Solving an inequality.

May 2010
254
8
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
Jun 2008
16,217
6,765
North Texas
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.
 
  • Like
Reactions: Ulysses
May 2010
254
8
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
Jun 2008
16,217
6,765
North Texas
I'm afraid I'm not following ... what is the original problem?
 
Similar Math Discussions Math Forum Date
Algebra
Algebra
Algebra
Algebra