# Solving an inequality.

#### 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.

#### skeeter

MHF Helper
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.

Ulysses

#### 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?

#### skeeter

MHF Helper
I'm afraid I'm not following ... what is the original problem?

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