Question

How do you solve `|x^2-1|<=3 and |x^2+1|>=0`, preferably by graphing?

Answer 1

`abs x <= 2`

Explanation:

The second inequality goes without saying.

`abs ( x^2 + 1 ) = x^2 +1 >= 1`

The first means

`- 3 <= x^2 - 1 <= 3 rArr x^2 <= 4 rArr -2 <= x <= 2.`
graph{(y -(x^2-4))((x-2)^2+y^2-0.001)((x+2)^2+y^2-0.001)=0[-2.1 2.1 -1 1]}

The x-intercept segments of the parabola `y = x^2 - 4` represent

the graph for `_2 <= x <= 2`. See plots at the ends.