连续求导两次,将上述条件代入得 (x0-a)2+f(x)-b2=p2 (x-a)+[f(x)-bf(x0)=0 l+|f"(x0)2+|f(x0)-bf"(x0)=0 解得 +[f(x0 f"(x0) b=∫(x0)+ l+[f"(x0) l1+f(x0) f"(x)连续求导两次，将上述条件代入得 2 2 0 2 0 (x − a) +[ f (x ) − b] =  (x0 − a) + [ f (x0 ) − b]f (x0 ) = 0 1 [ ( )] [ ( 0 ) ] ( 0 ) 0 2 + f  x0 + f x − b f  x = 解得 ( ) 1 [ ( )] ( ) 0 2 0 0 0 f x f x a x f x  +  = −  ( ) 1 [ ( )] ( ) 0 2 0 0 f x f x b f x  +  = + | ( )| [1 ( )] 0 2 3 0 2 f x f x  +   =