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y  =   x

       _____

      √[x - 3]

This function will not exist   on  ( -infinity , 3 ]

When x  approaches  3  from the right, y approaches infinity

Using Calculus, we can find the x value where the minimum occurs

y '  = (x - 3)^(-1/2)  + (-1/2)x(x - 3)^(-3/2)

Set this to 0

(x - 3)^(-1/2)  + (-1/2) x (x - 3)^(-3/2)  = 0       factor

(x - 3)^(-3/2)  [ (x - 3) - (1/2)x)  ]  = 0

(x - 3) - (1/2)x  =  0

(1/2)x -  3  =  0

(1/2)x  =  3

x  = 6

So....the minimum occurs at 

6 / √[6 - 3 ]   =   6 / √3

So.....the range  is    [ 6 / √3, infinity   )