Question

Is `f(x)=(-x^3+x^2-5x+6)/(x-2)` increasing or decreasing at `x=0`?

Answer 1

increasing at x = 0

Explanation:

To test if f(x) is increasing / decreasing :

• If f'(x) > 0 at x = 0 then increasing
• If f'(x) < 0 at x = 0 then decreasing

Differentiate f(x) using `color(blue)(" quotient rule")`

`f'(x) =((x-2)d/dx(-x^3+x^2-5x+6)-(-x^3+x^2-5x+6)d/dx(x-2))/(x-2)^2`

`=( (x-2)(-3x^2+2x-5) -(-x^3+x^2-5x+6) .1)/(x-2)^2`

`=( -3x^3+2x^2-5x+6x^2-4x+10+x^3-x^2+5x-6)/(x-2)^2`

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

# f'(0) = 4/4 = 1 > 0

hence f(x) is increasing at x = 0
graph{(-x^3+x^2-5x+6)/(x-2) [-10, 10, -5, 5]}