Critical number and classification of extrema

First compute the derivative: $f'(x) = -2(x-3)$. Setting $f'(x) = 0$ gives $x = 3$ as the only critical number, since $f'$ is defined everywhere.

Since $f(x) = 5-(x-3)^2$ is a downward-opening parabola, $f'(x) > 0$ for $x < 3$ and $f'(x) < 0$ for $x > 3$, so by the First Derivative Test $x=3$ is a local (and in fact absolute) maximum with $f(3) = 5$. As $x \to \pm\infty$, $(x-3)^2 \to \infty$, so $f(x) \to -\infty$; the function decreases without bound in both directions, so there is no absolute minimum.