Finding critical numbers

Differentiate $g(x) = x^4 - 4x^3$ to get $g'(x) = 4x^3 - 12x^2$. This is a polynomial, so it is defined everywhere, meaning the only critical numbers come from solving $g'(x) = 0$.

Factor: $g'(x) = 4x^2(x - 3)$. Setting this equal to zero gives $x = 0$ and $x = 3$ as the critical numbers.

The critical numbers $x=0$ and $x=3$ divide the real line into three intervals: $(-\infty, 0)$, $(0, 3)$, and $(3, \infty)$. Since $g'(x) = 4x^2(x-3)$ and $4x^2 \geq 0$ always, the sign of $g'(x)$ matches the sign of $(x-3)$.

On $(-\infty, 0)$, pick $x=-1$: $g'(-1) = 4(1)(-4) = -16 < 0$, so $g$ is decreasing. On $(0,3)$, pick $x=1$: $g'(1) = 4(1)(-2) = -8 < 0$, so $g$ is still decreasing. On $(3,\infty)$, pick $x=4$: $g'(4) = 4(16)(1) = 64 > 0$, so $g$ is increasing.

Near $x=0$: $g'(x) < 0$ on both sides (for $x$ slightly less than 0 and slightly greater than 0, up to 3), since $g'$ is negative on both $(-\infty,0)$ and $(0,3)$. There is no sign change, so $x=0$ is neither a local max nor a local min.

Near $x=3$: $g'(x) < 0$ just to the left (on $(0,3)$) and $g'(x) > 0$ just to the right (on $(3,\infty)$), so $g'$ changes from negative to positive. By the First Derivative Test, $x=3$ is a local minimum, with value $g(3) = 81 - 108 = -27$.

No, the derivative does not change sign at that critical number. That critical number is x equals zero, where the derivative is negative on both sides, staying negative throughout the interval before and after it. Because there is no sign change, the First Derivative Test confirms x equals zero is neither a local maximum nor a local minimum.