Solution provided by

Bunuel is self-explanatory. But I saw few posters were having issue in understanding how to get to \(x<−1\) or \(0<x<1\) from \(x(x−1)(x+1)<0\)

To break any equality of this type - First, we will find the trigger point. Trigger points are the point on which equality will give output = 0 .

So in this equation, we have 3 i.e. \(x=0\) , \(x=1\) and \(x=-1\)

Now we will mark these points on number line.

___-1____0_____1____

Here we are having 4 regions

1. \(x <-1\)

2. \(-1<x<0\)

3. \(0<x<1\)

4. \(x>1\)

Second step , test one value from each region and see if you get the value < 0

\(x(x−1)(x+1)<0\)

1. \(x <-1\) ---> Put x = -2 ; -6 < 0 ;

So this range is valid .

2. -\(1<x<0\) ---> Put x = -1/2 ; Some +ve < 0 ;

So this range is not valid.

3. \(0<x<1\) ---> Put x = 1/2 ; Some -ve value < 0 ;

So this range is valid .

4.\(x>1\) ---> Put x =2 ; SOme +ve value < 0 ;

So this range is not valid.

So from here we get 2 ranges \(x <-1\) or \(0<x<1\)

Same way try for Statement 2 .

Please +1 kudos If this post helps.