Let
x--------> the length side of the square base
h--------> the height of the box
we know that
the volume of the box is equal to
[tex]V=x^{2} *h\\ V=9\ m^{3}[/tex]
so
[tex]x^{2} *h=9\\\\h=\frac{9}{x^{2} }[/tex]
the surface area of the box is equal to
[tex]SA=area\ of\ the\ base+perimeter\ of\ base*height[/tex] (remember that the box is open)
area of the base=[tex](x^{2})\ m^{2}[/tex]
Perimeter of the base=[tex](4*x)\ m[/tex]
height=(h) m
[tex]h=\frac{9}{x^{2}}[/tex]
substitute
[tex]SA=x^{2} +4*x*h\\ \\ \\ SA=x^{2} +4*x*\frac{9}{x^{2} } \\ \\ SA=x^{2} +\frac{36}{x}[/tex]
we know that
the value of x can not be negative and the denominator can not be zero
therefore
the answer is
the domain of SA is x> 0
the domain is the interval-------------> (0,∞)