another inequality question

**thereddevils** · June 29th 2009, 04:47 AM

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 .

**I-Think** · June 29th 2009, 06:04 AM

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

**Soroban** · June 29th 2009, 06:38 AM

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: . $\left(0,\:1\!+\!\sqrt{2}\right)$

**thereddevils** · June 29th 2009, 06:39 AM

Originally Posted by I-Think

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 .

**thereddevils** · June 29th 2009, 06:41 AM

Originally Posted by Soroban

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 .

**Krizalid** · June 29th 2009, 07:29 AM

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 $x\in(0,1)\cup \left( 1,1+\sqrt{2} \right).$

**thereddevils** · June 29th 2009, 11:37 PM

Originally Posted by Krizalid

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 .

**Krizalid** · June 30th 2009, 01:57 PM

Of course graphical method should yield the same answer, then I think Soroban messed up somewhere.

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