Solve the inequality |x^2 - x| - |x| < 1 analytically.
If it is for either x = 0 or x = 1, why is the case 0 < x < 1 since that involves all the values between those numbers as well?
Sorry for my slowness but I have received no prior instruction in inequalities.
So far here is my attempt to solve, nearly positive there's a few mistakes:
Case 1: x < 0
$\displaystyle -x(-x-1) + x - 1 < 0 $
$\displaystyle x^2 + x + x - 1 < 0 $
$\displaystyle x^2 + 2x - 1 < 0 $
(used quadratic formula to solve x's)
$\displaystyle (x + 2.4142)(x - 0.4142) < 0 $
solution: 0.24142 < x < 0.4142
If I pay attention to the earlier restriction on x, would that be changed to 0.24142 < x < 0 ?
Case 2:
I don't even know where to begin to solve this one
Case 3: x > 1
$\displaystyle x(x-1) - x - 1 < 0 $
$\displaystyle x^2 - x - x - 1 < 0 $
$\displaystyle x^2 - 2x - 1 < 0 $
$\displaystyle (x + 0.4142)(x - 2.4142) < 0 $
solution: -0.4142 < x < 2.4142
which then becomes 1 < x < 2.4142
Put $\displaystyle x=0$ in that inequality, is it true? In the same fashion put $\displaystyle x=1,$ is it true?
In both cases, those values satisfy the inequality, so that's why I picked $\displaystyle x\le0$ instead of $\displaystyle x<0.$
As for the sake of the problem, you solved the first case wrong, 'cause for $\displaystyle x\le0$ the inequality becomes $\displaystyle x^{2}-x+x-1<0\implies -1<x<1.$ Since $\displaystyle x\le0,$ the first solution set is $\displaystyle (-1,0].$
For the second case, the inequality becomes $\displaystyle x(1-x)-x-1<0\implies x^{2}+1>0,$ which is obviously true, so the second solution set it's just $\displaystyle [0,1].$
Finally, having $\displaystyle x\ge1,$ the inequality becomes $\displaystyle x^{2}-x-x-1<0\implies (x-1)^2<2,$ and then $\displaystyle 1-\sqrt2<x<1+\sqrt2,$ so the third solution set is $\displaystyle \Big[1,1+\sqrt2\Big).$
Put these solutions set together and the original inequality verifies for each $\displaystyle -1<x<1+\sqrt2.$
Thank you, it's much clearer now. I'm just curious, when solving for the second case would it be possible to go about it without first factoring the original equation out into |x||x-1|-|x|-1 < 0? Under normal circumstances I doubt I could've thought of that myself