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

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(C) The function is defined as $f(x) = \frac{1}{\sqrt{x+|x|}}$.For the function to be defined,the expression inside the square root must be strictly g

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$(0, \infty)$

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(C) The function is defined as $f(x) = \frac{1}{\sqrt{x+|x|}}$.For the function to be defined,the expression inside the square root must be strictly greater than zero:$x + |x| > 0$.Case $1$: If $x > 0$,then $|x| = x$. Substituting this,we get $x + x = 2x > 0$,which implies $x > 0$.Case $2$: If $x Case $3$: If $x = 0$,then $x + |x| = 0 + 0 = 0$. Since the denominator cannot be zero,$x = 0$ is not in the domain.Therefore,the domain of the function is $(0, \infty)$.

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

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