Local maxima/minima and removable discontinuities

There are no minimas at the interval (-∞, ∞ for this function, yes. However, for any interval [ε, ∞] or [-∞, -ε], 0 < ε, there would be a local minima. Removing the infinities from the intervals will also give a local maxima.

Noah

This is correct ^. Except shame on noah for the bolded part

(x^3)/x is only defined from (-inf, 0) U (0,inf)

y = x^3/x

dy/dx = 2x which can only equal zero if x = 0. Since x=0 isn't in the domain of this function, it cannot be considered for maxima/minima. Since critical points have failed to find max/mins then the only other candidates for max/min are endpoints of the domain. But none of those endpoints are included, thus there are no points which are candidates for max/mins.

TL;DR: To be a local maximum or a local minimum of a function, you have to be in the domain of the function. So no, it doesn't have a min at x=0.