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Unformatted text preview: ) f ( c )  ( vu ). [Hint: Add and substract f ( c )cf ( c ) to the left side.] Fix > 0. Since f is dierentiable at c , we have that for any x I with  xc  < f ( x )f ( c ) xcf ( c ) < / 2 . That is  f ( x )f ( c )f ( c )( xc ) < 2 ( xc ) . We apply this with x = u so that c < u c and x = v with c v < c + . Then,  f ( v )f ( u )( vu ) f ( c )   f ( v )f ( c )( vc ) f ( c )  +  f ( u )f ( c )( uc ) f ( c )  2  vc  + 2  uc   vu  . 1...
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 Spring '10
 MetCalfe
 Calculus

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