How do solve #(x+1)^2/(x^2+2x-3)<=0# and write the answer as a inequality and interval notation?

1 Answer
Dec 13, 2016

Answer:

The answer is # -3< x <1 #
or #x in ]-3,1[#

Explanation:

Let's factorise the denominator

#x^2+2x-3=(x-1)(x+3)#

Let #f(x)=(x+1)^2/((x-1)(x+3))#

The domain of #f(x)# is #D_f(x)=RR-{-3,1}#

The numerator is #>0, AA x in D_f(x)#

Now, we can do the sign chart

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

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

#color(white)(aaaa)##x-1##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##+#

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

So,

# f(x)>=0 # when # x in ] -3,1 [ # or #-3< x < 1 #

Let # graph{y=(x+1)^2/((x-1)(x+3)) [-12.66, 12.65, -6.33, 6.33]} #