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December 21, 2014

December 21, 2014

Posted by **Meron** on Thursday, November 10, 2011 at 8:00am.

- Math -
**Reiny**, Thursday, November 10, 2011 at 8:42amIf you meant

(x^3+1)/(x^2-1)

then if x = -1, your expression becomes 0/0 which is "indeterminate".

notice that (x^3+1)/(x^2-1)

= (x+1)(x^2 + x + 1)/((x+1)(x-1))

= (x^2 + x + 1)/(x-1)

I don't know if you know anything about limits, but

the limit of the above , as x --> - 1

= (1 - 1 + 1)/(-2) = -1/2

There are actually no zeros for this graph,

as

http://www.wolframalpha.com/input/?i=%28x%5E3%2B1%29%2F%28x%5E2-1%29

will show you.

the expression actually reduces to

y = (x^2 + x + 1)/(x-1) , which has a vertical asymptote at x=1

the original function

y = (x^3 + 1)(/(x^2-1) looks exactly the same except it would have a "hole" or missing point at (-1, -1/2)

- Math -
**Meron**, Thursday, November 10, 2011 at 8:54amThank u so much!

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