Unformatted text preview: K > 0 be the constant given in Lemma 3.8.2 , such that d 2 −→ a f ( △ −→ x ) ≥ K b△ −→ x b 2 . Since ε ( −→ x ) → 0 as −→ x → −→ a , For b△ −→ x b su±ciently small, we have  ε ( −→ x )  < K 4 and hence f ( −→ x ) − f ( −→ a ) ≥ { K 2 − ε ( −→ x ) }b△ −→ x b 2 > K 4 b△ −→ x b 2 > or f ( −→ x ) > f ( −→ a ) For −→ x n = −→ a ( b△ −→ x b su±ciently small). The argument when d 2 −→ a f is negative defnite is analogous (Exercise 4a ). An analogous argument (Exercise 4b ) gives Lemma 3.8.4. If d 2 −→ a f takes both positive and negative values at the critical point −→ x = −→ a of f , then f does not have a local extremum at −→ x = −→ a ....
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 Fall '08
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 Calculus, Critical Point, Optimization, local extremum

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