Question 696354
You cannot take the square root of a -ve number, so {{{ x+5 >= 0 }}}
Hence, {{{ x >= -5 }}}


So the range is {{{ x >= -5 }}}



{{{ sqrt(x+5) }}} is smallest when {{{ x+5 = 0 }}}, ie {{{ x=-5 }}}, for all other values of x (subject to {{{ x >= -5 }}} ) then
{{{ x+5 > 0 }}} so {{{ sqrt(x+5)>0 }}}

so then {{{ sqrt(x+5)+3 }}} has a minimum values of 3 when {{{ x=-5 }}}


so the domain is {{{ f(x) >=3 }}} or if you prefer {{{ 3 <= f(x) <= infinity }}}