Verified Answer :

arthurpdc

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

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

Then, the interval is:

\(
\boxed{x\in~]-\infty,4]}\)

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