Q.The range of the function $f: R-\{3\}\to R$ defined by $f(x) = \dfrac{|x-3|}{(x-3)}$ is:

$|x|\ge 0$ for all real $x$ and covers every non-negative value, so $f(x)=|x|-5$ covers every value $\ge -5$, giving range $[-5,\infty)$.

For $f:\mathbb{R}\to\mathbb{R}$, $f(x)=|x|-5$.

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$x$ can be any real number, but $x^2$ can never be negative, so the range is restricted while the domain is not.

Domain: $f(x) = x^2$ is defined for every real input, so the domain is all of $\mathbb{R}$.

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Splitting by cases on the sign of $x-3$ shows $f(x)$ only ever equals $1$ or $-1$, so the range is $\{1,-1\}$.

For $x>3$: $x-3>0$, so $|x-3|=x-3$, giving $f(x) = \frac{x-3}{x-3} = 1$.

For $x<3$: $x-3<0$, so $|x-3|=-(x-3)=3-x$, giving $f(x) = \frac{3-x}{x-3} = \frac{-(x-3)}{x-3} = -1$.

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$|x|-x>0$ only when $x<0$ (it equals $0$ for $x\ge0$), so the domain is $(-\infty,0)$.

For $f(x)=\dfrac{1}{\sqrt{|x|-x}}$ to be defined (real and nonzero denominator), we need $|x|-x>0$.

Case $x\ge0$: $|x|=x$, so $|x|-x=0$, which is not $>0$. Excluded.

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