\textbf{Exercise 3 — 5 points}\\
Theme: function study\\
Parts A and B can be treated independently

\textbf{Part A}

The plane is equipped with an orthogonal coordinate system.\\
Below is represented the curve of a function $f$ defined and twice differentiable on $\mathbb{R}$, as well as that of its derivative $f'$ and its second derivative $f''$.

\begin{enumerate}
  \item Determine, by justifying your choice, which curve corresponds to which function.
  \item Determine, with the precision allowed by the graph, the slope of the tangent line to curve $\mathscr{C}_{2}$ at the point with abscissa 4.
  \item Give, with the precision allowed by the graph, the abscissa of each inflection point of curve $\mathscr{C}_{1}$.
\end{enumerate}

\textbf{Part B}

Let $k$ be a strictly positive real number.\\
We consider the function $g$ defined on $\mathbb{R}$ by:
$$g(x) = \frac{4}{1 + \mathrm{e}^{-kx}}$$

\begin{enumerate}
  \item Determine the limits of $g$ at $+\infty$ and at $-\infty$.
  \item Prove that $g'(0) = k$.
  \item By admitting the result below obtained with computer algebra software, prove that the curve of $g$ has an inflection point at the point with abscissa 0.
\end{enumerate}

\begin{center}
\begin{tabular}{ | l l | }
\hline
$\triangleright$ & Computer algebra \\
\hline
 & $g(x) = 4 / (1 + \mathrm{e}^{\wedge}(-kx))$ \\
1 &  \\
 & $\rightarrow g(x) = \frac{4}{\mathrm{e}^{-kx} + 1}$ \\
\hline
 & Simplify $(g''(x))$ \\
2 &  \\
 & $\rightarrow g''(x) = -4\mathrm{e}^{kx}(\mathrm{e}^{kx} - 1)\frac{k^{2}}{(\mathrm{e}^{kx} + 1)^{3}}$ \\
\hline
\end{tabular}
\end{center}