HELP! Inequality!

**allenfarmer** · July 2nd 2009, 11:03 AM

i have no problem solving inequalities but this one has me stumped, one over x is greater than x. For what real number is 1/x > x ?

**fcabanski** · July 2nd 2009, 11:22 AM

Do some manipulation:

1/x > x (multiply both sides by 1/x)

1 > x^2 (take the square root of both sides)

+, - 1 = x

Check intervals:

- infinity to - 1

(-2)

1/-2 > -2 YES

-1 to 1

1/2 works, but -1/2 doesn't...hmmm

1/(1/2) > 1/2

2 > 1/2 YES

1/(-1/2) > -1/2

-2 > -1/2 NO

1 to infinity

(2)

1/2 > 2 NO

Solution

X < -1

0 < x < 1

**Plato** · July 2nd 2009, 11:59 AM

Here is a second way to get that solution set.
$\frac{1}{x} > x\; \Rightarrow \;\frac{{1 - x^2 }}<br /> {x} > 0$

The soution set clear.

**fcabanski** · July 2nd 2009, 12:05 PM

Plato, doesn't that proceed to the same point?

(1-X^2)/x >0

1 - x^2 > 0

1 > x^2

etc.

**Plato** · July 2nd 2009, 12:11 PM

Originally Posted by fcabanski

Plato, doesn't that proceed to the same point?

Well yes it is. And I clearly said so.
But it is a far better way to proceed in that there is no squaring it deal with.

**fcabanski** · July 2nd 2009, 12:14 PM

I agree.

**dhiab** · July 2nd 2009, 01:05 PM

Originally Posted by fcabanski

Plato, doesn't that proceed to the same point?

(1-X^2)/x >0

1 - x^2 > 0

1 > x^2

etc.

Hello :This solution is Right if x positif

**HallsofIvy** · July 2nd 2009, 05:32 PM

A very general way of solving complicated inequalities: solve the equation first.

To solve $\frac{1}{x}> x$ , first look at $\frac{1}{x}= x$ . That's easy: multiplying both sides by x, $\frac{1}= x^2$ which has roots $x= \pm 1$ . The point of that is that functions can change form ">" to "<" only where they are "=" or where they are not continuous. The functions involved here are equal at x= 1 or x= -1 and not continuous at x= 0 where the left side is not defined. Now we need only check on point in each of the intervals $(-\inf, -1)$ , $(-1, 0)$ , $(0, 1)$ , and $(1, \infty)$ .

For example:
-2< -1 and if x= -2, $\frac{1}{x}= -\frac{1}{2}> -2= x$ .
$-2< -\frac{1}{2}< 0$ and if $x= -\frac{1}{2}$ , $\frac{1}{x}= -2< -\frac{1}{2}= x$ .
$0< \frac{1}{2}< 1$ and if $x= \frac{1}{2}$ , $\frac{1}{x}= 2> \frac{1}{2}= x$
1< 2 and if x= 2, $\frac{1}{x}= \frac{1}{2}< 2= x$

That is, $\frac{1}{x}> x$ for x< -1 or 0< x< 1.

**VonNemo19** · July 2nd 2009, 05:44 PM

This same question was asked here.

http://www.mathhelpforum.com/math-he...questions.html

**dhiab** · July 2nd 2009, 10:31 PM

Hello : THIS THE SOLUTION

remark : all solution set is open in 0 ; -1; 1, -infinity

**Prove It** · July 2nd 2009, 10:44 PM

Originally Posted by allenfarmer

i have no problem solving inequalities but this one has me stumped, one over x is greater than x. For what real number is 1/x > x ?

We are told:

$\frac{1}{x} > x$ .

First, we have to note that $x \neq 0$ .

If we are trying to solve this inequality, we need to take into account two possibilities: $x > 0$ and $x < 0$ .

Case 1: $x > 0$ .

$\frac{1}{x} > x$

$1 > x^2$

$x^2 < 1$

$|x| < 1$

$-1 < x < 1$ .

But since $x > 0$

$0 < x < 1$ .

Case 2: $x < 0$

$\frac{1}{x} >x$

$1 < x^2$

$x^2 > 1$

$|x| > 1$

$x < -1$ or $x > 1$ .

But since $x < 0$ we have $x < -1$ .

So putting everything together:

$x < -1$ or $0 < x < 1$ .

Thread: HELP! Inequality!

LinkBack

Thread Tools

Display

HELP! Inequality!

Similar Math Help Forum Discussions

Inequality with x in the denominator (rational inequality)

Proving the triangle inequality using the Cauchy-Schwartzs inequality

inequality

inequality

Inequality

Search Tags