Question

At where f(x)=|(x^2)-9| is differentiable?

Answer 1

Everywhere except x=3 and x=-3

Explanation:

Since `x^2-9` is a polynom it is differentiable everywhere. Then, in first approximation `|x^2-9|` is differentiable everywhere except when `x^2-9=0`, which is `x=+-3`.
Now we must calculate derivates at the points `-3^+, -3^-` and `3^+, 3^-`, and see if the function derivate is continuous.

If `x^2-9>0`, `f(x)=x^2-9`, and `f'(x)=2x`

If `x^2-9<0`, `f(x)=-x^2+9`, and `f'(x)=-2x`

When `x=3`:

`f'(3^-)=-2*3=-6`
`f'(3^+)=2*3=6`
The values are different, therefore f(x) is not differentiable in x=3.

The same conclusion could be obtained when x=-3.