#f(x) = sqrt(x^2+9/x^2+3)#
First note that if #x=0# then the denominator of #9/x^2# is zero, resulting in an undefined value. So #x=0# is not part of the domain.
For any #x != 0#, we have #x^2 > 0#, #9/x^2 > 0# and #3 > 0#, hence the radicand #x^2+9/x^2+3 > 0#, resulting in a well defined Real square root.
So the domain is #RR "\" { 0 } = (-oo, 0) uu (0, oo)#
To find the domain, let #y = f(x)# and solve for #x#, then analyse what values of #y# give solutions:
#y = sqrt(x^2+9/x^2+3)#
Square both sides:
#y^2 = x^2+9/x^2+3#
Let #t = x^2# (noting that we require #t >= 0#
#y^2 = t+9/t+3#
Subtract #y^2# from both ends to get:
#0 = t+9/t+(3-y^2)#
Multiply through by #t# and transpose to get:
#t^2+(3-y^2)t+9 = 0#
This is in the form #at^2+bt+c = 0# with #a=1#, #b=(3-y^2)# and #c=9#. It has discriminant #Delta# given by the formula:
#Delta = b^2-4ac = (3-y^2)^2-4(1)(9) = (3-y^2)^2-6^2#
So for our quadratic in #t# to have Real solutions (let alone positive ones), we require:
#(3-y^2)^2 >= 6^2#
Hence:
#3-y^2 <= -6#
Hence #y^2 >= 9#
Hence #y >= 3# (since we require #y > 0#)
If #y = 3# then our quadratic in #t# becomes:
#0 = t^2-6t+9 = (t-3)^2#
Hence #x^2 = t = 3# and #x = +-sqrt(3)#
Hence we can deduce that the range of #f(x)# is #[3, oo)#
graph{(y-sqrt(x^2+9/x^2+3))(y-3) = 0 [-10.12, 9.88, -1.56, 8.44]}
