Question

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

Answer 1

Increasing.

Explanation:

Find the first derivative at the given point. If the value is positive, the function is increasing. If the value is negative the function is decreasing.

Finding `f'(x)` will require the quotient rule:

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

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

`=(3x^3+13x^2+11x-3-(x^3+2x^2-x-2))/(x+3)^2`

`=(2x^3+11x^2+12x-1)/(x+3)^2`

Now, calculate the derivative at the point.

`f'(-2)=(2(-8)+11(4)+12(-2)-1)/(-2+3)^2`

`=(-16+44-24-1)/1`

`=3`

Since `3>0`, the function is increasing when `x=-2`.