Engineering Calculus Notes 372

Engineering Calculus Notes 372 - K > 0 be the...

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360 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION Using Lemma 3.8.2 and Taylor’s theorem (Proposition 3.7.2 ), we can show that a critical point with defnite Hessian is a local extremum. Proposition 3.8.3. Suppose f is a C 2 function and −→ a is a critical point for f where the Hessian form d 2 −→ a f is deFnite. Then f has a local extremum at −→ a : If d 2 −→ a f is positive deFnite, then f has a local minimum at −→ a ; If d 2 −→ a f is negative deFnite, then f has a local maximum at −→ a . Proof. The Fact that the quadratic approximation T 2 −→ a f ( −→ x ) has second order contact with f ( −→ x ) at −→ x = −→ a can be written in the Form f ( −→ x ) = T 2 −→ a f ( −→ x ) + ε ( −→ x ) b −→ x −→ a b 2 , where lim −→ x −→ a ε ( −→ x ) = 0 . Since −→ a is a critical point, d −→ a f ( −→ x ) = 0, so T 2 −→ a f ( −→ x ) = f ( −→ a ) + 1 2 d 2 −→ a f ( −→ x ) , or f ( −→ x ) f ( −→ a ) = 1 2 d 2 −→ a f ( −→ x ) + ε ( −→ x ) b△ −→ x b 2 . Suppose d 2 −→ a f is positive defnite, and let
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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 ( x ) 0 as x a , For b x b suciently small, we have | ( x ) | < K 4 and hence f ( x ) f ( a ) { K 2 ( x ) }b x b 2 > K 4 b x b 2 > or f ( x ) > f ( a ) For x n = a ( b x b suciently 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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