The set of all real values of x for which f(x | vedclass

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(A) For the function $f(x) = \sqrt{\frac{|x|-2}{|x|-3}}$ to be well-defined,the expression inside the square root must be non-negative and the denomin

Vedclass · Accepted Answer

$(-\infty, -3) \cup [-2, 2] \cup (3, \infty)$

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(A) For the function $f(x) = \sqrt{\frac{|x|-2}{|x|-3}}$ to be well-defined,the expression inside the square root must be non-negative and the denominator must not be zero.So,$\frac{|x|-2}{|x|-3} \geq 0$ and $|x|-3 
eq 0$.Let $t = |x|$,where $t \geq 0$. The inequality becomes $\frac{t-2}{t-3} \geq 0$.The critical points are $t=2$ and $t=3$.Testing intervals for $t$: $1$) If $0 \leq t  0$ (e.g.,$t=1$,$\frac{-1}{-2} = 0.5 > 0$).$2$) If $2 \leq t $3$) If $t > 3$,$\frac{t-2}{t-3} > 0$ (e.g.,$t=4$,$\frac{2}{1} = 2 > 0$).Thus,the solution for $t$ is $0 \leq t \leq 2$ or $t > 3$.Substituting back $|x| = t$,we get $0 \leq |x| \leq 2$ or $|x| > 3$.$|x| \leq 2 \implies x \in [-2, 2]$.$|x| > 3 \implies x \in (-\infty, -3) \cup (3, \infty)$.Combining these,the domain is $(-\infty, -3) \cup [-2, 2] \cup (3, \infty)$.

The set of all real values of $x$ for which $f(x)=\sqrt{\frac{|x|-2}{|x|-3}}$ is a well-defined function is

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