculate $f ^ { \prime } ( x )$, where $x$ belongs to the interval $[ 0 ; + \infty [$.\\
b. Show that the equation $f ^ { \prime } ( x ) = 0$ is equivalent to the equation: $\left( \mathrm { e } ^ { x } \right) ^ { 2 } - 4 \mathrm { e } ^ { x } - 1 = 0$.\\
c. By setting $X = \mathrm { e } ^ { x }$, show that the equation $f ^ { \prime } ( x ) = 0$ has as its unique real solution the number $\ln ( 2 + \sqrt { 5 } )$.

  \item The sign table of the derivative function $f ^ { \prime }$ of $f$ is given below:
\begin{center}
\begin{tabular}{ | c | c c c c | }
\hline
$x$ & 0 &  & $\ln ( 2 + \sqrt { 5 } )$ & $+ \infty$ \\
\hline
$f ^ { \prime } ( x )$ & - & 0 & + &  \\
\hline
\end{tabular}
\end{center}
a. Draw up the variation table of the function $f$.\\
b. Prove that the equation $f ( x ) = 0$ has a unique strictly positive solution which we denote by $\alpha$.

  \item Consider the following algorithm where the variables $a$, $b$ and $m$ are real numbers:
\begin{verbatim}
While $b - a > 0.1$ do:
    $m \leftarrow \frac { a + b } { 2 }$
    If $\mathrm { e } ^ { m } + \mathrm { e } ^ { - m } - 4 m - 2 > 0$, then:
        $b \leftarrow m$
    Else:
        $a \leftarrow m$
    End If
End While
\end{verbatim}
a. Before execution of this algorithm, the variables $a$ and $b$ contain respectively the values 2 and 3.\\
What do they contain at the end of the algorithm execution?\\
Justify the answer by reproducing and completing the table opposite with the different values taken by the variables at each step of the algorithm.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$m$ & $a$ & $b$ & $b - a$ \\
\hline
 & 2 & 3 & 1 \\
\hline
2.5 &  &  &  \\
\hline
$\ldots$ & $\ldots$ & $\ldots$ &  \\
\hline
 &  &  &  \\
\hline
\end{tabular}
\end{center}
b. How can we use the values obtained at the end of the algorithm in the previous question?

  \item The width of the Gateway Arch arc, expressed in metres, is equal to twice the strictly positive solution of the equation:
$$\left( E ^ { \prime } \right) : \mathrm { e } ^ { \frac { t } { 39 } } + \mathrm { e } ^ { - \frac { t } { 39 } } - 4 \frac { t } { 39 } - 2 = 0$$
Give a bound for the height of the Gateway Arch.
\end{enumerate}