# Find the absolute maximum and minimum values of in Visual Studio .NET Generating QR Code ISO/IEC18004 in Visual Studio .NET Find the absolute maximum and minimum values of

EXAMPLE Find the absolute maximum and minimum values of
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f (x) = x 3 5x 2 + 3x + 1 on [0, 1] and nd the arguments at which these values are achieved. The function is continuous everywhere; in particular, on [0, 1]. Since f (x) = 3x 2 10x + 3 is de ned for all x, the only critical numbers are the solutions of 3x 2 10x + 3 = 0 (3x 1)(x 3) = 0 3x 1 = 0 or x 3=0 x=3 x=3 3x = 1 or 1 or x= 3
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Hence, the only critical number in the open interval (0, 1) is 1 . Now construct Table 14-2: 3 f 1 3 1 5 1 3 1 2 1 +1= +1+1 5 +3 3 3 3 27 9 1 15 14 40 +2=2 = = 27 27 27 27 =
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f 0 = 03 5 0
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+3 0 +1=1
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2 f 1 = 13 5 1 + 3 1 + 1 = 1 5 + 3 + 1 = 0
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The absolute maximum is the largest value in the second column, 40 , and it is achieved at x = 1 . The absolute minimum is the 27 3 smallest value, 0, which is achieved at x = 1.
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Table 14-2 x 1/3 0 1 f (x) 40/27 1 0
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Solved Problems
14.1 Justify the tabular method for locating the absolute maximum and minimum of a function on a closed interval.
By the extreme-value theorem (Theorem 14.2), a function f continuous on [a, b] must have an absolute maximum and an absolute minimum on [a, b]. Let p be an argument at which the absolute maximum is achieved. Case 1: p is one of the endpoints, a or b. Then f (p) will be one of the values in our table. In fact, it will be the largest value in the table, since f (p) is the absolute maximum of f on [a, b]. Case 2: Case 3: p is not an endpoint and f (p) is not de ned. Then p is a critical number and will be one of the numbers c1 , c2 , . . . , ck in our list. Hence, f (p) will appear as a tabulated value, and it will be the largest of the tabulated values.
p is not an endpoint and f (p) is de ned. Since p is the absolute maximum of f on [a, b], f (p) f (x) for all x near p. Thus, f has a relative maximum at p, and Theorem 14.1 gives f (p) = 0. But then p is a critical number, and the conclusion follows as in Case 2. A completely analogous argument shows that the method yields the absolute minimum.
14.2 Find the absolute maximum and minimum of each function on the given interval:
(a) f (x) = 2x 3 5x 2 + 4x 1 on [ 1, 2] (b) f (x) = x2 + 3 x+1 on [0, 3]
(a) Since f (x) = 6x 2 10x + 4, the critical numbers are the solutions of: 6x 2 10x + 4 = 0 3x 2 5x + 2 = 0 (3x 2)(x 1) = 0 3x 2 = 0 or x 1=0 x=1 x=1 3x = 2 or 2 or x= 3
CHAP. 14]
MAXIMUM AND MINIMUM PROBLEMS
Thus, the critical numbers are 2 and 1, both of which are in ( 1, 2). Now construct Table 14-3: 3 f 2 3 =2 2 3 2 2 2 5 +4 3 3 3
16 60 72 27 1 16 20 8 + 1= + = 27 9 3 27 27 27 27 27
f 1 =2 1
5 1
+4 1 1=2 5+4 1=0
f 1 = 2 1 f 2 =2 2
5 1
+ 4 1 1 = 2 5 4 1 = 12
5 2
+ 4 2 1 = 16 20 + 8 1 = 3
Thus, the absolute maximum is 3, achieved at x = 2, and the absolute minimum is 12, achieved at x = 1. Table 14-3 x 2/3 1 1 2 f (x) 1/27 0 12 min 3 max
(b) f (x) = =