中值定理证明函数f(x)在【0,1】连续,在(0,1)可导,f(0)=0,且在(0,1)内f(x)!=0.证明至少存在一点ξ∈(0,1)使得3f'(ξ)/f(ξ) = 4f'(1-ξ)/f(1-ξ)
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![中值定理证明函数f(x)在【0,1】连续,在(0,1)可导,f(0)=0,且在(0,1)内f(x)!=0.证明至少存在一点ξ∈(0,1)使得3f'(ξ)/f(ξ) = 4f'(1-ξ)/f(1-ξ)](/uploads/image/z/8795528-8-8.jpg?t=%E4%B8%AD%E5%80%BC%E5%AE%9A%E7%90%86%E8%AF%81%E6%98%8E%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%E3%80%900%2C1%E3%80%91%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%EF%BC%880%2C1%EF%BC%89%E5%8F%AF%E5%AF%BC%2Cf%EF%BC%880%EF%BC%89%3D0%2C%E4%B8%94%E5%9C%A8%EF%BC%880%2C1%EF%BC%89%E5%86%85f%EF%BC%88x%EF%BC%89%21%3D0.%E8%AF%81%E6%98%8E%E8%87%B3%E5%B0%91%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9%CE%BE%E2%88%88%280%2C1%29%E4%BD%BF%E5%BE%973f%27%28%CE%BE%29%2Ff%28%CE%BE%29+%3D+4f%27%281-%CE%BE%29%2Ff%281-%CE%BE%29)
中值定理证明函数f(x)在【0,1】连续,在(0,1)可导,f(0)=0,且在(0,1)内f(x)!=0.证明至少存在一点ξ∈(0,1)使得3f'(ξ)/f(ξ) = 4f'(1-ξ)/f(1-ξ)
中值定理证明
函数f(x)在【0,1】连续,在(0,1)可导,f(0)=0,且在(0,1)内f(x)!=0.证明至少存在一点ξ∈(0,1)使得
3f'(ξ)/f(ξ) = 4f'(1-ξ)/f(1-ξ)
中值定理证明函数f(x)在【0,1】连续,在(0,1)可导,f(0)=0,且在(0,1)内f(x)!=0.证明至少存在一点ξ∈(0,1)使得3f'(ξ)/f(ξ) = 4f'(1-ξ)/f(1-ξ)
设g(x) = [f(x)]^3[f(1-x)]^4
则,g(x)在[0,1]连续,在(0,1)可导.
g(0) = [f(0)]^3[f(1)]^4 = 0,
g(1) = [f(1)]^3[f(0)]^4 = 0 = g(0).
g'(x) = 3[f(x)]^2f'(x)[f(1-x)]^4 + 4[f(x)]^3[f(1-x)]^3f'(1-x)(-1)
= [f(x)]^2[f(1-x)]^3{3f'(x)f(1-x) - 4f(x)f'(1-x)}
由罗尔中值定理,至少存在一点ξ∈(0,1)使得
g'(ξ)=[f(ξ)]^2[f(1-ξ)]^3{3f'(ξ)f(1-ξ) - 4f(ξ)f'(1-ξ)} = 0,
但由于,在(0,1)内f(x)!=0.
因此,[f(ξ)]^2[f(1-ξ)]^3 不等于0,
故,必有,
3f'(ξ)f(1-ξ) - 4f(ξ)f'(1-ξ) = 0,
成立.
也就是,
3f'(ξ)f(1-ξ) = 4f(ξ)f'(1-ξ),
因此,有
3f'(ξ)/f(ξ) = 4f'(1-ξ)/f(1-ξ).
所以命题得证.