How do you solve #(x+4)/(1-x)<=0# using a sign chart?

1 Answer
Oct 17, 2016

Answer:

#x in {(-oo,-4],(+1,+oo)}#
(see below for method using sign chart)

Explanation:

First consider the "boundary values" of #x# for #(x+4)/(1-x)#.

The "boundary values" are the values of #x# for which the numerator or denominator become equal to #0#.

For this example
#color(white)("XXX")x+4=0 rarr x=-4#
and
#color(white)("XXX")1-x=0 rarr x=1#

So our boundary values are #{-4,+1}#

and teh sign table looks like
#{: (,"||",(-oo,-4),"|",-4,"|",(-4:+1),"|",+1,"|",(+1:+oo),"|"), (,,,,,,,,,,,), (x+4,"||",-ve,"|",0,"|",+ve,"|",+ve,"|",+ve,"|"), (1-x,"||",+ve,"|",+ve,"|",+ve,"|",0,"|",-ve,"|"), (,,,,,,,,,,,), ((x+4)/(1-x),"||",-ve,"|",0,"|",+ve,"|","undef'n","|",-ve,"|"), (,,,,,,,,,,,), ((x+4)/(1-x)<=0,"||","True","|","True","|","False","|","undef'n","|","True","|") :}#

So #(x+4)/(1-x)<=0color(white)("X")# for #x<=-4color\(white)("X")# or #color(white)("X")x>1#