For what values of #x# is #(x+2)/abs(x+2) <= 0# ?

1 Answer
Apr 8, 2017

Answer:

# x in (-2,-oo)#

Explanation:

Note that for #(x+2)/(abs(x+2))# to be defined #x!=-2#

Substitute #k=x+2#

#(x+2)/(abs(x+2)) = -1# (with the restriction that #x!=-2#)
#rArr k=-abs(k)#
This will be true for any non-negative value of #k# (except for the case that would result in #x=-2#)

That is #k <=0 # (excluding any case where #x=-2#)

#rArr x+2 <=0 rarr x <= -2 #(excluding #x=-2#)

#rArr x < -2color(white)("XX")#or equivalently#color(white)("XX")x in (-2,-oo)#