Let f ( x ) = x 2 + x g ′ ( 1 ) + g ′′ ( 2 ) and g ( x ) = f ( 1 ) x 2 + x f ′ ( x ) + f ′′ ( x ) , then f ( 3 ) + g ( 3 ) =

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Let g ′ ( 1 ) = a , g ′′ ( 2 ) = b , then f ( x ) = x 2 + a x + b , f ( 1 ) = 1 + a + b ​ g ( x ) = ( 1 + a + b ) x 2 + x ( 2 x + a ) + 2 = ( 3 + a + b ) x 2 + a x + 2 ​ g ′ ( x ) = 2 x ( 3 + a + b ) + a ; g ′ ( 1 ) = a ⇒ a + b + 3 = 0 ; g ′′ ( x ) = 2 ( 3 + a + b ) = 0 = b ​ ⇒ a = − 3 , b = 0 ; f ( x ) = x 2 − 3 x , g ( x ) = − 3 x + 2 ; f ( 3 ) + g ( 3 ) = 0 − 7 = − 7 ​

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