Is #f(x)=(-x^2+3x+2)/(x^2-1)# increasing or decreasing at #x=2#?

1 Answer
Jun 20, 2016

#f(x)# is decreasing at #x=2#

Explanation:

Whether a function #f(x)# is increasing or decreasing at, say #x=a#, is indicated by the value of #(df)/(dx)# at #x=a#. If it is positive the function is increasing and if it is negative, the function is decreasing.

As #f(x)=(-x^2+3x+2)/(x^2-1)#, using quotient rule

#(df)/(dx)=((-2x+3)(x^2-1)-2x(-x^2+3x+2))/(x^2-1)^2#

= #(-2x^3+2x+3x^2-3+2x^3-6x^2-4x)/(x^2-1)^2#

= #(-3x^2-2x-3)/(x^2-1)^2#

and at #x=2#, we have #(df)/(dx)=(-3xx2^2-2xx2-3)/(2^2-1)^2=-19/9#

Hence as #(df)/(dx)# is negative at #x=2#, #f(x)# is decreasing at #x=2#

graph{(-x^2+3x+2)/(x^2-1) [-8.62, 11.38, -4.16, 5.84]}