作业帮一条涉及中值定理的高数题,已知函数f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=λb,f(b)=λa,其中λ为不等于0的常数,证明存在ξ∈(a,b),使得f(ξ)=λξ存在两个不同点α β∈(s,b),使得f‘（α）*f’（β）=λ&
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/16 12:01:40
![一条涉及中值定理的高数题,已知函数f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=λb,f(b)=λa,其中λ为不等于0的常数,证明存在ξ∈(a,b),使得f(ξ)=λξ存在两个不同点α β∈(s,b),使得f‘（α）*f’（β）=λ&](/uploads/image/z/4055773-13-3.jpg?t=%E4%B8%80%E6%9D%A1%E6%B6%89%E5%8F%8A%E4%B8%AD%E5%80%BC%E5%AE%9A%E7%90%86%E7%9A%84%E9%AB%98%E6%95%B0%E9%A2%98%2C%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%28a%2Cb%29%E5%86%85%E5%8F%AF%E5%AF%BC%2C%E4%B8%94f%28a%29%3D%CE%BBb%2Cf%28b%29%3D%CE%BBa%2C%E5%85%B6%E4%B8%AD%CE%BB%E4%B8%BA%E4%B8%8D%E7%AD%89%E4%BA%8E0%E7%9A%84%E5%B8%B8%E6%95%B0%2C%E8%AF%81%E6%98%8E%E5%AD%98%E5%9C%A8%CE%BE%E2%88%88%28a%2Cb%29%2C%E4%BD%BF%E5%BE%97f%28%CE%BE%29%3D%CE%BB%CE%BE%E5%AD%98%E5%9C%A8%E4%B8%A4%E4%B8%AA%E4%B8%8D%E5%90%8C%E7%82%B9%CE%B1+%CE%B2%E2%88%88%28s%2Cb%29%2C%E4%BD%BF%E5%BE%97f%E2%80%98%EF%BC%88%CE%B1%EF%BC%89%2Af%E2%80%99%EF%BC%88%CE%B2%EF%BC%89%3D%CE%BB%26)
一条涉及中值定理的高数题,已知函数f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=λb,f(b)=λa,其中λ为不等于0的常数,证明存在ξ∈(a,b),使得f(ξ)=λξ存在两个不同点α β∈(s,b),使得f‘（α）*f’（β）=λ&
一条涉及中值定理的高数题,
已知函数f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=λb,f(b)=λa,其中λ为不等于0的常数,证明存在ξ∈(a,b),使得f(ξ)=λξ存在两个不同点α β∈(s,b),使得f‘（α）*f’（β）=λ²（注意是导数）
一条涉及中值定理的高数题,已知函数f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=λb,f(b)=λa,其中λ为不等于0的常数,证明存在ξ∈(a,b),使得f(ξ)=λξ存在两个不同点α β∈(s,b),使得f‘（α）*f’（β）=λ&
let g(x) = f(x) - λξ, 那么g(a) = λ(b-a), g(b) = λ(a-b), so g(a)= - g(b),
存在ξ∈(a,b),使得g(ξ)=0, that is , f(ξ)=λξ.
let h(x) = f(x)/λ, h(a) = b>h(b) = a, h(ξ)=ξ∈(a,b).
h‘（α）= [h(ξ)-h(a)]/(ξ-a) =(ξ-h(a))/(ξ-a) ,
h’（β）= [h(ξ)-h(b)]/(b-ξ) = (ξ-a)/(b-ξ),
so f‘（α）*f’（β）=λ²
.
g(x) = f(x) - λξ, 那么g(a) = λ(b-a), g(b) = λ(a-b), so g(a)= - g(b),
存在ξ∈(a,b),使得g(ξ)=0, that is , f(ξ)=λξ.
let h(x) = f(x)/λ, h(a) = b>h(b) = a, h(ξ)=ξ∈(a,b).
h‘（α）= [h(ξ)-h(a)]/(ξ-a) =(ξ-h(a))/(ξ-a) ,
h’（β）= [h(ξ)-h(b)]/(b-ξ) = (ξ-a)/(b-ξ),