设函数f(x)在闭区间[0 a]上连续,在(0 a)内可导,且f(0,a)=0,证明:在(0,a)内至少存在一点s,使f(s)+sf’(s)=0 更正 :且f(0 a)=0改为 f(a)=0
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![设函数f(x)在闭区间[0 a]上连续,在(0 a)内可导,且f(0,a)=0,证明:在(0,a)内至少存在一点s,使f(s)+sf’(s)=0 更正 :且f(0 a)=0改为 f(a)=0](/uploads/image/z/8714116-28-6.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E5%9C%A8%E9%97%AD%E5%8C%BA%E9%97%B4%5B0+a%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%EF%BC%880+a%EF%BC%89%E5%86%85%E5%8F%AF%E5%AF%BC%2C%E4%B8%94f%EF%BC%880%2Ca%EF%BC%89%3D0%2C%E8%AF%81%E6%98%8E%EF%BC%9A%E5%9C%A8%EF%BC%880%2Ca%EF%BC%89%E5%86%85%E8%87%B3%E5%B0%91%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9s%2C%E4%BD%BFf%EF%BC%88s%EF%BC%89%EF%BC%8Bsf%E2%80%99%EF%BC%88s%EF%BC%89%EF%BC%9D0+%E6%9B%B4%E6%AD%A3+%EF%BC%9A%E4%B8%94f%EF%BC%880+a%EF%BC%89%3D0%E6%94%B9%E4%B8%BA+f%EF%BC%88a%EF%BC%89%3D0)
设函数f(x)在闭区间[0 a]上连续,在(0 a)内可导,且f(0,a)=0,证明:在(0,a)内至少存在一点s,使f(s)+sf’(s)=0 更正 :且f(0 a)=0改为 f(a)=0
设函数f(x)在闭区间[0 a]上连续,在(0 a)内可导,且f(0,a)=0,
证明:在(0,a)内至少存在一点s,使f(s)+sf’(s)=0
更正 :且f(0 a)=0改为 f(a)=0
设函数f(x)在闭区间[0 a]上连续,在(0 a)内可导,且f(0,a)=0,证明:在(0,a)内至少存在一点s,使f(s)+sf’(s)=0 更正 :且f(0 a)=0改为 f(a)=0
思路:看到题目的模样,有两个联想
1.( sf(s) )' = 0,
2.罗尔定理.
证明:
构造函数g(x) = xf(x),易知g'(x) = f(x) + xf'(x)
由题知,g(x)在[0,a]上连续,在(0,a)可导,而且
g(0) = g(a) = 0
于是,由罗尔定理,在(0,a)内至少存在一点s,使g'(s) = 0
证毕.
设F(x)=x*f(x)
F(0)=0且F(a)=0,
运用罗尔定理
对F(x)求导即F‘(x)=0;
F(x)=f(x)+xf'(x)=0
证毕
令F(x)=xf(x),则F(x)在闭区间[0 a]上连续,在(0,a)内可导,且F(0)=F(a)=0,根据罗尔定理,在(0,a)内至少存在一点s,使F'(S)=0。而F'(S)=f(s)+sf’(s)。证毕