Is (0,0) a Point on the Graph of y=x^-1?
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Tinyboss
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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.
Homework Helper
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Yes, if I converted the power to a fraction as so: [tex]\frac{1}{\frac{1}{x}}[/tex] then I'd be relying on that rule, but what about if I used the rule of powers, i.e. [tex]\frac{1}{x^a}=x^{-a}[/tex] So simply, [tex]\frac{1}{x^{-1}}=x^{-(-1)}=x[/tex]
It just seems to me that only sometimes this is undefined, depending on how you treat the problem.
Sort of like [tex]\sqrt{x^2}=|x|[/tex] while [tex](\sqrt{x})^2=x[/tex] and defined for only [itex]x\geq 0[/itex]
It just seems to me that only sometimes this is undefined, depending on how you treat the problem.
Sort of like [tex]\sqrt{x^2}=|x|[/tex] while [tex](\sqrt{x})^2=x[/tex] and defined for only [itex]x\geq 0[/itex]
Tinyboss
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That rule explicitly requires [tex]x\ne0[/tex].
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