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)
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).