[tex]f'(x)=(\frac{x+2}{x-1})'=-\frac{3}{(x-1)^2}\\
\text{ Trebuie determinat } x_0 \text{ pentru care}\\ f'(x_0)=m_d=-3\\
-\frac{3}{(x_0-1)^2}=-3 \Rightarrow (x_0-1)^2=1\Rightarrow x_0=0\vee x_0=2[/tex]
In concluzie ar fi 2 astefel de puncte A(0,-2), B(2,4)
