Calculus + Algebra = Calgebra!
Let be a monic quartic polynomial such that ,
, , ,
Then evaluate :
The answer is 252.
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1 solution
f ( 1 ) = 1 1 = ( 1 2 + 1 ) 2 + ( 1 + 5 ) ( 1 − 1 ) + 7 f ( 2 ) = 2 3 = ( 2 2 + 1 ) 2 + ( 2 + 5 ) ( 2 − 1 ) + 7 f ( 3 ) = 1 2 3 = ( 3 2 + 1 ) 2 + ( 3 + 5 ) ( 3 − 1 ) + 7 f ( 4 ) = 3 2 3 = ( 4 2 + 1 ) 2 + ( 4 + 5 ) ( 4 − 1 ) + 7 H e n c e , f ( x ) = ( x 2 + 1 ) 2 + ( x + 5 ) ( x − 1 ) + 7 = x 4 + 3 x 2 + 4 x + 3 f ′ ( x ) = 4 x 3 + 6 x + 4 a n d f ′ ′ ( x ) = 1 2 x 2 + 1 x → 1 lim ( f ′ ( x ) . f ′ ′ ( x ) ) = f ′ ( 1 ) f ′ ′ ( 1 ) = ( 1 4 ) ( 1 8 ) = 2 5 2