
Kenneth S. answered 03/14/16
Tutor
4.8
(62)
Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018
I shall assume that the function with which you are dealing is y = 1/(x+3) + 3. Its domain is all Reals, except x ≠ -3, because that value would mean an attempt at division by zero, which is undefined.
The graph of y can never equal 3, so its range is all Reals except 3 because, over its domain, the term 1/(x+3) will always be something that causes the y value to be 3 ± some non-zero value.
(Another method of showing this involves finding its inverse; the inverse will have x ≠ 3, and since domain & range of inverse become range & inverse of y, this shows that the range of y is all Reals except 3.)