How do you solve and write the following in interval notation: #(x-2)/(x+1) <=0#?

1 Answer
Jun 30, 2017

Answer:

The solution is #x in (-1,2]#

Explanation:

Let #f(x)=(x-2)/(x+1)#

We build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##-1##color(white)(aaaaaa)##2##color(white)(aaaaaaa)##+oo#

#color(white)(aaaa)##x+1##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aa)##0##color(white)(aaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aa)##0##color(white)(aaa)##+#

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

Therefore,

#f(x)<=0# when #x in (-1,2]#

graph{(x-2)/(x+1) [-14.24, 14.24, -7.12, 7.12]}