设f(x)在[0,a]连续,在(0,a)内可导,且f(a)=0,证明;存在一点c属于(0,a),使c^2f(c)+2cf(c)=0

题目

设f(x)在[0,a]连续,在(0,a)内可导,且f(a)=0,证明;存在一点c属于(0,a),使c^2f(c)+2cf(c)=0
是c^2f(c)+2cf'(c)=0,我确定希望那位大师帮帮忙

答案

应该是c²f'(c)+2cf(c) = 0吧.设g(x) = x²f(x),则g(c)在[0,a]连续,在(0,a)可导,且g(0) = 0 = g(a).由罗尔定理,存在c∈(0,a)使g'(c) = 0,即有c²f'(c)+2cf(c) = 0.不可能是c²f(c)+2cf'(c) = 0.反例...