# another inequality question

• Jun 29th 2009, 04:47 AM
thereddevils
another inequality question
Solve this inequality .

$\displaystyle |x-2|<\frac{1}{x}$

case 1 :

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

i got $\displaystyle 0<x<1$and $\displaystyle x>1$

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

i got $\displaystyle x<-0.414$or $\displaystyle 0<x<2.414$

How can i combine the results to get the solution set ?? THanks for your explaination .
• Jun 29th 2009, 06:04 AM
I-Think
The only valid solution set is $\displaystyle 0<x<2.414$

Graph the inequality and you'll see why.
$\displaystyle \frac{1}{x}$ takes on only negative values for all negative x
• Jun 29th 2009, 06:38 AM
Soroban
Hello, thereddevils!

I-Think is absolutely correct . . .

Quote:

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

The graph of $\displaystyle g(x) \:=\:\tfrac{1}{x}$ is a hyperbola in the 1st and 3rd quadrants.
Code:

                |               \ |*        /               \|        /                 \ *    /                 |\ *  /                 | \  oP                 |  \ /      *   - - - - - - - + - * - - - - -     *          |           *    |             *  |               * |                 |               *|                 |

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

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

• Jun 29th 2009, 06:39 AM
thereddevils
Quote:

Originally Posted by I-Think
The only valid solution set is $\displaystyle 0<x<2.414$

Graph the inequality and you'll see why.
$\displaystyle \frac{1}{x}$ takes on only negative values for all negative x

But $\displaystyle |1-2|<\frac{1}{1}$

1<1 ???

Or maybe there's a simpler approach to this question .. anyone ?? Thanks .
• Jun 29th 2009, 06:41 AM
thereddevils
Quote:

Originally Posted by Soroban
Hello, thereddevils!

I-Think is absolutely correct . . .

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

The graph of $\displaystyle g(x) \:=\:\tfrac{1}{x}$ is a hyperbola in the 1st and 3rd quadrants.
Code:

                |               \ |*        /               \|        /                 \ *    /                 |\ *  /                 | \  oP                 |  \ /      *   - - - - - - - + - * - - - - -     *          |           *    |             *  |               * |                 |               *|                 |
The graphs intersect at: $\displaystyle P\left(1\!+\!\sqrt{2},\:\sqrt{2}\!-\!1\right)$

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

Thanks a lot Soroban , but is it possible for me to solve it without using the graphical method ?? Just out of curiousity .
• Jun 29th 2009, 07:29 AM
Krizalid
Of course it's possible and quite straightforward:

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

Eeach one of this must be solved separately.

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

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

Almost full solution set will be given by $\displaystyle (0,\infty )\cap \left( 1-\sqrt{2},1+\sqrt{2} \right)=\left( 0,1+\sqrt{2} \right),$ and since we stated that it's $\displaystyle x\ne1$ our full solution set is actually $\displaystyle x\in(0,1)\cup \left( 1,1+\sqrt{2} \right).$
• Jun 29th 2009, 11:37 PM
thereddevils
Quote:

Originally Posted by Krizalid
Of course it's possible and quite straightforward:

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

Eeach one of this must be solved separately.

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

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

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

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 .
• Jun 30th 2009, 01:57 PM
Krizalid
Of course graphical method should yield the same answer, then I think Soroban messed up somewhere.