How do you solve #(x(4-x))/(x+2)>=0# using a sign chart?

1 Answer
Nov 28, 2017

The solution is #x in (-oo, -2) uu [0, 4]#

Explanation:

Let

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

Let's build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-2##color(white)(aaaaaa)##0##color(white)(aaaaaa)##4##color(white)(aaaaa)##+oo#

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

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

#color(white)(aaaa)##4-x##color(white)(aaaaaaa)##+##color(white)(aa)##||##color(white)(aa)##+##color(white)(aa)####color(white)(aaa)##+##color(white)(aa)##0##color(white)(aa)##-#

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

Therefore,

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

graph{(x(4-x))/(x+2) [-38.63, 43.6, -7.9, 33.2]}