To find whether #f(x)=(-x^2+5x-2)/(x^2-1)# is increasing or decreasing at #x=-3#, we should find the value of #(df(x))/(dx)# at #x=3#.
#(df(x))/(dx)=((x^2-1)(-2x+5)-(-x^2+5x-2)(2x))/(x^2-1)^2# or
= #((-2x^3+2x+5x^2-5)-(-2x^3+10x^2-4x))/(x^2-1)^2# or
= #(-5x^2+6x-5)/(x^2-1)^2# and at #x=-3#, we have
#f'(3)=(-5xx(-3)^2+6xx(-3)-5)/((-3)^2-1)^2#
= #(-45-18-5)/64=-68/64#
As #(df(x))/(dx)# is negative at #x=-3#
#f(x)=(-x^2+5x-2)/(x^2-1)# is decreasing.
Note that in #(-5x^2+6x-5)#
= #-5(x^2-(6/5)x+(3/5)^2-(3/5)^2+1)# or
= #-5{(x-3/5)^2+16/25}# and hence negative for all #x#
Hence it is always decreasing.
graph{(-x^2+5x-2)/(x^2-1) [-10, 10, -5, 5]}
