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twitter

  • 2 years ago

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

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  1. twitter
    • 2 years ago
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    Your answer is not making any sense.

  2. Denebel
    • 2 years ago
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    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
    • 2 years ago
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    hope that makes sense

  4. ghazi
    • 2 years ago
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    \[-1<\frac{ x-1}{ x+1 } <1\]

  5. twitter
    • 2 years ago
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    @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
    • 2 years ago
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    @twitter :) multiply both sides by x+1 and solve it

  7. ghazi
    • 2 years ago
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    \[(x-1)<(x+1)\]

  8. twitter
    • 2 years ago
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    @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
    • 2 years ago
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    hmmm

  10. satellite73
    • 2 years ago
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    \[\left| \frac{x-1}{x+1} \right| <1\] \[ \frac{x-1}{x+1} <1\] 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
    • 2 years ago
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    \[\frac{x-1}{x+1}-1<0\] \[\frac{-2}{x+1}<0\] \[x>-1\] for the first one

  12. satellite73
    • 2 years ago
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    oh this is ancient history nvm

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

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