利用Roll定理构造函数设函数f(x)在区间[0,1]上连续,在(0,1)内可导,且f(0)=f(1)=0,f(1/2)=1,试证:(1)存在η∈(0.5,1)使得f(η)=η(2)对任意实数λ,必存在ξ∈(0,η),使得f'(ξ)-λ[f(ξ)-ξ]=1(这里的η是第一问的
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![利用Roll定理构造函数设函数f(x)在区间[0,1]上连续,在(0,1)内可导,且f(0)=f(1)=0,f(1/2)=1,试证:(1)存在η∈(0.5,1)使得f(η)=η(2)对任意实数λ,必存在ξ∈(0,η),使得f'(ξ)-λ[f(ξ)-ξ]=1(这里的η是第一问的](/uploads/image/z/15033251-11-1.jpg?t=%E5%88%A9%E7%94%A8Roll%E5%AE%9A%E7%90%86%E6%9E%84%E9%80%A0%E5%87%BD%E6%95%B0%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5B0%2C1%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%280%2C1%29%E5%86%85%E5%8F%AF%E5%AF%BC%2C%E4%B8%94f%280%29%3Df%281%29%3D0%2Cf%281%2F2%29%3D1%2C%E8%AF%95%E8%AF%81%EF%BC%9A%281%29%E5%AD%98%E5%9C%A8%CE%B7%E2%88%88%280.5%2C1%29%E4%BD%BF%E5%BE%97f%28%CE%B7%29%3D%CE%B7%282%29%E5%AF%B9%E4%BB%BB%E6%84%8F%E5%AE%9E%E6%95%B0%CE%BB%2C%E5%BF%85%E5%AD%98%E5%9C%A8%CE%BE%E2%88%88%280%2C%CE%B7%29%2C%E4%BD%BF%E5%BE%97f%27%28%CE%BE%29-%CE%BB%5Bf%28%CE%BE%29-%CE%BE%5D%3D1%EF%BC%88%E8%BF%99%E9%87%8C%E7%9A%84%CE%B7%E6%98%AF%E7%AC%AC%E4%B8%80%E9%97%AE%E7%9A%84)
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