Question

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

Answer 1

Strictly speaking it is neither. `f(1)=0` is a constant.

Explanation:

However, `f'(x) = 2(x-2)(x-1)+(x-2)^2`,

so `f'(1)=1`.

Note that `f'` is continuous near `1`. So, `f'(x)` is positive near `1` and `f(x)` is increasing near `1`.

Note
For a number `a`, the phrase "near `a`" means "in some open interval containing `a`"