# Solving an inequality.

• May 26th 2010, 03:39 PM
Ulysses
Solving an inequality.
Hi, I must solve this inequality. How must I proceed?

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

I don't know how to, couse I cant do this: $\displaystyle 1<x^2$ so... I know its stupid, but... im a little bit stupid I think.
• May 26th 2010, 03:46 PM
skeeter
Quote:

Originally Posted by Ulysses
Hi, I must solve this inequality. How must I proceed?

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

I don't know how to, couse I cant do this: $\displaystyle 1<x^2$ so... I know its stupid, but... im a little bit stupid I think.

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

$\displaystyle \frac{1}{x^2} - 1 < 0$

$\displaystyle \frac{1 - x^2}{x^2} < 0$

critical values are $\displaystyle x = 1$ , $\displaystyle x = -1$ , and $\displaystyle x = 0$

check the original inequality with a value from each of the four intervals defined by the critical values of x ... if the value makes the original inequality true, then all values of x in that interval make the inequality true.

if false ... all values of x in that interval will not work.
• May 26th 2010, 04:03 PM
Ulysses
I've realized that I could "pass" the $\displaystyle x^2$ couse it won't change signs.

But now I have the second derivative $\displaystyle \displaystyle\frac{2}{x^3}>0$ and I must analize its sign for the concave intervals. So, I know the solution actually, but if I proceed the same way than before I can't "pass" the $\displaystyle x^3$ couse it would change the sign.

How should I think this?
• May 26th 2010, 04:49 PM
skeeter
I'm afraid I'm not following ... what is the original problem?