图解法原理 7= C4o-CALC40-CA f(Ca) r1=f(C)= T T A y=f(x)= C4o_x KCA) T -1/ CA1 CA 0 1 0 1 1 1 1 ( ) A A A A A A C C C C r f C  − − = = 0 1 1 1 ( ) A A A A C C r f C   = = − -1/ CA1 CA0 CA rA f(CA ) 0 ( ) CA x y f x   = = − 图解法原理 CA1