Question

Is `f(x)=(x^2-4x-2)/(x^2+1)` increasing or decreasing at `x=-3`?

Answer 1

`f(x)` is increasing at `x=-3.`

Explanation:

If `f'(-3)>0, f(x)` is increasing at that point.

If `f'(-3)<0, f(x)` is decreasing at that point.

So, first, let's determine `f'(x)` using the Quotient Rule:

`f'(x)=((x^2+1)d/dx(x^2-4x-2)-(x^2-4x-2)d/dx(x^2+1))/(x^2+1)^2`

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

There's not really much of a need for simplification since we're evaluating the derivative at a point only to determine whether it is positive or negative.

`f'(-3)=((9+1)(2(-3)-4)-(81+12-2)(-6))/(10^2)=7/50>0`

`f(x)` is increasing at `x=-3.`