Home > Algebra > Absolute Value Equations

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

Answer 1

# x in (-2,-oo)#

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)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Eli Assouline

The inequality (x+2)/abs(x+2) <= 0 holds when x < -2.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 3

Genesis Allen

The inequality (x + 2) / |x + 2| ≤ 0 holds true when x ∈ (-∞, -2] U {0}.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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