Advanced Calculus fi..

146 Advanced Calculus, Fifth EditionHorizontal~nflect~onpolntmlnFigure 2.15Critical points of function y = f (x).These two rules cover most cases of interest. It can of course happen thatf "(xo) = 0 at a critical point, in which case one must consider the higher derivatives.A repeated application of the previous reasoning gives the following generalrule:Let f '(xO) = 0, f ll(xO) = 0, ..., f ( n)(~o) = 0, but f ("+')(xo)# 0; then f (x) has arelative maximum at xo if n is odd and f '"+"(xo) < 0; f (x) has a relative minimumat xo if n is odd and f '"+"(xo) > 0; f (x) has neither relative maximum nor relativeminimum at xo but a horizontal inflection point at xo if n is even.The preceding discussion has been restricted to points xo within the intervalof definition of f (x) and to relative maxima and minima. The notion of relativemaximum and minimum can readily be extended to the endpoints a and b, andrules can be formulated in terms of derivatives. However, the interest in these pointsarises mainly in connection with the absolute maximum and minimum of f (x) onthe interval a 5 x 5 b. A function f (x) is said to have an absolute maximum Mfor a certain set of values of x if f (xo) = M for some xo of the set given andf (x) 5 M for all x of the set; an absolute minimum is defined similarly, with thecondition f (x) 1 M replacing the condition f (x) 5 M. The following theorem isthen fundamental.THEOREM If f (x) is continuous in the closed interval a 5 x I b, then f (x)has an absolute minimum MI and an absolute maximum M2 on this interval.The proof of this theorem requires a more profound analysis of the real numbersystem; see Section 2.23.It should be remarked that the inclusion of the end-values is essential for thistheorem, as the simple example y = x for 0 < x < 1 illustrates; here the functionhas no minimum or maximum on the set given. For the set 0 5 x 5 1, the absoluteminimum is 0, for x = 0, and the absolute maximum is 1, for x = 1. The exampley = tanx for -3712 < x < n/2 illustrates a function with no absolute minimumor maximum; in this case, adjoining values at the endpoints x = h /2 does nothelp.A function f (x) is said to be bounded for a given set of x if there is a constant Ksuch that I f (x)l 5 K on this set. The theorem above implies that iff (x) is continuousin the closed interval a 5 x 5 b, then f (x) is bounded in this interval; for K canbe chosen as the larger of JM1 1, JM2J. For example, let y = sin x for 0 5 x 5 n.