For x in mathbb{R},if f(x) = sqrt{log_{10}lef | vedclass

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(B) For the function $f(x) = \sqrt{\log_{10}\left(\frac{3-x}{x}\right)}$ to be defined,the expression inside the square root must be non-negative and

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$\left(0, \frac{3}{2}\right]$

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(B) For the function $f(x) = \sqrt{\log_{10}\left(\frac{3-x}{x}\right)}$ to be defined,the expression inside the square root must be non-negative and the argument of the logarithm must be positive.$1$. Condition for the logarithm argument: $\frac{3-x}{x} > 0$.Solving this inequality,we get $x \in (0, 3)$.$2$. Condition for the square root: $\log_{10}\left(\frac{3-x}{x}\right) \geq 0$.This implies $\frac{3-x}{x} \geq 10^0$,so $\frac{3-x}{x} \geq 1$.$\frac{3-x}{x} - 1 \geq 0 \implies \frac{3-x-x}{x} \geq 0 \implies \frac{3-2x}{x} \geq 0$.Multiplying by $-1$ (and reversing the inequality),we get $\frac{2x-3}{x} \leq 0$.The critical points are $x = 0$ and $x = \frac{3}{2}$. Testing intervals,we find $x \in (0, \frac{3}{2}]$.Combining both conditions: $(0, 3) \cap (0, \frac{3}{2}] = (0, \frac{3}{2}]$.Thus,the domain of $f$ is $\left(0, \frac{3}{2}\right]$.

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y = f(x) = x^2$

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

For $x \in \mathbb{R}$,if $f(x) = \sqrt{\log_{10}\left(\frac{3-x}{x}\right)}$,then the domain of $f$ is

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