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Math Help - another inequality question

  1. #1
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    another inequality question

    Solve this inequality .

     <br />
|x-2|<\frac{1}{x}<br />

    case 1 :

    -\frac{1}{x}<x-2

    i got 0<x<1 and x>1

    case 2 : x-2<\frac{1}{x}

    i got x<-0.414 or 0<x<2.414

    How can i combine the results to get the solution set ?? THanks for your explaination .
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  2. #2
    Senior Member I-Think's Avatar
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    The only valid solution set is 0<x<2.414<br />

    Graph the inequality and you'll see why.
    \frac{1}{x} takes on only negative values for all negative x
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  3. #3
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    Hello, thereddevils!

    I-Think is absolutely correct . . .


    Solve this inequality: . |x-2| \:<\:\tfrac{1}{x}
    The graph of f(x) \:=\:|x-2| is a V-shaped graph with its vertex at (2,0).

    The graph of g(x) \:=\:\tfrac{1}{x} is a hyperbola in the 1st and 3rd quadrants.
    Code:
                    |
                  \ |*        /
                   \|        /
                    \ *     /
                    |\ *   /
                    | \   oP
                    |  \ /      *
      - - - - - - - + - * - - - - -
        *           |
              *     |
                 *  |
                  * |
                    |
                   *|
                    |

    The graphs intersect at: P\left(1\!+\!\sqrt{2},\:\sqrt{2}\!-\!1\right)

    We see that f(x) < g(x) on the interval: . \left(0,\:1\!+\!\sqrt{2}\right)





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  4. #4
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    Quote Originally Posted by I-Think View Post
    The only valid solution set is 0<x<2.414<br />

    Graph the inequality and you'll see why.
    \frac{1}{x} takes on only negative values for all negative x
    But |1-2|<\frac{1}{1}

    1<1 ???


    Or maybe there's a simpler approach to this question .. anyone ?? Thanks .
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, thereddevils!

    I-Think is absolutely correct . . .

    The graph of f(x) \:=\:|x-2| is a V-shaped graph with its vertex at (2,0).

    The graph of g(x) \:=\:\tfrac{1}{x} is a hyperbola in the 1st and 3rd quadrants.
    Code:
                    |
                  \ |*        /
                   \|        /
                    \ *     /
                    |\ *   /
                    | \   oP
                    |  \ /      *
      - - - - - - - + - * - - - - -
        *           |
              *     |
                 *  |
                  * |
                    |
                   *|
                    |
    The graphs intersect at: P\left(1\!+\!\sqrt{2},\:\sqrt{2}\!-\!1\right)

    We see that f(x) < g(x) on the interval: . \left(0,\:1\!+\!\sqrt{2}\right)




    Thanks a lot Soroban , but is it possible for me to solve it without using the graphical method ?? Just out of curiousity .
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  6. #6
    Math Engineering Student
    Krizalid's Avatar
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    Of course it's possible and quite straightforward:

    \left| x-2 \right|<\frac{1}{x}\Rightarrow -\frac{1}{x}<x-2<\frac{1}{x}.

    Eeach one of this must be solved separately.

    First inequality: x-2+\frac{1}{x}=\frac{x^{2}-2x+1}{x}=\frac{(x-1)^{2}}{x}>0. In order so that this works, we require that x>0 and, of course x\ne1.

    Second inequality: x-2-\frac{1}{x}=\frac{x^{2}-2x-1}{x}=\frac{(x-1)^{2}-2}{x}<0\implies\left| x-1 \right|<\sqrt{2}. But why we're allowed to do this, because in the original inequality we need to have x>0, otherwise there's no solution to this. How can a negative number be greater than a positive one? That's why we have x>0; thus the solution to this remaining inequality is 1-\sqrt{2}<x<1+\sqrt{2}.

    Almost full solution set will be given by (0,\infty )\cap \left( 1-\sqrt{2},1+\sqrt{2} \right)=\left( 0,1+\sqrt{2} \right), and since we stated that it's x\ne1 our full solution set is actually x\in(0,1)\cup \left( 1,1+\sqrt{2} \right).
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    Of course it's possible and quite straightforward:

    \left| x-2 \right|<\frac{1}{x}\Rightarrow -\frac{1}{x}<x-2<\frac{1}{x}.

    Eeach one of this must be solved separately.

    First inequality: x-2+\frac{1}{x}=\frac{x^{2}-2x+1}{x}=\frac{(x-1)^{2}}{x}>0. In order so that this works, we require that x>0 and, of course x\ne1.

    Second inequality: x-2-\frac{1}{x}=\frac{x^{2}-2x-1}{x}=\frac{(x-1)^{2}-2}{x}<0\implies\left| x-1 \right|<\sqrt{2}. But why we're allowed to do this, because in the original inequality we need to have x>0, otherwise there's no solution to this. How can a negative number be greater than a positive one? That's why we have x>0; thus the solution to this remaining inequality is 1-\sqrt{2}<x<1+\sqrt{2}.

    Almost full solution set will be given by (0,\infty )\cap \left( 1-\sqrt{2},1+\sqrt{2} \right)=\left( 0,1+\sqrt{2} \right), and since we stated that it's x\ne1 our full solution set is actually x\in(0,1)\cup \left( 1,1+\sqrt{2} \right).
    THanks a lot Krizalid .. yeah that makes sense .

    But how come the answer got from graphical method and analytical method is different ?? Shouldnt they be the same ?
    The answer got from graphical method includes 1 which makes the inequality invalid .
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  8. #8
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    Krizalid's Avatar
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    Of course graphical method should yield the same answer, then I think Soroban messed up somewhere.
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