How do you solve #(x+3)/(x^2-5x+6)<=0#?

1 Answer
Jan 24, 2017

Answer:

The answer is #x in ]-oo,-3] uu ]-1,6[#

Explanation:

Let's factorise the denominator

#x^2-5x+6=(x+1)(x-6)#

Let #f(x)=(x+3)/((x+1)(x-6))#

The domain of #f(x)# is #D_f(x)=RR-{-1,6}#

Now, we can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaaaa)##-1##color(white)(aaaaaaaaa)##6##color(white)(aaaaaa)##+oo#

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

#color(white)(aaaa)##x+1##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaa)##||##color(white)(aaaa)##+##color(white)(aaa)##||##color(white)(aaaa)##+#

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

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

Therefore,

#f(x)<=0# when #x in ]-oo,-3] uu ]-1,6[#