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#
