Is ((2x)3^x)/(x+1) < 0 Solvable for x?

[hedgefire7]

Can Someone please solve this inequality!

Homework Statement

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

Thanks is advance!

Homework Equations

The Attempt at a Solution

 

[NewtonianAlch]

Read the sticky thread:

1) Did you show your work? Homework helpers will not assist with any questions until you've shown your own effort on the problem. Remember, we help with homework, we don't do your homework. We already passed those classes, it's your turn to do so.

 

[symbolipoint]

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

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

 

[hedgefire7]

1. Homework Statement

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

Thanks is advance!
2. Homework 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

 

[symbolipoint]

Is the question solvable? Does it have a solution?

 

[hedgefire7]

Yeah there a solution provided on the appendix

 

[symbolipoint]

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

 

[symbolipoint]

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.

 

[hedgefire7]

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 :)

 

[symbolipoint]

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 :)

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).

 

[HallsofIvy]

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?

 

[geniusboy1]

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)

 

[Mentallic]

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)

And yet you got it wrong

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