设函数f(x)在区间[0,1]上连续,在(0,1)上可导,且f(0)=f(1)=0,f(1/2)=1,求证:⑴存在η属于(1/2,1),使f(η)=η⑵对λ属于R,存在ξ属于(0,η),使f'(ξ)-λ(f(ξ)-ξ)=1已知函数f(x)在区间[0,1]上连续,在(0,1)上可导
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![设函数f(x)在区间[0,1]上连续,在(0,1)上可导,且f(0)=f(1)=0,f(1/2)=1,求证:⑴存在η属于(1/2,1),使f(η)=η⑵对λ属于R,存在ξ属于(0,η),使f'(ξ)-λ(f(ξ)-ξ)=1已知函数f(x)在区间[0,1]上连续,在(0,1)上可导](/uploads/image/z/5349996-36-6.jpg?t=%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%EF%BC%880%2C1%EF%BC%89%E4%B8%8A%E5%8F%AF%E5%AF%BC%2C%E4%B8%94f%280%29%3Df%281%29%3D0%2Cf%281%2F2%29%3D1%2C%E6%B1%82%E8%AF%81%EF%BC%9A%E2%91%B4%E5%AD%98%E5%9C%A8%CE%B7%E5%B1%9E%E4%BA%8E%281%2F2%2C1%29%2C%E4%BD%BFf%28%CE%B7%29%3D%CE%B7%E2%91%B5%E5%AF%B9%CE%BB%E5%B1%9E%E4%BA%8ER%2C%E5%AD%98%E5%9C%A8%CE%BE%E5%B1%9E%E4%BA%8E%280%2C%CE%B7%29%2C%E4%BD%BFf%27%28%CE%BE%29-%CE%BB%28f%28%CE%BE%29-%CE%BE%29%3D1%E5%B7%B2%E7%9F%A5%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%EF%BC%880%2C1%EF%BC%89%E4%B8%8A%E5%8F%AF%E5%AF%BC)
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