Question

What is the answer to iii) ??

Answer 1

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:

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