What is the answer to iii) ??
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:\(\boxed{x\in~]-\infty,4]}\)
