The domain and range of f(x) = frac{1}{sqrt{| | vedclass

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(C) For the domain $A$, we require $|x| - x^2 > 0$. Since $|x|^2 = x^2$, this is $|x| - |x|^2 > 0$, which implies $|x|(1 - |x|) > 0$. This holds when

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$(-1, 0) \cup (0, 1) \cup [1, \infty)$

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(C) For the domain $A$, we require $|x| - x^2 > 0$. Since $|x|^2 = x^2$, this is $|x| - |x|^2 > 0$, which implies $|x|(1 - |x|) > 0$. This holds when $0 For the range $B$, let $y = \frac{1}{\sqrt{|x| - x^2}}$. As $x 	o 0$, $|x| - x^2 	o 0^+$, so $y 	o \infty$. As $|x| 	o 1$, $|x| - x^2 	o 0^+$, so $y 	o \infty$. The maximum value of $|x| - x^2$ occurs at $|x| = 1/2$, giving $1/2 - 1/4 = 1/4$. The minimum value of the denominator is $0$ (exclusive) and the maximum value is $1/4$. Thus, $\sqrt{|x| - x^2} \in (0, 1/2]$. Therefore, $y \in [2, \infty)$, so $B = [2, \infty)$. Finally, $A \cup B = ((-1, 0) \cup (0, 1)) \cup [2, \infty)$.

The domain and range of $f(x) = \frac{1}{\sqrt{|x| - x^2}}$ are $A$ and $B$ respectively. Then $A \cup B =$

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