How do you solve #(x^2-4)/(3-x)>=0# using a sign chart?

1 Answer
Nov 19, 2016

Answer:

The answer is #x in] -oo,-2 ] uu [2, 3[#

Explanation:

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

The domain is #D_f=RR-{3} #

Let's do the sign chart

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

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

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

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

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

Therefore, #f(x)>=0#

when #x in] -oo,-2 ] uu [2, 3[#

graph{(x^2-4)/(3-x) [-34.51, 30.44, -18.83, 13.64]}