Найдите область определения функции у=√х+1 + √3х-х²

[latex]y= \sqrt{x+1}+ \sqrt{3x-x^2}\\ \left \{ {{x+1 \geq 0} \atop {3x-x^2 \geq 0}} \right.\\ \left \{ {{x \geq -1} \atop {x(3-x) \geq 0}} \right. [/latex] _____________-1 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\               -                               +                          - _________________0\\\\\\\\\\\\\\\\\\3____________ x∈[0;3] - область определения функции