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anonymous
 one year ago
If f(x)= x+1^1 and g(x)=x2 , what is the domain of f(x) divided by g(x)
anonymous
 one year ago
If f(x)= x+1^1 and g(x)=x2 , what is the domain of f(x) divided by g(x)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0honestly I don't know, I want to say all values of x but im not sure :(

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Can you write out \(\dfrac{f(x)}{g(x)}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no I don't know how sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you guys help me pleaseeee

hybrik
 one year ago
Best ResponseYou've already chosen the best response.0x+1^1, So your saying the whole quantity by 1, or just "1"

carolinar7
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ x+1 }{ x2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah the exponent is 1

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1its the inverse \((x+1)^{1}=\dfrac{1}{x+1}\)

carolinar7
 one year ago
Best ResponseYou've already chosen the best response.0The exponent is only on 1 not (x+1)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1So you have \(\dfrac{(x+1)^{1}}{x2}=\dfrac{1}{(x+1)(x2)}\). Now what numbers force us to divide by \(0\)?

carolinar7
 one year ago
Best ResponseYou've already chosen the best response.0\[Did she mean (x+1)^{1}\]

hybrik
 one year ago
Best ResponseYou've already chosen the best response.0The zeroes of (x+1)(x2)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1yes. there is no reason to raise something to 1

hybrik
 one year ago
Best ResponseYou've already chosen the best response.0Also known as (x+1)=0, (x2)=0, Solve for X on both of them @idgm_idontgetmath

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1please let them figure something out man. you are not helping if you give thm everything....

hybrik
 one year ago
Best ResponseYou've already chosen the best response.0But im trying to show him to solve zeroes??????

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@campbell_st can you help?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1ok... here is what I'll do, if you are happy that \[\frac{f(x)}{g(x)} = \frac{1}{(x +1)(x2)}\] you need to solve x + 1 = 0 and x  2 = 0 can you do that

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1so here is an image of f(x)/g(x)

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1the 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...

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1so the regions start with  infinity to the 1st asymptote between the asymptotes and the right asymptote to infinity... this will be the domain

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If im understanding right the solution is B then
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