Răspuns:
[tex]\overline{S}=\begin{cases}0, m > \frac{1}{e}\\ 1, m = \frac{1}{e}\\ 2, 0<m<\frac{1}{e}\\ 1, m \leq 0\end{cases}[/tex]
Explicație pas cu pas:
[tex]\displaystyle \textrm{Fie functia} f : (-1,+\infty)\to \mathbb{R}, f(x) = ln(x+1)\\\textrm{Fie } x \textrm{ punctul pentru care } f(x) \textrm{ este tangenta la functia } m(x+1), m > 0\\ \implies \textrm{Avem urmatorul sistem: }\\\\ \begin{cases}-1 = x - \frac{f(x)}{f'(x)}\\ f'(x) = [m(x+1)]'\end{cases} \\\\ \iff \begin{cases}x+1 = \frac{ln(x+1)}{\frac{1}{x+1}}\\ \frac{1}{x+1} = m\end{cases} \\ \\ \iff \begin{cases}x+1 = (x+1)ln(x+1)\\ \frac{1}{x+1}=m\end{cases}\\\\ \iff\begin{cases}ln(x+1) = 1\\ \frac{1}{x+1} = m\end{cases}\\\\ \iff\begin{cases}x = e-1\\ \frac{1}{x+1} = m\end{cases}\\\\ \implies x = e-1, m = \frac{1}{e}\\ \\ \textrm{Cum } \lim\limits_{x\to\infty}m(x+1) > \lim\limits_{x\to\infty}f(x)\iff m > 0\\ \\ \implies \overline{S}=\begin{cases}0, m > \frac{1}{e}\\ 1, m = \frac{1}{e}\\ 2, 0<m<\frac{1}{e}\\ 1, m \leq 0\end{cases}[/tex]
