a necessary (but not sufficient) condition for f to have a local maximum or minimum at a, is that f'(a) = 0.
in the case at hand, f'(x) = 1/(x-1)^2, which is never 0. so if f has a maximum or minimum, it must be global.
if ±infinity doesn't count, f has no minimum. f(x) gets very small as |x| gets large, but it is never 0. it has a horizontal asymptote (the x-axis).
f(x) is unbounded in any neighborhood of 1, and undefined at 1, it has a vertical asymptote at x = 1.
if we changed variables, by letting u = x-1, f(u) = 1/u. so the graph of f looks just like the graph of g(x) = 1/x, but "shifted right by 1".