Is (0,0) a Point on the Graph of y=x^-1?

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Discussion Overview

The discussion revolves around whether the point (0,0) is on the graph of the function y=x^-1, particularly focusing on the implications of defining the function at x=0 and the rules of powers involved in the simplification of the expression.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions if x=0 is undefined in the context of the function y=\frac{1}{x^{-1}}, expressing uncertainty about the existence of the point (0,0).
  • Another participant asserts that the function is not defined at x=0, explaining that simplification to y=x relies on multiplying by x/x, which is undefined at x=0.
  • A different viewpoint suggests that using the rule of powers could lead to treating the function differently, proposing that \frac{1}{x^{-1}} can be interpreted as x, but acknowledges the complexity of when this is defined.
  • One participant emphasizes that the rule used to simplify explicitly requires x to be non-zero.

Areas of Agreement / Disagreement

Participants express disagreement regarding the treatment of the function at x=0, with no consensus reached on whether (0,0) can be considered a point on the graph.

Contextual Notes

Participants highlight the dependence on definitions and rules of powers, particularly regarding the undefined nature of expressions at x=0.

Mentallic
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If I'm asked to graph this function: y=\frac{1}{x^{-1}}

Is x=0 undefined? Obviously by the rule of powers, this equation is the same as y=x, but I'm unsure if the point (0,0) exists in this equation or not.
 
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That function is not defined at x=0. To simplify it to x, you rely on the fact that you can multiply by 1=x/x. But x/x isn't defined when x=0, so you can't use simplification to get around the undefinedness at 0.
 
Yes, if I converted the power to a fraction as so: \frac{1}{\frac{1}{x}} then I'd be relying on that rule, but what about if I used the rule of powers, i.e. \frac{1}{x^a}=x^{-a} So simply, \frac{1}{x^{-1}}=x^{-(-1)}=x

It just seems to me that only sometimes this is undefined, depending on how you treat the problem.

Sort of like \sqrt{x^2}=|x| while (\sqrt{x})^2=x and defined for only x\geq 0
 
That rule explicitly requires x\ne0.
 
Ahh yes, of course!

Thanks tinyboss :smile:
 

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