Question

(a) Explain your strategy in solving inequalities of the form \( \frac{f(x)}{g(x)} < c. \)
(b) Solve \(\left| \frac{x-1}{x+1} \right| <1\).

Answer 1

Your answer is not making any sense.

Answer 2

Because whatever is in absolute value can be positive and negative (you wouldn't know), you have to think about it in both cases. For example, |x| = 4 can mean x=4 or x=-4 because the abs. value turns the number positive.
so ((x-1)/(x+1)) < 1 and ((x-1)/(x+1)) > -1
-1 < ((x-1)/(x+1)) < 1

Answer 3

hope that makes sense

Answer 4

\[-1<\frac{ x-1}{ x+1 } <1\]

Answer 5

@ghazi still don't make sense. What is the answer?
@Denebel Yes I know. But how do you solved it? Both the denominator and dividend is a function.

Answer 6

@twitter :) multiply both sides by x+1 and solve it

Answer 7

\[(x-1)<(x+1)\]

Answer 8

@ghazi yeah? how do you solved that? I am not impressed with your half work. I bet you are plain wrong though. X on both sides is just going to cancel each other and besides you deleted your first response because it was plain stupid. You are not helping.

Answer 9

hmmm

Answer 10

\[\left| \frac{x-1}{x+1} \right| <1\]
\[ \frac{x-1}{x+1} <1\] or \[ \frac{x-1}{x+1} >-1\]
you cannot multiply by \(x+1\) because you do not know if it is positive or negative
start with
\[\frac{x-1}{x+1}-1<0\] solve that one, then
\[\frac{x-1}{x+1}+1>0\] solve that one, then take the intersection

Answer 11

\[\frac{x-1}{x+1}-1<0\]
\[\frac{-2}{x+1}<0\]
\[x>-1\] for the first one

Answer 12

oh this is ancient history
nvm

Answer 13

\[\frac{x-1}{x+1}+1>0 \]\[\frac{2x}{x+1}>0\]\[x>0\] Therefore \( x \) is greater than zero.
Thanks @satellite73 ! I am your fan now! ;))