$\displaystyle \frac{x^3-3x^2+1}{x^3}$ Would the domain of this function be y defined for all x and for all x not equal to 0
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All x not equal to 0! right you are sir! And do you know why?
Last edited by VonNemo19; May 17th 2009 at 01:01 PM. Reason: forgot something
Originally Posted by VonNemo19 All x not equal to 0! right you are sir! And do you know why? 0 can't be used with an exponent?
Originally Posted by anon_404 0 can't be used with and exponent? That's only true for exponents < 0 What's $\displaystyle \frac 1{0}$?
Originally Posted by derfleurer That's only true for exponents < 0 What's $\displaystyle \frac 1{0}$? 0?
If $\displaystyle \frac {10}5 = 2$ than working backwards $\displaystyle 2 * 5 = 10$ So if $\displaystyle \frac 10 = 0$ than how is it that $\displaystyle 0 * 0 = 1$?
not possible to /0?
In basic arithmetic division by zero is meaningless. So for a function like $\displaystyle y = \frac 1x$, we simply say that @ x = 0, y is undefined. This is also the case in your original equation.
How can you divide by nothing? See what I'm driving at?
i think the domain would be x
Yes, the intent was for what values of our domain is our range defined.
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