Little help on Inequalities
- Thread starter jaymaster
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How did you get that? Please share your work - so that we know where/how to help you!
1/(x+1) > 1
Multiply through by (x+1)2
(X+1)2/(x+1) > (x+1)2
x(x+1) < 0
This is where I got stucked
HallsofIvy
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I'm no sure why you did that. The initial problem was simpler than what you wound up with. The original inequality is \(\displaystyle \frac{1}{x+ 1}< 1\) which can only be true if the denominator is larger than 1: x+ 1> 1 so x> 0.
If you really wanted to solve "x(x+ 1)< 0[/tex]" use the fact that the product of two numbers is negative if and only if one number is positive and the other negative. Here that means either x< 0 and x+ 1> 0 or x> 0 and x+ 1< 0. If x+ 1> 0, then x> -1 so the first pair give -1< x< 0. if x+ 1< 0 then x< -1. But in that case x cannot be larger than 0 so the solution to that question (not your original question) is -1< x< 0.
If you really wanted to solve "x(x+ 1)< 0[/tex]" use the fact that the product of two numbers is negative if and only if one number is positive and the other negative. Here that means either x< 0 and x+ 1> 0 or x> 0 and x+ 1< 0. If x+ 1> 0, then x> -1 so the first pair give -1< x< 0. if x+ 1< 0 then x< -1. But in that case x cannot be larger than 0 so the solution to that question (not your original question) is -1< x< 0.
Hello, jaymaster!
. . but I agree with it.
We have: .\(\displaystyle \dfrac{1}{x+1} \:>\:1\)
Multiply by \(\displaystyle (x+1).\)
We must consider two cases.
[1] \(\displaystyle (x+1)\) is negative.
. . That is: .\(\displaystyle x+1\,<\,0 \quad\Rightarrow\quad x \,<\,\text{-}1\)
Then: .\(\displaystyle 1 \:<\: x+1 \quad\Rightarrow\quad 0 \,<\,x \quad\Rightarrow\quad x \,>\,0\)
We have: .\(\displaystyle x\,<\,\text{-}1\,\text{ and }\,x \,>\,0\)
. . But this is impossible.
[2] \(\displaystyle (x+1)\) is positive.
. . That is: .\(\displaystyle x+1 \:>\:0 \quad\Rightarrow\quad x \:>\:\text{-}1 \quad\Rightarrow\quad \text{-}1 \,< x\)
Then: .\(\displaystyle 1 \:>\: x+1 \quad\Rightarrow\quad 0 \,>\,x \quad\Rightarrow\quad x\,<\,0\)
We have: .\(\displaystyle \text{-}1 \,<\,x\,\text{ and }\,x\,<\,0\)
Therefore: .\(\displaystyle \boxed{\text{-}1 \:<\: x \:<\:0}\)
I don't know how you got your answer,
. . but I agree with it.
We have: .\(\displaystyle \dfrac{1}{x+1} \:>\:1\)
Multiply by \(\displaystyle (x+1).\)
We must consider two cases.
[1] \(\displaystyle (x+1)\) is negative.
. . That is: .\(\displaystyle x+1\,<\,0 \quad\Rightarrow\quad x \,<\,\text{-}1\)
Then: .\(\displaystyle 1 \:<\: x+1 \quad\Rightarrow\quad 0 \,<\,x \quad\Rightarrow\quad x \,>\,0\)
We have: .\(\displaystyle x\,<\,\text{-}1\,\text{ and }\,x \,>\,0\)
. . But this is impossible.
[2] \(\displaystyle (x+1)\) is positive.
. . That is: .\(\displaystyle x+1 \:>\:0 \quad\Rightarrow\quad x \:>\:\text{-}1 \quad\Rightarrow\quad \text{-}1 \,< x\)
Then: .\(\displaystyle 1 \:>\: x+1 \quad\Rightarrow\quad 0 \,>\,x \quad\Rightarrow\quad x\,<\,0\)
We have: .\(\displaystyle \text{-}1 \,<\,x\,\text{ and }\,x\,<\,0\)
Therefore: .\(\displaystyle \boxed{\text{-}1 \:<\: x \:<\:0}\)
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Thanks a lot . In the textbook they gave the answer as .\(\displaystyle x\,<\,\text{-}1\,\text{ and }\,x \,>\,0\)
which I know is Impossible
Thanks for the further explanation
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John Marsh
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Get More Help 
Given 1/(x+1) > 1
1/(x+1)-1>0
[(-1)(x+1)+1*1]/(x+1)>0
[-(x+1)+1]/(x+1)>0
-x/(x+1)>0
x/(x+1)<0
x positive for x>0
x negative for x<0
x zero for x=0
x+1 positive for x>-1
x+1 negative for x<-1
x+1 zero for x=-1
Hence -1<x<0
1/(x+1)-1>0
[(-1)(x+1)+1*1]/(x+1)>0
[-(x+1)+1]/(x+1)>0
-x/(x+1)>0
x/(x+1)<0
x positive for x>0
x negative for x<0
x zero for x=0
x+1 positive for x>-1
x+1 negative for x<-1
x+1 zero for x=-1
Hence -1<x<0