Se rezolva ecuatia:
[tex]\frac{x+1}{x^2+x+1}= y \\\\\\\ y(x^2+x+1)=x+1 \\\\ yx^2+yx+y-x-1=0 \\\\ yx^2+x(y-1)+y-1=0 \\\\ x \in R \ \ \ ===> \ \ \ \ \Delta \geq 0 \\\\\\ \Delta= (y-1)^2-4y(y-1) \\\\=(y-1)[(y-1)-4y] \\\\ = (y-1)(-3y-1) \\\\ = -3y^2-y+3y+1 \\\\ \underline{\Delta = -3y^2+2y+1}[/tex]
Se va rezolva inecuatia unde vom avea nevoie de a calcula un nou discriminant (Δ') pentru:
[tex]-3y^2+2y+1 \geq 0 \\\\\\ \Delta'= 4-4*(-3)*1= 16 \\\\ y_1=\frac{-2+4}{-6} = \frac{2}{-6} \to -\frac{1}{3} \\\\ y_2= \frac{-2-4}{-6}= \frac{-6}{-6} \to 1[/tex]
Se realizeaza un tabel de semn pentru inecuatia data:
y | -∞ -1/3 1 +∞
-3y²+2y+1 |--------------O+++++++++O-----------
Avem nevoie doar de valorile pozitive, deci:
[tex]y \in [-\frac{1}{3};1] \\\\\\ \hbox{ Imaginea functiei date este \boxed{Im(f)=[-\frac{1}{3};1]}}[/tex]
