The domain of sqrt{|x|-x} is | vedclass

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(C) For the function $f(x) = \sqrt{|x|-x}$ to be defined,the expression inside the square root must be non-negative: $|x| - x \geq 0$ $|x| \geq x$ We

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$(-\infty, \infty)$

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(C) For the function $f(x) = \sqrt{|x|-x}$ to be defined,the expression inside the square root must be non-negative: $|x| - x \geq 0$ $|x| \geq x$ We know that for any real number $x$,the absolute value $|x|$ is always greater than or equal to $x$ (i.e.,$|x| \geq x$ is true for all $x \in R$). If $x \geq 0$,then $|x| = x$,so $x - x = 0 \geq 0$. If $x  0$ (since $x$ is negative). Thus,the inequality holds for all real numbers. Therefore,the domain is $(-\infty, \infty)$.

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

The domain of $\sqrt{|x|-x}$ is

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$y = f(x) = x^2$

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