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

1. carolinar7

what do you think

2. anonymous

honestly I don't know, I want to say all values of x but im not sure :(

3. zzr0ck3r

Can you write out $$\dfrac{f(x)}{g(x)}$$

4. anonymous

no I don't know how sorry

5. anonymous

can you guys help me pleaseeee

6. hybrik

x+1^-1, So your saying the whole quantity by -1, or just "1"

7. carolinar7

$\frac{ x+1 }{ x-2 }$

8. anonymous

yeah the exponent is -1

9. zzr0ck3r

its the inverse $$(x+1)^{-1}=\dfrac{1}{x+1}$$

10. carolinar7

no

11. carolinar7

The exponent is only on 1 not (x+1)

12. zzr0ck3r

So you have $$\dfrac{(x+1)^{-1}}{x-2}=\dfrac{1}{(x+1)(x-2)}$$. Now what numbers force us to divide by $$0$$?

13. carolinar7

$Did she mean (x+1)^{-1}$

14. hybrik

The zeroes of (x+1)(x-2)

15. zzr0ck3r

yes. there is no reason to raise something to 1

16. hybrik

Also known as (x+1)=0, (x-2)=0, Solve for X on both of them @idgm_idontgetmath

17. carolinar7

x=-1,2

18. zzr0ck3r

please let them figure something out man. you are not helping if you give thm everything....

19. hybrik

But im trying to show him to solve zeroes??????

20. zzr0ck3r

Not you.

21. anonymous

@carolinar7

22. anonymous

@vera_ewing

23. anonymous

@juanpabloJR

24. anonymous

@campbell_st can you help?

25. campbell_st

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

26. campbell_st

so here is an image of f(x)/g(x)

27. campbell_st

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...

28. campbell_st

so the regions start with - infinity to the 1st asymptote between the asymptotes and the right asymptote to infinity... this will be the domain

29. anonymous

If im understanding right the solution is B then

30. campbell_st

that's correct

31. anonymous

awww thanks