Fig. 14-3 it is clear that f has neither a relative maximum nor a relative minimum at x = 0. in .NET framework

Print QR Code 2d barcode in .NET framework Fig. 14-3 it is clear that f has neither a relative maximum nor a relative minimum at x = 0.

Fig. 14-3 it is clear that f has neither a relative maximum nor a relative minimum at x = 0.
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EXAMPLE Consider the function f (x) = x3 . Because f (x) = 3x2 , f (x) if and only if x = 0. But from the graph of f in
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In 23, a method will be given that often will enable us to determine whether a relative extremum actually exists when f (c) = 0.
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14.2 ABSOLUTE EXTREMA Practical applications usually call for nding the absolute maximum or absolute minimum of a function on a given set. Let f be a function de ned on a set E (and possibly at other points, too), and let c belong to E . Then f is said to achieve an absolute maximum on E at c if f (x) f (c) for all x in E . Similarly, f is said to achieve an absolute minimum on E at d if f (x) f (d) for all x in E . If the set E is a closed interval [a, b], and if the function f is continuous over [a, b] (see Section 10.3), then we have a very important existence theorem (which cannot be proved in an elementary way). Theorem 14.2: (Extreme-Value Theorem): Any continuous function f over a closed interval [a, b] has an absolute maximum and an absolute minimum on [a, b].
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MAXIMUM AND MINIMUM PROBLEMS
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EXAMPLES
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(a) Let f (x) = x + 1 for all x in the closed interval [0, 2]. The graph of f is shown in Fig. 14-4(a). Then f achieves an absolute maximum on [0, 2] at x = 2; this absolute maximum value is 3. In addition, f achieves an absolute minimum at x = 0; this absolute minimum value is 1. (b) Let f (x) = 1/x for all x in the open interval (0, 1). The graph of f is shown in Fig. 14-4(b). f has neither an absolute maximum nor an absolute minimum on (0, 1). If we extended f to the half-open interval (0, 1], then there is an absolute minimum at x = 1, but still no absolute maximum. x + 1 if 1 x < 0 (c) Let f (x) = 0 if x = 0 x 1 if 0 < x 1 See Fig. 14-4(c) for the graph of f . f has neither an absolute maximum nor an absolute minimum on the closed interval [ 1, 1]. Theorem 14.2 does not apply, because f is discontinuous at 0.
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Fig. 14-4
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Critical Numbers To actually locate the absolute extrema guaranteed by Theorem 14.2, it is useful to have the following notion. De nition: A critical number of a function f is a number c in the domain of f for which either f (c) = 0 or f (c) is not de ned. EXAMPLES
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(a) Let f (x) = 3x 2 2x + 4. Then f (x) = 6x 2. Since 6x 2 is de ned for all x, the only critical numbers are given by 6x 2 = 0 6x = 2 2 1 x= = 6 3 Thus, the only critical number is 1 . 3
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(b) Let f (x) = x 3 x 2 5x + 3. Then f (x) = 3x 2 2x 5, and since 3x 2 2x 5 is de ned for all x, the only critical numbers are the solutions of 3x 2 2x 5 = 0 (3x 5)(x + 1) = 0 3x 5 = 0 or x + 1 = 0 x = 1 x = 1 3x = 5 or 5 or x= 3 Hence, there are two critical numbers, 1 and 5 . 3 (c) Let f (x) = |x|. Thus, f (x) = x if 0 x x if x < 0. We already know from the example in Section 13.1 that f (0) is not de ned. Hence, 0 is a critical number. Since Dx (x) = 1 and Dx ( x) = 1, there are no other critical numbers.
Method for Finding Absolute Extrema Let f be a continuous function on a closed interval [a, b]. Assume that there are only a nite number of critical numbers c1 , c2 , . . . , ck of f inside [a, b]; that is, in (a, b). (This assumption holds for most functions encountered in calculus.) Tabulate the values of f at these critical numbers and at the endpoints a and b, as in Table 14-1. Then the largest tabulated value is the absolute maximum of f on [a, b], and the smallest tabulated value is the absolute minimum of f on [a, b]. (This result is proved in Problem 14.1.) Table 14-1 x c1 c2 . . . ck a b f (x) f (c1 ) f (c2 ) . . . f (ck ) f (a) f (b)
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