How do you solve and write the following in interval notation: #(x(x-2))/ (x^2-1x-30)<= 0#?

1 Answer
Jun 30, 2017

Answer:

The solution is #x in (-5,0] uu[2,6)#

Explanation:

Let's factorise the denominator

#(x^2-x_30)=(x-6)(x+5)#

Let #f(x)=(x(x-2))/((x-6)(x+5)#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaaaa)##0##color(white)(aaaaaaa)##2##color(white)(aaaaaa)##6##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+5##color(white)(aaaa)##-##color(white)(aaa)##||##color(white)(aa)##+##color(white)(aa)##0##color(white)(aa)##+##color(white)(aa)##0##color(white)(aa)##+##color(white)(aa)##||##color(white)(aa)##+#

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

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

#color(white)(aaaa)##x-6##color(white)(aaaa)##-##color(white)(aaa)##||##color(white)(aa)##-##color(white)(aa)##0##color(white)(aa)##-##color(white)(aa)##0##color(white)(aa)##-##color(white)(aa)##||##color(white)(aa)##+#

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

Therefore,

#f(x)>=0# when #x in (-5,0] uu[2,6)#

graph{(x(x-2))/((x-6)(x+5)) [-10, 10, -5, 4.995]}