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Math Help - inequalities

  1. #1
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    inequalities

    Sketch on the same coordinate axes , the graphs for y=\frac{1}{|x|} and 4y=3|x|-4

    Hence , find the set of values of x such that 3|x|-4>\frac{1}{|x|}

    I managed to sketch y=\frac{1}{|x|} and y=\frac{3}{4}|x|-1

    Is \frac{3}{4}|x|-1>\frac{1}{|x|} the same as 3|x|-4>\frac{1}{|x|} ??

    Zz i am confused . Thanks !
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  2. #2
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    Quote Originally Posted by thereddevils View Post

    Hence , find the set of values of x such that 3|x|-4>\frac{1}{|x|}

    I managed to sketch y=\frac{1}{|x|} and y=\frac{3}{4}|x|-1

    Is \frac{3}{4}|x|-1>\frac{1}{|x|} the same as 3|x|-4>\frac{1}{|x|} ??

    Zz i am confused . Thanks !
    no, they can't be equal. It seems you multiplied by 4, but you need to do it to the entire inequality, not just one side, so the inequality becomes 3\left| x \right|-4>\frac{4}{\left| x \right|}.

    now i am confused, because dunno if you want to solve this inequality or the first one you wrote.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    no, they can't be equal. It seems you multiplied by 4, but you need to do it to the entire inequality, not just one side, so the inequality becomes 3\left| x \right|-4>\frac{4}{\left| x \right|}.

    now i am confused, because dunno if you want to solve this inequality or the first one you wrote.
    Thanks so i will need to sketch y=\frac{1}{|x|} and 4y=3|x|-4 and find their intersection ?
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  4. #4
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    no, you can solve it analitically, can you do that?
    Last edited by Krizalid; September 17th 2009 at 05:03 AM.
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    no, you can sove it analitically, can you do that?
    Yes . But i am not sure how to do it ?
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  6. #6
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    Quote Originally Posted by thereddevils View Post
    Thanks so i will need to sketch y=\frac{1}{|x|} and 4y=3|x|-4 and find their intersection ?
    You need to find values of x so that the graph of y= 3|x|+ 4 is above the graph of y= 1/|x|.
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    You need to find values of x so that the graph of y= 3|x|+ 4 is above the graph of y= 1/|x|.

    why is it y=3|x|+4 , what happen to 4y=3|x|-4 ?? Thanks .
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  8. #8
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    i gotta work now.

    when i get back i'll tell you how to make it.
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  9. #9
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    Quote Originally Posted by thereddevils View Post

    Hence , find the set of values of x such that 3|x|-4>\frac{1}{|x|}
    is this the inequality you need to solve?
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  10. #10
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    Quote Originally Posted by Krizalid View Post
    is this the inequality you need to solve?
    yes
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  11. #11
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    okay, sorry for the delay.

    here's the solution:

    obviously it's x\ne0 and besides |x|>0, so we rewrite the inequality as 3|x|^2-4|x|-1>0. This is not factorable as we want, so with some of algebra we get 3\left| x \right|^{2}-4\left| x \right|-1=\frac{9\left| x \right|^{2}-12\left| x \right|-3}{3}=\frac{\left( 3\left| x \right|-2 \right)^{2}-7}{3}>0, hence <br />
\left( 3\left| x \right|-2+\sqrt{7} \right)\left( 3\left| x \right|-2-\sqrt{7} \right)>0, and the first factor is positive, so we require that 3\left| x \right|-2-\sqrt{7}>0\implies x>\frac{2+\sqrt{7}}{3}\,\,\vee \,\,x<-\frac{2+\sqrt{7}}{3}.
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