设g(x)=(x-b)f(x), åg(x)å¨[a,b]è¿ç», å¨(a,b)å¯å¯¼, ä¸g(a)=(a-b)f(a), g(b)=0.
ç±Lagrangeä¸å¼å®ç, åå¨(a,b)å
ç¹t, 使g'(t)=(g(a)-g(b))/(a-b)=f(a).
èg'(x)=((x-b)f(x))'=f(x)+(x-b)f'(x), äºæ¯æf(a)=f(t)+(t-b)f'(t).
æ´çå³å¾f'(t)=(f(t)-f(a))/(b-t).
温馨提示:答案为网友推荐,仅供参考