[tex]f(x)= \frac{x+1}{x^2+2} =y \\ \\ x+1=y(x^2+2) \\ \\ x+1-y*x^2-2y=0 \\ \\ y*x^2-x+2y-1=0 \\ \\ DELTA\ \textgreater \ 0 \\ \\ DELTA=1-4y(2y-1)=1-8y^2+4y\ \textgreater \ 0 \\ \\ 8y^2-4y-1\ \textless \ 0 \\ \\ 8y^2-4y-1=0 \\ \\ DELTA=16+4*8=16+32=48 \\ \\ y1= \frac{4+4 \sqrt{3} }{16} = \frac{1+ \sqrt{3} }{4} ;y2= \frac{1- \sqrt{3} }{4} \\ \\ Se~face~tabel~iar~y~apartine~intervalului~( \frac{1- \sqrt{3} }{4} , \frac{1+ \sqrt{3} }{4} ). \\ \\ Imf=( \frac{1- \sqrt{3} }{4} , \frac{1+ \sqrt{3} }{4} ).[/tex]