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Thread: Inequalities with a variable in the denominator

  1. #1
    Member Chokfull's Avatar
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    Inequalities with a variable in the denominator

    I'm past this, but couldn't remember the rule....I have
    $\displaystyle
    \frac {5} {(x+1)^3} < N
    $
    Then there is a rule for solving this.... You can't multiply both sides by$\displaystyle (x+1)^3$,because you don't know if $\displaystyle x$ is negative or positive. The book says you should get
    $\displaystyle
    {\sqrt [3]{5/|N|}}
    $
    Thanks!
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by Chokfull View Post
    I'm past this, but couldn't remember the rule....I have
    $\displaystyle
    \frac {5} {(x+1)^3} < N
    $
    Then there is a rule for solving this.... You can't multiply both sides by$\displaystyle (x+1)^3$,because you don't know if $\displaystyle x$ is negative or positive. The book says you should get
    $\displaystyle
    {\sqrt [3]{5/|N|}}
    $
    Thanks!
    With these types of equalities you have a couple of chouces.

    As you said you cannot multiply both sides by$\displaystyle (x+1)^3$, however you can multiply both by$\displaystyle (x+1)^6$, as you know this is a positive number.

    You can also make both sides into fractions and add or subtract to rearrange.

    For example,$\displaystyle \frac {5} {(x+1)^3} <\frac{N(x+1)^3}{(x+1)^3}$, leading to $\displaystyle \frac {5} {(x+1)^3} - \frac{N(x+1)^3}{(x+1)^3}<0$

    Can you solve this from here?
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  3. #3
    Member Chokfull's Avatar
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    Quote Originally Posted by craig View Post
    With these types of equalities you have a couple of chouces.

    As you said you cannot multiply both sides by$\displaystyle (x+1)^3$, however you can multiply both by$\displaystyle (x+1)^6$, as you know this is a positive number.

    You can also make both sides into fractions and add or subtract to rearrange.

    For example,$\displaystyle \frac {5} {(x+1)^3} <\frac{N(x+1)^3}{(x+1)^3}$, leading to $\displaystyle \frac {5} {(x+1)^3} - \frac{N(x+1)^3}{(x+1)^3}<0$

    Can you solve this from here?
    OK, but how does that get an absolute value? I'll fiddle with it for now but don't see how it gets $\displaystyle {\sqrt [3]{5/|N|}}
    $
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