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Math Help - Inequality question

  1. #1
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    Inequality question

    Hey guys, again i need help with the 2nd part of the question:

    Find the range of values of x for which satisfy the inequality 1/(x-1) > 2/(x+2). Hence deduce the values of x for 1/(1-e^x) > 2/(1+2e^x).

    Thanks!
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  2. #2
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    Quote Originally Posted by margaritas View Post
    Hey guys, again i need help with the 2nd part of the question:

    Find the range of values of x for which satisfy the inequality 1/(x-1) > 2/(x+2). Hence deduce the values of x for 1/(1-e^x) > 2/(1+2e^x).

    Thanks!
    \frac{1}{x - 1} > \frac{2}{x + 2}

    \frac{1}{x - 1} - \frac{2}{x + 2} > 0

    \frac{(x + 2) - 2(x + 1)}{(x - 1)(x + 2)} > 0

    \frac{-x}{(x - 1)(x + 2)} > 0

    Now, for what values of x is this true? Critical values of x are where the numerator or denominator is 0, so let's split the real line into 4 segements and check:
    ( -\infty, -2): \frac{-x}{(x - 1)(x + 2)} > 0 (Check!)

    ( -2, 0): \frac{-x}{(x - 1)(x + 2)} < 0 (No!)

    ( 0,  1): \frac{-x}{(x - 1)(x + 2)} > 0 (Check!)

    ( 1, \infty): \frac{-x}{(x - 1)(x + 2)} < 0 (No!)

    So the solution set is:
    (-\infty, -2) \cup (0, 1)

    Can you do the rest?

    -Dan
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  3. #3
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    Erh yes I got the answer for the first part but I still can't do the 2nd part.
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  4. #4
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    Hi:

    Using the identical procedure as did Dan, I get solution set (Ln(0.25), 0).

    Regards,

    Rich B.
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