Equation A————> $y=x$ , It has domain of Real Numbers.
Equation B————> $y=\frac{1}{(1/x)}$, It has domain of $R-${$0$}.
Context: I solved $y=\frac{1}{ \sqrt{ \frac{x-1}{x-2}} }$ and calculated domain $(- \infty ,1) \cup(2,\infty).$ I wonder the domain will change if the expression is simplified i.e.(x-2) taken to numerator, so the question is can we simplify the given expression first and then find the domain or we are not allowed to alter the expression in question while finding the domain?
Graph from wolfram for $y=\frac{1}{ \sqrt{ \frac{x-1}{x-2}} }$
https://www.wolframalpha.com/input/?i=f%28x%29%3Dsqrt%281%2F%28%28x-1%29%2F%28x-2%29%29%29
Graph from wolfram for $ y=\sqrt{ \frac{x-2}{x-1}}$
https://www.wolframalpha.com/input/?i=sqrt%28%28x-2%29%2F%28x-1%29%29
Best Answer
No, you can't. When you study a function defined as follows: $$f(x)=\frac{1}{\sqrt{g(x)}}$$ you have first to study when the denominator is defined or in other words when: $$g(x)\neq 0$$ and then when the square root is defined: $$g(x)\geq0$$ At this point, you can bind the two conditions into a system and find the domain of $f(x)$.
For example the domain of $f(x)=\sqrt{ \frac{x-2}{x-1}}$ is $x<1 \vee x\geq 2$ and this $f(x)$ can be obtained by $g(x)=\frac{1}{f(x)}$. If you are given $g(x)$ you can notice that the domain of $g(x)$ doesn't include $x=2$ which is a part of the domain of $f(x)$.