Hello, jaymaster!
I found this question in my textbook.
I tried solving it but I got a different answer. Different from what?
. . \(\displaystyle \dfrac{1}{x+1}\: >\: 1\)
My own answer was: \(\displaystyle -1\,<\,x\,<\,0\)
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:
.\(\displaystyle \boxed{\text{-}1 \:<\: x \:<\:0}\)