Question

Is `f(x)=(-x^2-5x-2)/(x^2+1)` increasing or decreasing at `x=-3`?

Answer 1

Increasing.

Explanation:

Find the value of the derivative at `x=-3`. If it is positive, the function is increasing. If it's negative, the function is decreasing (at that point).

To find the derivative of the function, use the quotient rule.

`f'(x)=((x^2+1)d/dx[-x^2-5x-2]-(-x^2-5x-2)d/dx[x^2+1])/(x^2+1)^2`

Each of these derivatives can be found through the power rule:

`d/dx[-x^2-5x-2]=-2x-5`

`d/dx[x^2+1]=2x`

Plugging these back in yields

`f'(x)=((x^2+1)(-2x-5)+(x^2+5x+2)(2x))/(x^2+1)^2`

Distribution and simplification of the numerator gives a derivative of

`f'(x)=(5x^2+2x-5)/(x^2+1)^2`

We can now find the value of the derivative at `x=-3:`

`f'(-3)=(5(9)-6-5)/(9+1)^2=34/100=17/50`

Since this is `>0`, the function is increasing at `x=-3`.

We can check the graph of `f(x):`

graph{(-x^2-5x-2)/(x^2+1) [-5, 5, -5.89, 3]}