Let us solve the following rational inequality.
#f(x)={x+1}/{x^2+x-6} le 0#
Set the numerator equal to zero, and solve for #x#.
#x+1=0 => x=-1#
(Note: #f(-1)=0#)
Set the denominator equal to zero, and solve for #x#.
#x^2+x-6=(x+3)(x-2)=0 => x=-3,2#
(Note: #f(-3)# and #f(2)# are undefined.)
Using #x=-3,-1,2# above to split the number line into open intervals:
#(-infty,-3), (-3,-1),(-1,2)#, and #(2,infty)#
Using sample numbers #x=-4,-2,0,3# for each interval above, respectively, we can determine the sign of (LHS).
#f(-4)=-2<0 => f(x)<0# on #(-infty,-3)#
#f(-2)=1/4>0 => f(x)>0# on #(-3,-1)#
#f(0)=-1/6<0 => f(x)<0# on #(-1,2)#
#f(3)=2/3>0 => f(x)>0# on #(2,infty)#
Hence, #f(x) le 0# on #(-infty,-3)cup[-1,2)#.
(Note: #-1# is included since #f(-1)=0#.)
The graph of #y=f(x)# looks like:
I hope that this was helpful.
