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# note8 - Math021 week 8 Example 8.1 Find all the relative...

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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 0 ( x ) = x ( e x - 2) > 0 x > ln 2 or x < 0 = 0 x = 0 or ln 2 < 0 0 < x < ln 2 f has a relative minimum at ln 2 and a relative maximum at 0. Concavity Definition 8.2 The graph of a function f is concave up (down) if f 0 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 0 ( x ) = - xe - x 2 / 2 , f 00 ( x ) = ( x 2 - 1) e - x 2 / 2 . Thus, f 0 ( x ) = - x ( e x 2 / 2 ) > 0 x < 0 = 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 > 0 x > 1 or x < - 1 = 0 x = 1 or - 1 < 0 - 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

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Theorem 8.6 (Second Derivative Test) Suppose that c is a critical point of a function f . 1. If f 00 ( c ) > 0 , f has a relative minimum at c . 2. If f 00 ( c ) < 0 , 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 0 is increasing over an interval including c . But we know that f 0 ( c ) = 0. Therefore, f 0 ( x ) 0 when x is a bit less than c and f 0 ( 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. Example 8.7 Find all the relative maximum and minimum of the function f ( x ) = 2 x 3 + 3 x 2 - 12 x + 7 . solution: Since f 0 ( x ) = 6( x + 2)( x - 1). The critical points of f are - 2 and 1. Now, f 00 ( x ) = 12 x + 6 so that f 00 ( - 2) = - 18 < 0, f 00 (1) = 18 > 0. Hence, f has a relative maximum at - 2 and a relative minimum at 1.
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