How do you solve and write the following in interval notation: #(x-4) (x+3) (2-x)<=0#?

1 Answer
Apr 15, 2017

Answer:

The solution is #x in [-3,2] uu [4,+oo)#

Explanation:

Let

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

We build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaa)##2##color(white)(aaaaa)##4##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x+3##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

#color(white)(aaaa)##x-4##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-#

Therefore,

#f(x)<=0# when #x in [-3,2] uu [4,+oo)#