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anonymous
 4 years ago
(a) Explain your strategy in solving inequalities of the form \( \frac{f(x)}{g(x)} < c. \)
(b) Solve \(\left \frac{x1}{x+1} \right <1\).
anonymous
 4 years ago
(a) Explain your strategy in solving inequalities of the form \( \frac{f(x)}{g(x)} < c. \) (b) Solve \(\left \frac{x1}{x+1} \right <1\).

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Your answer is not making any sense.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Because 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 ((x1)/(x+1)) < 1 and ((x1)/(x+1)) > 1 1 < ((x1)/(x+1)) < 1

Ghazi
 4 years ago
Best ResponseYou've already chosen the best response.0\[1<\frac{ x1}{ x+1 } <1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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.

Ghazi
 4 years ago
Best ResponseYou've already chosen the best response.0@twitter :) multiply both sides by x+1 and solve it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\left \frac{x1}{x+1} \right <1\] \[ \frac{x1}{x+1} <1\] or \[ \frac{x1}{x+1} >1\] you cannot multiply by \(x+1\) because you do not know if it is positive or negative start with \[\frac{x1}{x+1}1<0\] solve that one, then \[\frac{x1}{x+1}+1>0\] solve that one, then take the intersection

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x1}{x+1}1<0\] \[\frac{2}{x+1}<0\] \[x>1\] for the first one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh this is ancient history nvm

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x1}{x+1}+1>0 \]\[\frac{2x}{x+1}>0\]\[x>0\] Therefore \( x \) is greater than zero. Thanks @satellite73 ! I am your fan now! ;))
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