By Kedlaya K.S.
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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. If the interval is open or infinite at either end, one must also check the limiting behavior of the function there. A special case of what we have just said is independently useful: if a function is positive at the left end of an interval and has nonnegative derivative on the interval, it is positive on the entire interval.
1 Quick reference Here’s a handy reference guide to the techniques we’ve introduced. 2 Additional problems Here is an additional collection of problems covering the entire range of techniques we have introduced, and one or two that you’ll have to discover for yourselves! 2 1. Let x, y, z > 0 with xyz = 1. Prove that x + y + z ≤ x2 + y 2 + z 2 . 2. The real numbers x1 , x2 , . . , xn belong to the interval [−1, 1] and the sum of their cubes is zero. Prove that their sum does not exceed n/3. 3. (IMO 1972/2) Let x1 , .
Equivalently, f is convex if its restriction to any line is convex. Of course, we say f is concave if −f is convex. The analogue of the second derivative test for convexity is the Hessian criterion. A symmetric matrix M (that is, one with Mij = Mji for all i, j) is said to be positive definite if M x·x > 0 for all nonzero vectors x, or equivalently, if its eigenvalues are all real and positive. (The first condition implies the second because all eigenvalues of a symmetric matrix are real. ) Theorem 27 (Hessian test).
A is less than B by Kedlaya K.S.