The domain of the function f(x) = frac{1}{sqr | vedclass

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(A) For the function $f(x) = \frac{1}{\sqrt{|x| - x}}$ to be defined,the expression inside the square root must be strictly greater than zero.Therefor

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$( -\infty, 0 )$

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(A) For the function $f(x) = \frac{1}{\sqrt{|x| - x}}$ to be defined,the expression inside the square root must be strictly greater than zero.Therefore,we require $|x| - x > 0$.This inequality can be rewritten as $|x| > x$.Case $1$: If $x \geq 0$,then $|x| = x$. The inequality becomes $x - x > 0$,which simplifies to $0 > 0$. This is false for all $x \geq 0$.Case $2$: If $x  0$,which simplifies to $-2x > 0$,or $x Since the condition $x < 0$ satisfies the inequality for all negative values,the domain of the function is $(-\infty, 0)$.

The domain of the function $f(x) = \frac{1}{\sqrt{|x| - x}}$ is

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