Advanced Calculus fi..

Chapter 2 Differential Calculus of Functions of Several Variables 14514. The proof at the end of Section 2.15 uses properties of the double integral. Justify thefollowing:a) If g(x, y) is continuous on a square region R: xo 5 x 5 xo + h, yo 5 y 5 yo + h,and g(x, y) > 0 in R, thenb) If g(x, y) is continuous in domain D andfor every region R as in (a) contained in D, then g(x, y) GO. [Hint: Supposeg(xo, yo) > 0. Then by continuity, g(x, y) > 0 in some region R (Problem 7 followingSection 2.4). Hence by (a) the double integral over R could not be 0. If g(xo, yo) < 0,then apply the same reasoning to -g(x, y).]2.19 MAXIM AND MINIMA OF FUNCTIONSOF SEVERAL VARIABLESWe first recall the basic facts concerning maxima and minima of functions of onevariable. Let y = f (x) be defined and differentiable in a closed interval a _( x _( b,and let xo be a number between a and b: a < xo < b. The function f (x) is said tohave a relative maximum at xo iff (x) 5 f (xo) for x sufficiently close to xo. It followsfrom the very definition of the derivative that iff '(xo) > 0, then f (x) > f (xo) for allx > xo and sufficiently close to xo; similarly, if f '(xo) < 0, then f (x) > f (xo) forall x < xo and sufficiently close to xo. Hence at a relative maximum, necessarilyf '(xo) = 0. A relative minimum of f (x) is defined by the condition f (x) f (xo)for all x sufficiently close to xo. A reasoning such as the preceding enables one toconclude that at a relative minimum off (x), necessarily f '(xo) = 0.. The points xo at which fl(xo) = 0 are termed the critical points of f (x).Although every relative maximum and minimum occurs at a critical point, a criticalpoint need not give either maximum or minimum. This is illustrated by the functiony = x3, - 1 ( x _( 1, at x = 0. This function has a critical point at x = 0, butthe point is neither maximum nor minimum and is an example of what is termed ahorizontal injection point. This is illustrated in Fig. 2.15.Let xo be a critical point, so that f '(xo) = 0, and let it be assumed that f "(xo) > 0.Then f (x) has a relative minimum at xo. For by the Mean Value theorem whenx > xo, f (x)- f (xo) = f '(XI)(x - XO), where xo t xl < x. Since f "(xo) > 0,fl(x,) > 0 for XI =. xo and XI sufficiently close to xo. Hence f (x)- f (xo) > 0 forx > xo and x sufficiently close to xo. Similarly, f (x)- f (xo) > 0 for x < xo and xsufficiently close to xo. Therefore f (x) has a relative minimum at xo. An analogousreasoning applies when f "(xo) < 0. Accordingly, one can state the rules:,If f '(xo) = 0 and f "(xo) > 0, then f (x) has a relative minimum at xo; iff '(xo) = 0and f "(xo) < 0, then f (x) has a relative maximum at xo.