[tex]f(x)=\dfrac3{x+2}-\sqrt{x-3}[/tex]
Two immediate observations:
(1) Clearly, we can't have [tex]x=-2[/tex], since that would make both the rational and the square root terms undefined.
(2) If [tex]x[/tex] is real, then [tex]\sqrt x[/tex] only exists if [tex]x\ge0[/tex]. This means that for [tex]\sqrt{x-3}[/tex] to exist, we require [tex]x-3\ge0\implies x\ge3[/tex].
If [tex]x\ge3[/tex], then we don't have to worry about [tex]x[/tex] ever taking on a value of -2.
