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

1 Answer
Aug 9, 2018

#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.