Restrictions of denominators with negative exponents.

[sickle]

what are the restrictions (if any) on y=1/x^(-2)?
is it equal to just x^2 so no restrictions or what
it is have hole at x=0?

or do you change too all positive exponents before looking at restrictions?

what exactly are the exponents rules that govern this stuff

so confused

 

[Stephen Tashi]

The function is undefined at x = 0, so it has a hole in its graph there.

People apply the rules for simplifying expression mechanically without thinking, but what you are supposed to do is think about restrictions that would make the simplification invalid. You're supposed to say (1/x)^-2 = 1/ ((1/x)^2) PROVIDED 1/x is a real number to begin with. You make a mental note that the simplification is invalid if x = 0.

I really like it when manipulating symbols does away with having to do detailed thinking, but unfortunately, it never is able to completely do away with the need to think in words.

 

[JThompson]

Yes, $\frac{1}{x^{-2}}=x^{2}$. By restrictions, do you mean domain restrictions? No it does not have a hole at x=0, since $x^{2}$ does not. In general it is a good idea to convert the denominator to non-negative exponents before analyzing where the equation is defined, but this is mostly because negative exponents in the denominator are stylistically confusing. If you didn't convert, the analysis would go something like...
$\frac{1}{x^{-2}}$ is defined everywhere that $x^{-2}\neq 0$, but $x^{-2} = \frac{1}{x}$ which is never equal to zero, so $\frac{1}{x^{-2}}$ is defined everywhere.

That's a fine analysis, but it is more work than just looking at $x^{2}$.
What sort of exponential rules are you wondering about? Those that govern negative exponents, or exponentials in general?

 

[HallsofIvy]

I disagree. $1/x^{-2}$ is equal to $x^2$ for all x except 0, just as $(x^2- 4)/(x+ 2)= x- 2$ for all x except 2.

The formula given, $1/x^{-2}$ is NOT defined at x= 0 because you cannot calculate $x^{-2}$ if x= 0.

 

[sickle]

Yes, $\frac{1}{x^{-2}}=x^{2}$. By restrictions, do you mean domain restrictions? No it does not have a hole at x=0, since $x^{2}$ does not. In general it is a good idea to convert the denominator to non-negative exponents before analyzing where the equation is defined, but this is mostly because negative exponents in the denominator are stylistically confusing. If you didn't convert, the analysis would go something like...
$\frac{1}{x^{-2}}$ is defined everywhere that $x^{-2}\neq 0$, but $x^{-2} = \frac{1}{x}$ which is never equal to zero, so $\frac{1}{x^{-2}}$ is defined everywhere.

That's a fine analysis, but it is more work than just looking at $x^{2}$.
What sort of exponential rules are you wondering about? Those that govern negative exponents, or exponentials in general?

yeah the exponent rules that we learn in elementary rules (but those were for numbers, not functions), so i was wondering when you can use them and if they ever cause any restrictions.

 

[JThompson]

I disagree. $1/x^{-2}$ is equal to $x^2$ for all x except 0, just as $(x^2- 4)/(x+ 2)= x- 2$ for all x except 2.

The formula given, $1/x^{-2}$ is NOT defined at x= 0 because you cannot calculate $x^{-2}$ if x= 0.

I was taught to remove negative exponents when determining the domain of a function, but it is entirely possible my teacher was incorrect.
If this were a composite function, say $f(x)=x^{-1}$ and $g(x)=\frac{1}{x^2}$, then the domain of $g(f(x))=\frac{1}{(x^{-1})^{2}}$ would definitely be $x\neq 0$, since the domain restriction on $f(x)$ is retained. But composite functions are a separate issue.

However I acknowledge that your experience with the subject is greater than mine, and so I defer to you.