Domain of f(x) depends of second term only, since the first term is a constant. The second term involves square root function in x; As square root functions are defined only for real values greater than or equal to zero, x >/= 0
Thus domain of the function is [0, inf) ==> 0 </= x < inf
Since x >/= 0, for all 'x', x + √x >/= 0
==> -(x + √x) </= 0
==> (6 + √6) - (x + √x) </= (6 + √6)
Thus range of f(x) is (-inf, 6 + √6], that is -inf < f(x) </= (6 + √6)
Answers & Comments
Verified answer
Grouping conveniently, f(x) = (6 + √6) - (x + √x)
Domain of f(x) depends of second term only, since the first term is a constant. The second term involves square root function in x; As square root functions are defined only for real values greater than or equal to zero, x >/= 0
Thus domain of the function is [0, inf) ==> 0 </= x < inf
Since x >/= 0, for all 'x', x + √x >/= 0
==> -(x + √x) </= 0
==> (6 + √6) - (x + √x) </= (6 + √6)
Thus range of f(x) is (-inf, 6 + √6], that is -inf < f(x) </= (6 + √6)