Finding critical numbers of p(x)

Apply the quotient rule: $p'(x) = \dfrac{2x(x-1) - (x^2-4)(1)}{(x-1)^2}$. Expanding the numerator gives $2x^2 - 2x - x^2 + 4 = x^2 - 2x + 4$... wait, let's recheck: $2x(x-1) = 2x^2 - 2x$, and subtracting $(x^2-4)$ gives $2x^2 - 2x - x^2 + 4 = x^2 - 2x + 4$. So $p'(x) = \dfrac{x^2 - 2x + 4}{(x-1)^2}$. Actually re-examining: this quadratic has discriminant $4 - 16 < 0$, so it has no real roots — this suggests the correct simplified form should be checked again carefully to identify true critical points.

Recomputing carefully: $p'(x) = \dfrac{2x(x-1) - (x^2-4)}{(x-1)^2} = \dfrac{2x^2-2x-x^2+4}{(x-1)^2} = \dfrac{x^2-2x+4}{(x-1)^2}$. Setting the numerator equal to zero, $x^2 - 2x + 4 = 0$, has discriminant $(-2)^2-4(1)(4) = 4-16=-12<0$, so there are no real solutions — meaning $p'(x)$ is never zero. Since $p'(x)$ is undefined only at $x=1$, which is excluded from the domain, this function actually has no critical numbers in its domain.

The numerator $x^2 - 2x + 4 = (x-1)^2 + 3$ is always positive for every real $x$, and the denominator $(x-1)^2$ is also always positive for $x \neq 1$. Therefore $p'(x) > 0$ for all $x$ in the domain, meaning $p$ is increasing on $(-\infty, 1)$ and on $(1, \infty)$.

Since $p'(x)$ never changes sign (it is always positive), there is no point where the derivative switches from positive to negative or vice versa. By the First Derivative Test, this means $p(x)$ has no local maximum or local minimum anywhere.

No, it does not. Because the derivative's numerator is always positive and the denominator is always positive, the derivative never switches from positive to negative or negative to positive. By the First Derivative Test, a sign change is required for a local extremum, so with no sign change anywhere, the function has no local maxima or minima.