Respuesta :

arthurpdc
Since there is an expression in the square root, it must be non-negative. So:

[tex]f(x)=\sqrt{12-3x}\\\\
\Longrightarrow 12-3x\geq0\\\\
3x\leq12\\\\
x\leq\dfrac{12}{3}\\\\
x\leq4[/tex]

Then, the interval is:

[tex]
\boxed{x\in~]-\infty,4]}[/tex]