If you were given:$$f(x) = \frac{1}{\frac{1}{x}}$$ would the domain be (-∞, ∞) or (-∞, 0)U(0, ∞)? This is a question for all such examples where you must multiply by the reciprocal in some way.

Question

anonymous · Answer

I think it would be the latter, $(-\infty,0)\cup(0,\infty)$. With the way the function is defined, $x=0$ yields an indeterminate result.

This despite $\dfrac{1}{\frac{1}{x}}=x$ for $x\not=0$.

It's like with the following function:
$$f(x)=\frac{x}{x}$$
For nonzero $x$, $f(x)=1$. The plot of $f(x)$ would be a horizontal line with a hole at $x=0$.

For your function, the plot would be that of $f(x)=x$ with a hole at $x=0$, I think.

anonymous · Answer

Ah. That makes sense. I was always wondered whether you simplified algebraically first or if you took the domain of the original. Same concept of removable discontinuities and such. Thank you :)

anonymous · Answer

For confirmation, WolframAlpha agrees: http://www.wolframalpha.com/input/?i=1%2F%281%2Fx%29 You're welcome.