# Can Someone please solve this inequality

1. Oct 5, 2011

### hedgefire7

Can Someone please solve this inequality!!

1. The problem statement, all variables and given/known data

((2x)3^x)/(x+1) < 0

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 5, 2011

### NewtonianAlch

Re: Can Someone please solve this inequality!!

3. Oct 5, 2011

### symbolipoint

Re: Can Someone please solve this inequality!!

((2x)3^x)/(x+1) < 0

The two basic possibilities are:
(2x)3x<0 AND x+1>0
OR
(2x)3x>0 AND x+1<0

You see that the positivity or negativity of x IN THE EXPONENT does not affect the positivity or the negativity of the numerator expression. On the other hand, the positivity or negativity of x still does affect the positivity or negativity of the numerator expression.

4. Oct 5, 2011

### hedgefire7

Re: Can Someone please solve this inequality!!

1. The problem statement, all variables and given/known data

((2x)3^x)/(x+1) < 0

2. Relevant equations

3. The attempt at a solution

((2x)3^x)/(x+1) < 0
((2x)3^x) < (x+1)
I was thinking along the lines of factoring the left side and grouping the "x" variables together...the exponent is what is causing me grief...not sure on how to tackle that
my next line is:
(2x)(3^x)-(x) < 1
get stuck here :s

5. Oct 5, 2011

### symbolipoint

Re: Can Someone please solve this inequality!!

Is the question solvable? Does it have a solution?

6. Oct 5, 2011

### hedgefire7

Re: Can Someone please solve this inequality!!

Yeah there a solution provided on the appendix

7. Oct 5, 2011

### symbolipoint

Re: Can Someone please solve this inequality!!

Was my post, #3, any use for you?

8. Oct 5, 2011

### symbolipoint

Re: Can Someone please solve this inequality!!

By little bit more examining, there is a critical point for x values, being that x cannot be -1.

Look again at the suggestion of the conditions to analyze. You should find that one of those conditions will give you a solution while the other condtion will not give you a solution. Note that x cannot be -1, and the expression is 0 when x is 0. Your two critical points of the number line interval will be at x at -1 and at 0.

9. Oct 5, 2011

### hedgefire7

Re: Can Someone please solve this inequality!!

Yeah...after some more playing around I have noticed that x cannot be -1 or 0...Just not sure if there is a way to come to that point algebraically...Thank you :)

10. Oct 5, 2011

### symbolipoint

Re: Can Someone please solve this inequality!!

The variable, x, CAN be zero, but will it satisfy the inequality?

The way to find the critical values of x is to examine the expression of interest, and look for any restrictions on the allowable x values and look for any other values which would be critical (such as where is the left-hand member actually equal to zero).

11. Oct 5, 2011

### HallsofIvy

Staff Emeritus
Re: Can Someone please solve this inequality!!

Symbolicpoint said it all in his first response that you do not seem to understand. At least, you have not done anything with it.

A fraction is negative if and only if its numerator and denominator are of opposite signs. So you have two possible cases:

a) $(2x)3^x> 0$ and $x+1< 0$
An exponential, like $3^x$ is always positive so the numerator is positive only for x> 0. x+ 1< 0 only for x< -1. It is impossible for both of those to be true so this case is impossible.

b) $(2x)3^x< 0$ and $x+ 1> 0$
Again $3^x$ is always positive so you must have x< 0. What do x< 0 and x+1> 0 give you?

12. Nov 24, 2011

### geniusboy1

Re: Can Someone please solve this inequality!!

this is too easy.....
here x belons to(- ∞,0) - {-1}

(hope you have a little idea about set theory and the notations used in it)

13. Nov 24, 2011

### Mentallic

Re: Can Someone please solve this inequality!!

And yet you got it wrong :yuck:

Please don't necro threads, if you take a look, the last post before yours was about 7 weeks ago.