b)
[tex]f'(x) = \displaystyle\frac{4x^2-4x-2}{(2x-1)^2}\\f'(x) = 0\Rightarrow 4x^2-4x-2 = 0\Leftrightarrow 2x^2 - 2x - 1 = 0\\\Delta = 4+8 = 12\Rightarrow\sqrt\Delta = 2\sqrt3\\\displaystyle x_{1,2} = \frac{2\pm2\sqrt3}{4} = \frac{1 \pm \sqrt3}{2} \in \left(\frac12,\infty\right)\Rightarrow x = \frac{1 + \sqrt3}{2}[/tex]
Am făcut mai jos tabelul. Din tabel, rezultă că:
[tex]f\,s \nearrow \text{ pe } \displaystyle\left(\frac{1+\sqrt3}2,\infty\right)\\\text{Cum } (3,\infty) \subset \displaystyle\left(\frac{1+\sqrt3}2,\infty\right)\Rightarrow f\, s \nearrow \text{ pe } (3, \infty).[/tex]
c)
[tex]\text{Fie } A(x_A, y_A) \in G_f \text{ \c si fie dreapta } t \text{ tangenta \^in punctul } A \text{ la } G_f.\\t || Ox\Leftrightarrow m_t = m_{Ox}\\Ox\text{ are ecua\c tia: } y = 0\Rightarrow m_{Ox} = 0\\t \text{ are ecua\c tia:} t\,:\, y - y_A = f'(x_A)(x - x_A)\Rightarrow m_t = f'(x_A)\\\\\Rightarrow f'(x_A) = 0 \overset{\text{conform b)}}{\Rightarrow} \displaystyle x_A = \frac{1+\sqrt3}2[/tex]
[tex]y_A = f(x_A) = \displaystyle \frac{2\left(\frac{1+\sqrt3}2\right)^2 + 1}{2\frac{1+\sqrt3}2 - 1} = \frac{2\cdot\frac{1 + 2\sqrt3 + 3}{4}+1}{1+\sqrt3-1}\\\\= \frac{2 + \sqrt3 + 1}{\sqrt3} = \frac{3 + \sqrt3}{\sqrt3} = \frac{3\sqrt3 + 3}{3} = 1 + \sqrt3 \\\Rightarrow A\left(\frac{1+\sqrt3}2,\, 1+\sqrt 3\right)[/tex]
