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(D) For $f(x)$ to be a real-valued function,the expression inside the square root must be non-negative: $\frac{[x]-1}{[x]^2-[x]-6} \ge 0$. Let $n = [x

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$[-1, 2) \cup [4, \infty)$

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(D) For $f(x)$ to be a real-valued function,the expression inside the square root must be non-negative: $\frac{[x]-1}{[x]^2-[x]-6} \ge 0$. Let $n = [x]$. Then $\frac{n-1}{n^2-n-6} \ge 0$,which simplifies to $\frac{n-1}{(n-3)(n+2)} \ge 0$. Using the wavy curve method,the inequality holds for $n \in (-2, 1] \cup (3, \infty)$. Since $n$ is an integer $(n = [x])$,the possible values for $n$ are $\{-1, 0, 1, 4, 5, 6, \dots\}$. If $n = -1$,then $-1 \le x If $n = 0$,then $0 \le x If $n = 1$,then $1 \le x Combining these,we get $[-1, 2)$. If $n \ge 4$,then $x \ge 4$. Thus,the domain is $[-1, 2) \cup [4, \infty)$.

$A$. $f(x)=\frac{\|x+2\|}{x+2}, x \neq-2$	$1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=\|[x]\|, x \in R$	$2$. $Z$
$C$. $h(x)=\|x-[x]\|, x \in R$	$3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$	$4$. $[0, 1)$
	$5$. $\{-1, 1\}$

The set of all real values of $x$ such that $f(x) = \sqrt{\frac{[x]-1}{[x]^2-[x]-6}}$ is a real-valued function is

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