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By Kiran Kedlaya

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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.

X5 be positive reals such that (x2i+1 − xi+3 xi+5 )(x2i+2 − xi+3 xi+5 ) ≤ 0 for i = 1, . . , 5, where xn+5 = xn for all n. Prove that x1 = · · · = x5 . 4. (USAMO 1979/3) Let x, y, z ≥ 0 with x + y + z = 1. Prove that 1 x3 + y 3 + z 3 + 6xyz ≥ . 4 34 5. (Taiwan, 1995) Let P (x) = 1 + a1 x + · · · + an−1 xn−1 + xn be a polynomial with complex coefficients. Suppose the roots of P (x) are α1 , α2 , . . , αn with |α1 | > 1, |α2 | > 1, . . , |αj | > 1 and |αj+1 | ≤ 1, |αj+2 | ≤ 1, . . , |αn | ≤ 1.

The first condition implies the second because all eigenvalues of a symmetric matrix are real. ) Theorem 27 (Hessian test). A twice differentiable function f (x1 , . . , xn ) is convex in a region if and only if the Hessian matrix Hij = ∂2 ∂xi ∂xj 30 is positive definite everywhere in the region. Note that the Hessian is symmetric because of the symmetry of mixed partials, so this statement makes sense. Proof. The function f is convex if and only if its restriction to each line is convex, and the second derivative along a line through x in the direction of y is (up to a scale factor) just Hy · y evaluated at x.

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