If f is a real-valued function from A onto B | vedclass

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Question

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

Vedclass · Accepted Answer

$\phi$

Vedclass · Answer

(A) Given $f(x) = \frac{1}{\sqrt{|x - |x||}}$.For the function to be defined,the expression inside the square root must be strictly positive: $|x - |x|| > 0$.If $x \ge 0$,then $|x| = x$,so $|x - x| = 0$,which is not $> 0$.If $x  0$.Thus,the domain $A = (-\infty, 0)$.For $x \in (-\infty, 0)$,$f(x) = \frac{1}{\sqrt{-2x}}$. As $x$ ranges from $(-\infty, 0)$,$-2x$ ranges from $(0, \infty)$,and $\sqrt{-2x}$ ranges from $(0, \infty)$.Therefore,the range $B = (0, \infty)$.Finally,$A \cap B = (-\infty, 0) \cap (0, \infty) = \phi$.

If $f$ is a real-valued function from $A$ onto $B$ defined by $f(x) = \frac{1}{\sqrt{|x - |x||}}$,then $A \cap B = $

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