Let f : U → R, U ⊆ R n is open. if a → ∈ U is local Maxima or local minima, then a → is a critical point.a proof assumes ( ∇ f ) ( a → ) ≠ 0 → t , and therefore for small enough h ∈ R, h &gt; 0 we get: f ( a → + h ( ( ∇ f ) ( a → ) ) t ) &gt; f ( a → )which is in contradiction to maxima.why is have to be larger the f ( a → )

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Let   f  :  U  →  R,   U  ⊆      R    n   is open. if             a      →        ∈  U is local Maxima or local minima, then             a      →        is a critical point.a proof assumes   (  ∇  f  )  (            a      →        )  ≠                    0        →                           t      , and therefore for small enough   h  ∈  R,   h  &amp;gt;  0 we get:  f  (            a      →        +  h  (  (  ∇  f  )  (            a      →        )      )          t        )  &amp;gt;  f  (            a      →        )which is in contradiction to maxima.why is have to be larger the   f  (            a      →        )

martinmommy26nv8 · Accepted Answer

Formally, using Taylor&#039;s theorem:  f  (            a      →        +  h  (  (  ∇  f  )  (            a      →        )      )    t    )  =  f  (            a      →        )  +  h  ‖  (  ∇  f  )  (            a      →        )      ‖    2    +      R    2    (            a      →        ,  h  (  (  ∇  f  )  (            a      →        )      )    t    )where                     R        2            (                        a          →                    ,                        x          →                    )              ‖                        x          →                            ‖        2              →  0 as             x      →        →  0. Because of the vanishing remainder, you can choose   h small enough so that the right-hand side is larger than   f  (            a      →        ).

Let f:U rarr R, U sube Rn is open. if vec а in U is local Maxima or local minima, then vec а is a critical point.

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