Question #d7eb4

1 Answer
Narad T.
Dec 4, 2017

The solution is #x in (-oo,-2/3] uu (2,+oo)#

Explanation:

Let #f(x)=(3x+2)/(2x-4)#

Build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-2/3##color(white)(aaaa)####color(white)(aaaaaa)##2##color(white)(aaaa)####color(white)(aaaa)##+oo#

#color(white)(aaaa)##3x+2##color(white)(aaaa)##-##color(white)(aaaa)##0##color(white)(aa)####color(white)(aa)##+##color(white)(aaaa)####color(white)(aaaaaaa)##+#

#color(white)(aaaa)##2x-4##color(white)(aaaa)##-##color(white)(aaaa)####color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)####color(white)(aa)##+#

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

Therefore,

#f(x)>=0# when #x in (-oo,-2/3] uu (2,+oo)#

graph{(3x+2)/(2x-4) [-18.02, 18.03, -9.01, 9.01]}

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