西安毛子律技大学ax2+bx+c, x>0f(x)=x≤0g(x),b= g-(0)c = g(0)3)利用 f"(0)= f(0)，而g(x)- g-(0)f"(O) = limg"(0)x-0X-0(2ax +b)-b*(0) = lim=2ax-0x-→0+得a==g"(0)1414 ( 0 ) − b = g  3) 利用 ( 0 ) ( 0 ), − + f  = f  0 ( ) (0) (0) lim0 −  −   = − → − − x g x g f x (0) − = g 0 (2 ) (0) lim0 −+ −  = → + + x ax b b f x = 2a 而 得 ( 0 ) 21 − a = g  c = g ( 0 ) f ( x ) = , 0 2 ax + bx + c x  g ( x ) , x  0