A is less than B by Kedlaya K.S.

By Kedlaya K.S.

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Xn ) = λ (x1 , . . , xn ). ∂xi ∂xi Putting these conditions together with the constraint on g, one may be able to solve and thus put restrictions on the locations of the extrema. ) It is even more critical here than in the one-variable case that the Lagrange multiplier condition is a necessary one only for an interior extremum. Unless one can prove that the given function is convex, and thus that an interior extremum must be a global one, one must also check all boundary situations, which is far from easy to do when (as often happens) these extend to infinity in some direction.

We say f is continuous at x if limt→0 f (x + t) = f (x). If y is any vector and x is in the domain of f , we say the directional derivative of f along x in the direction y exists and equals fy (x) if f (x + ty) − f (x) fy (x) = lim . t→0 t If f is written as a function of variables x1 , . . , xn , we call the directional derivative along the i-th standard basis vector the partial derivative of f with respect to i and denote it by ∂f . In other words, the partial derivative is the derivative of f as a function of x i along, ∂xi regarding the other variables as constants.

F (t) ≤ f (x)) for all x in some open interval containing t. Theorem 25. If t is a local extremum for f and f is differentiable at t, then f (t) = 0. Corollary 26 (Rolle). If f is differentiable on the interval [a, b] and f (a) = f (b) = 0, then there exists x ∈ [a, b] such that f (x) = 0. So for example, to find the extrema of a continuous function on a closed interval, it suffices to evaluate it at • all points where the derivative vanishes, • all points where the derivative is not defined, and • the endpoints of the interval, since we know the function has global minima and maxima, and each of these must occur at one of the aforementioned points.

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