## Alexis1994_13 4 years ago (1+x)/(1-x)>0 What's next? :S

1. heromiles

multiply both sides by the denominator

2. amogh

you cannot do that as 1-x can be negative, You can do that be wavy curve method, Or common sense, If a ratio has to be +ve, both numerator n denominator have to be of the same sign, 1+x >0 , x>-1 1-x >0 , x<1 common soln. (-1,1) 1+x<0 , x<-1 1-x<0 , x>1 common soln. (nothing!) Therefore, the ans is (-1,1)

3. joemath314159

if you multiply both sides by the denominator you would get 1 + x > 0(1 - x) which is 1 +x > 0

4. heromiles

For some reason, I have to do this on paper, not in my head

5. satellite73

you cannot multiply both sides by a variable in an inequality

6. amogh

@satellite: you can if you know that its gonna be +ve or -ve

7. Alexis1994_13

I guess amogh it's right. Thanks guys.

8. amogh

np

9. satellite73

??

10. satellite73

11. heromiles

Okay, I give up

12. joemath314159

Satellite and amogh have the right idea.

13. Alexis1994_13

f(x) ε (-1,1)

14. heromiles

I neglected to use critical points and the intervals between them

15. satellite73

and it is easy enough to solve. for one thing you can just think of \[(1+x)(1-x)> 0\] which is a parabola facing down. therefore it is positive between the zeros and negative outside of them. the zeros are -1 and 1 so it is positive between those two numbers and negative outside of them

16. Alexis1994_13

Very usefull site! Thank ya all :)

17. satellite73

or you can say you have two factors , 1+x which is positive if x > -1 negative for x < -1 1-x which is positive if x < 1 and negative is x > 1 then what happens when you divide? if you are between -1 and 1 both are positive, and so the quotient will be as well