How do you solve #|x^2-1|<=3 and |x^2+1|>=0#, preferably by graphing?
1 Answer
Aug 9, 2018
Explanation:
The second inequality goes without saying.
The first means
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
the graph for