If f(x)= x+1^-1 and g(x)=x-2 , what is the domain of f(x) divided by g(x)

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If f(x)= x+1^-1 and g(x)=x-2 , what is the domain of f(x) divided by g(x)

Mathematics
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what do you think
honestly I don't know, I want to say all values of x but im not sure :(
Can you write out \(\dfrac{f(x)}{g(x)}\)

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no I don't know how sorry
can you guys help me pleaseeee
x+1^-1, So your saying the whole quantity by -1, or just "1"
\[\frac{ x+1 }{ x-2 }\]
yeah the exponent is -1
its the inverse \((x+1)^{-1}=\dfrac{1}{x+1}\)
no
The exponent is only on 1 not (x+1)
So you have \(\dfrac{(x+1)^{-1}}{x-2}=\dfrac{1}{(x+1)(x-2)}\). Now what numbers force us to divide by \(0\)?
\[Did she mean (x+1)^{-1}\]
The zeroes of (x+1)(x-2)
yes. there is no reason to raise something to 1
Also known as (x+1)=0, (x-2)=0, Solve for X on both of them @idgm_idontgetmath
x=-1,2
please let them figure something out man. you are not helping if you give thm everything....
But im trying to show him to solve zeroes??????
Not you.
@campbell_st can you help?
ok... here is what I'll do, if you are happy that \[\frac{f(x)}{g(x)} = \frac{1}{(x +1)(x-2)}\] you need to solve x + 1 = 0 and x - 2 = 0 can you do that
so here is an image of f(x)/g(x)
1 Attachment
the curve continues to the left.. to- infinity and to the right to infinity however the solutions to x + 1 = 0 and x -2 = 0 create vertical asymptotes... this is where the cruve doesn't exist but you can see that the curve exists between the asymptotes so there are 3 regions where the curve exists or can be drawn...
so the regions start with - infinity to the 1st asymptote between the asymptotes and the right asymptote to infinity... this will be the domain
If im understanding right the solution is B then
that's correct
awww thanks

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