f(x)在【0,1】上二阶可导,且f(0)=0,f(1)=1,f(x)在0的导数等于1,在1的导数等于2求证f(a)的二阶导=f(a)
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![f(x)在【0,1】上二阶可导,且f(0)=0,f(1)=1,f(x)在0的导数等于1,在1的导数等于2求证f(a)的二阶导=f(a)](/uploads/image/z/2558612-20-2.jpg?t=f%28x%29%E5%9C%A8%E3%80%900%2C1%E3%80%91%E4%B8%8A%E4%BA%8C%E9%98%B6%E5%8F%AF%E5%AF%BC%2C%E4%B8%94f%280%29%3D0%2Cf%281%29%3D1%2Cf%28x%29%E5%9C%A80%E7%9A%84%E5%AF%BC%E6%95%B0%E7%AD%89%E4%BA%8E1%2C%E5%9C%A81%E7%9A%84%E5%AF%BC%E6%95%B0%E7%AD%89%E4%BA%8E2%E6%B1%82%E8%AF%81f%28a%29%E7%9A%84%E4%BA%8C%E9%98%B6%E5%AF%BC%3Df%28a%29)
f(x)在【0,1】上二阶可导,且f(0)=0,f(1)=1,f(x)在0的导数等于1,在1的导数等于2求证f(a)的二阶导=f(a)
f(x)在【0,1】上二阶可导,且f(0)=0,f(1)=1,f(x)在0的导数等于1,在1的导数等于2求证f(a)的二阶导=f(a)
f(x)在【0,1】上二阶可导,且f(0)=0,f(1)=1,f(x)在0的导数等于1,在1的导数等于2求证f(a)的二阶导=f(a)
(1) 由f(x)在x = 0处连续,且lim{x → 0} f(x)/x存在,
有f(0) = lim{x → 0} f(x) = lim{x → 0} f(x)/x · lim{x → 0} x = 0.
又由lim{x → 0} f(x)/x = 1 > 0,根据极限保序性,存在0 < s < 1使f(s)/s > 0,进而有f(s) > 0.
同理,由lim{x → 1} f(x)/(x-1) = 2可得f(1) = 0,且存在0 < t < 1使f(t) < 0.
而f(x)在[0,1]连续,由介值定理,存在ξ ∈ (0,1)使f(ξ) = 0.
(2) 考虑函数g(x) = f(x)e^(-x),则g(x)在[0,1]连续且可导.
并有g(0) = g(ξ) = g(1) = 0.
在区间[0,ξ]上由Rolle定理,存在α ∈ (0,ξ),使g'(α) =0,即有f'(α)-f(α) = 0.
同理,存在β ∈ (ξ,1),使g'(β) =0,即有f'(β)-f(β) = 0.
再考虑函数h(x) = (f'(x)-f(x))e^x,则h(x)同样在[0,1]连续并可导.
又h(α) = h(β) = 0,在[α,β]上由Rolle定理,存在η ∈ (α,β)使h'(η) = 0.
代回h(x)的定义式即得f"(η)-f(η) = 0,也即f"(η) = f(η).