## twitter 3 years ago (a) Explain your strategy in solving inequalities of the form $$\frac{f(x)}{g(x)} < c.$$ (b) Solve $$\left| \frac{x-1}{x+1} \right| <1$$.

2. Denebel

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

3. ghazi

hope that makes sense

4. ghazi

$-1<\frac{ x-1}{ x+1 } <1$

@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.

6. ghazi

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

7. ghazi

$(x-1)<(x+1)$

@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.

9. satellite73

hmmm

10. satellite73

$\left| \frac{x-1}{x+1} \right| <1$ $\frac{x-1}{x+1} <1$ 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

11. satellite73

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

12. satellite73

oh this is ancient history nvm

$\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! ;))