1. ## HELP! Inequality!

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 ?

2. 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

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

The soution set clear.

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

(1-X^2)/x >0

1 - x^2 > 0

1 > x^2

etc.

5. 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.

6. I agree.

7. 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

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

To solve $\displaystyle \frac{1}{x}> x$, first look at $\displaystyle \frac{1}{x}= x$. That's easy: multiplying both sides by x, $\displaystyle \frac{1}= x^2$ which has roots $\displaystyle x= \pm 1$. 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 $\displaystyle (-\inf, -1)$, $\displaystyle (-1, 0)$, $\displaystyle (0, 1)$, and $\displaystyle (1, \infty)$.

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

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

9. This same question was asked here.

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

10. Hello : THIS THE SOLUTION

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

11. 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:

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

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

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

Case 1: $\displaystyle x > 0$.

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

$\displaystyle 1 > x^2$

$\displaystyle x^2 < 1$

$\displaystyle |x| < 1$

$\displaystyle -1 < x < 1$.

But since $\displaystyle x > 0$

$\displaystyle 0 < x < 1$.

Case 2: $\displaystyle x < 0$

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

$\displaystyle 1 < x^2$

$\displaystyle x^2 > 1$

$\displaystyle |x| > 1$

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

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

So putting everything together:

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