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Unformatted text preview: Math021, week 8 Example 8.1 Find all the relative maximum and minimum of the function f ( x ) = xe x- e x- x 2 . solution: Since f ( x ) = x ( e x- 2) > x > ln2 or x < = 0 x = 0 or ln2 < 0 0 < x < ln2 f has a relative minimum at ln2 and a relative maximum at 0. Concavity Definition 8.2 The graph of a function f is concave up (down) if f is increas- ing (decreasing). Corollary 8.3 If f 00 > ( < )0 , the graph of f is concave up. Definition 8.4 f has a point of inflection at c if f changes its concavity at c . Example 8.5 Find the points of inflection of f ( x ) = e- x 2 / 2 and hence sketch its graph. solution: f ( x ) =- xe- x 2 / 2 , f 00 ( x ) = ( x 2- 1) e- x 2 / 2 . Thus, f ( x ) =- x ( e x 2 / 2 ) > x < = 0 x = 0 < 0 0 < x f has a relative maximum at 0. Moreover, f 00 ( x ) = ( x- 1)( x + 1) e- x 2 / 2 > x > 1 or x <- 1 = 0 x = 1 or- 1 <- 1 < x < 1 Thus, f has points of inflection at- 1 and 1. The graph of f is concave up on [1 , + ∞ ) and (-∞ ,- 1] and it is concave down on [- 1 , 1]. Finally, note that the graph of f has a horizontal asymptote y = 0. 1 Theorem 8.6 (Second Derivative Test) Suppose that c is a critical point of a function f . 1. If f 00 ( c ) > , f has a relative minimum at c . 2. If f 00 ( c ) < , f has a relative maximum at c . 3. If f 00 ( c ) = 0 , f may have a relative maximum, minimum or neither of them. proof: If f 00 ( c ) > 0, f 00 ( x ) > 0 when x is close to c . Thus, f is increasing over an interval including c . But we know that f ( c ) = 0. Therefore, f ( x ) ≤ 0 when x is a bit less than c and f ( x ) ≥ 0 when x is a bit greater than c . f has a relative minimum at c . We may deal with the remaining case in which f 00 ( c ) < 0 in a similar way....
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This note was uploaded on 02/13/2012 for the course MATH 021 taught by Professor Luxnu during the Fall '10 term at HKUST.
- Fall '10