Question

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

Answer 1

f(x) is decreasing at x = 1

Explanation:

To find if the function is increasing / decreasing at x = 1

Require to check the value of f'(x) at x = 1.

• If f')x) > 0 then increasing

• If f'(x) < 0 then decreasing

differentiate using `color(blue)(" quotient rule ")`

`f'(x) =( (x^2+3x) d/dx (-x^3-x^2-x+2) - (-x^3-x^2-x+2) d/dx(x^2+3x))/(x^2+3x)^2`

`= ((x^2+3x)(-3x^2-2x-1) - (-x^3-x^2-x+2)(2x+3))/(x^2+3x)^2`

`=( -3x^4-2x^3-x^2-9x^3-6x^2-3x-(-2x^4-3x^3-2x^3-3x^2-2x^2-3x+4x+6))/(x^2+3x)^2`

`=( -3x^4-11x^3-7x^2-3x+2x^4+5x^3+5x^2-4x-6)/(x^2+3x)^2`

`=( -x^4-6x^3-2x^2-7x-6)/(x^2+3x)^2`

`f'(1) = (-1-6-2-7-6)/(1+3)^2 = -22/16 = -11/8`

now f'(1) < 0 hence f(x) is decreasing at x = 1

The graph of f(x) illustrates decreasing at x = 1

graph{(-x3-x^2-x+2)/(x^2+3x) [-10, 10, -5, 5]}