How do you solve #x/(x-2)>=0#?

2 Answers
Mar 21, 2018

Answer:

The solution is #x in (-oo, 0] uu(2, +oo)#

Explanation:

Let #f(x)=x/(x-2)#

Build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##0##color(white)(aaaaaaaa)##2##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aaaa)##0##color(white)(aaaa)##+##color(white)(aaaaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaa)##-##color(white)(aaaa)####color(white)(aaaaa)##-##color(white)(aa)##||##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaa)##0##color(white)(aaaa)##-##color(white)(aa)##||##color(white)(aa)##+#

Therefore,

#f(x)>=0# when ##

graph{x/(x-2) [-10, 10, -5, 5]}

Mar 21, 2018

Answer:

# (-oo, 0]# U #(2, +oo)#

Explanation:

#x /(x - 2)≥0#

#x /(x - 2)≥0" : is true if" {("either", x ≥0 and x - 2 > 0),("or",x ≤ 0 and x - 2 < 0):}#

#x ≥0 and x - 2 > 0#
# x > 2#

#x ≤ 0 and x - 2 < 0#
#x ≤ 0 #

Answer: #x ≤ 0# OR # x > 2#
In interval notation: # (-oo, 0]# U #(2, +oo)#