Question

Over what intervals is `f(x)=ln(9-x^2)` increasing and decreasing?

Answer 1

It is increasing on `(-3,0)` and decreasing on `(0,3)`

Explanation:

`f(x)=ln(9-x^2)`

Observe first that `f` is defined (real valued) only when `9-x^2` is positive.

So the domain of `f` is `(-3,3)`

`f'(x) = (-2x)/(9-x^2)`

Recall that the domain of `f` includes only values of `x` for which `9-x^2` is positive.
Therefore, the sign of `f'` on the domain of `f` is the same as that of `-2x`, which is

positive for `-3 < x < 0`, so `f` is increasing on `(-3,0)`, and

negative for `0 < x < 3`, so `f` is decreasing on `(0,3)`.