If f(x)=(x+1)^−1 and g(x)=x−2, what is the domain of f(x)÷g(x)?

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If f(x)=(x+1)^−1 and g(x)=x−2, what is the domain of f(x)÷g(x)?

Algebra
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well rewrite it as \[f(x) = \frac{1}{x + 1}\] then if you divide by g(x) its \[\frac{1}{(x + 1)} \div (x -2)~~~or ~~~\frac{1}{x +1} \times \frac{1}{x -2}\] so the domain is all real x, except there are 2 excepts.... you need to find them....
(−∞,−1] and [2,∞)?

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well you have the correct values for the exclusions.... but if you look at the domain you stated, you could have x = 0.... so I'd say all real x, except \[x \neq -1 or 2\]
i have the answer choice "all values of x" , (∞,−1),(−1,2), and (2,∞) , (−∞,2) and (2,∞) , & (−∞,−1] and [2,∞)
so looking at the problem... things don't match my thinking... as x = 1 is aslo between the values x = -1 and x = 2 substitute x = 1 and you still get a numerical value... so I'll let someone else who is smarter help you.
That's okay, I appreciate your help anyways
How about "If f(x)=2x^2−x−6 and g(x)=x^2−4, what's f(x)÷g(x)?
here is a graph of the curves...
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well for this question... its simple, start by factoring \[\frac{2x^2 - x - 6}{x^2 - 4} = \frac{(2x + 3)(x - 2)}{(x +2)(x-2)}\] and hopefully you can finish it from here.
Thanks!

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