How do you solve #1/3 |1/2x-2| >= 2#?

Question

1208 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-18T13:39:40

The solution is #x in (-oo,-8] uu[16,+oo)#

The point to consider is

#1/2x-2=0#

#=>#, #1/2x=2#

#=>#, #x=4#

There are #2# intervals to consider

#I_1=(-oo,4)# and #I_2=(4,+oo)#

In the first interval

#1/3(-1/2x+2)-2>=0#

#-1/6x+2/3-2>=0#

#1/6x<=-4/3#

#x<=-8#

This solution #in I_1#

In the second interval

#1/3(1/2x-2)-2>=0#

#1/6x-2/3-2>=0#

#1/6x>=8/3#

#x>=16#

This solution #in I_2#

The solution is #x in (-oo,-8] uu[16,+oo)#

graph{1/3|1/2x-2|-2 [-13.97, 18.06, -6.42, 9.6]}