Question

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

Answer 1

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]}