Question

x / |1-x^2| > 1 / 2 solve the inequalities. someone??

Answer 1

x different always the values of -/+1

Answer 2

x<1

Answer 3

the answer is [ 0.41, 1) U (1, 2.41]
how come??

Answer 4

@jhonyy9 @kumarsajan

Answer 5

\[\frac{x}{|1-x^2|}>\frac{1}{2}\]
\[2x>|1-x^2|\]
now \[|1-x^2|=1-x^2 \]on the intervals where \(1-x^2>0\) i.e. if \(x<-1\) or \(x>1\)

Answer 6

where the number line?? must use or not?

Answer 7

ok this might be somewhat clearer, maybe not.
lets start with \(2x>|1-x^2|\) or maybe
\[|(1-x)(1+x)|<2x\]
now this can't happen if x is negative, because the absolute value is always positive, so we have to only check on two intervals, \((0,1)\) and \((1,\infty)\)

Answer 8

okok. i'll try.

Answer 9

ok the. i understand that x can't be negative. so than. we need to put it up. not trying to make it less then or equal to zero. right?

Answer 10

well i made a mistake. if \(0<x<1\) we have \(|1-x^2|=1-x^2\) so solve
\[1-x^2<2x\] get
\[x^2+2x-1>0\]
\[x>\sqrt{2}-1\]

Answer 11

right, and you have to solve two inequalities depending on whether x is in the interval \((0,1)\) or \((1,\infty)\)

Answer 12

on \((1,\infty)\) you have \(|1-x^2|=x^2-1\) so you have to solve
\[x^2-1<2x\] or
\[x^2-2x-1<0\]