Is #f(x)=(-7x^3-x^2-2x+2)/(x^2+3x)# increasing or decreasing at #x=1#?

1 Answer
Feb 21, 2018

It is decreasing at #x=1#

Explanation:

Whether a function #f(x)# is increasing or decreasing depends on value of #f'(x)# at that point. If #f'(x)>0# i.e. it is positive, it is increasing and if #f'(x)<0# i.e. it is negative, it is decreasing.

Here #f(x)=(-7x^3-x^2-2x+2)/(x^2+3x)#

and using quotient formula, we have

#f'(x)=((x^2+3x)(-21x^2-2x-2)-(-7x^3-x^2-2x+2)(2x+3))/(x^2+3x)^2#

and #f'(1)=((1+3)(-21-2-2)-(-7-1-2+2)(2+3))/(1+3)^2#

= #(4*(-25)-(-8)*5)/16#

= #(-100+40)/16=-60/16=-3.75#

Hence #f(x)=(-7x^3-x^2-2x+2)/(x^2+3x)# is decreasing at #x=1#

graph{(-7x^3-x^2-2x+2)/(x^2+3x) [-4.58, 5.42, -4.1, 0.9]}