For f(x)=sin left(frac{1}{|x| sqrt{x^2-1}}rig | vedclass

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(B) Given function $f(x) = \sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)$.For the domain,the expression inside the square root must be strictly positiv

Vedclass · Accepted Answer

$R-[-1, 1]$ and $[-1, 1]$,respectively

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(B) Given function $f(x) = \sin \left(\frac{1}{|x| \sqrt{x^2-1}}ight)$.For the domain,the expression inside the square root must be strictly positive: $x^2 - 1 > 0$,which implies $x^2 > 1$,so $x \in (-\infty, -1) \cup (1, \infty)$,or $x \in R - [-1, 1]$.Also,the denominator must not be zero: $|x| \sqrt{x^2 - 1} 
eq 0$,which is satisfied for all $x$ in the domain $R - [-1, 1]$.Thus,the domain is $R - [-1, 1]$.Since the argument $	heta = \frac{1}{|x| \sqrt{x^2-1}}$ can take any value in $(0, \infty)$ as $x$ varies in its domain,the function $\sin(	heta)$ will oscillate between $-1$ and $1$.Therefore,the range is $[-1, 1]$.

For $f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)$,the domain and range of $f(x)$ in $R$ are:

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