设f(x)在[a,b]上连续可导,a>0 .证明:存在ξ,η∈(a,b),使得f'(ξ)=[(a+b)/2η]f‘(η)RT
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![设f(x)在[a,b]上连续可导,a>0 .证明:存在ξ,η∈(a,b),使得f'(ξ)=[(a+b)/2η]f‘(η)RT](/uploads/image/z/1136863-55-3.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%E5%8F%AF%E5%AF%BC%2Ca%3E0+.%E8%AF%81%E6%98%8E%EF%BC%9A%E5%AD%98%E5%9C%A8%CE%BE%2C%CE%B7%E2%88%88%28a%2Cb%29%2C%E4%BD%BF%E5%BE%97f%27%28%CE%BE%29%3D%5B%28a%2Bb%29%2F2%CE%B7%5Df%E2%80%98%28%CE%B7%29RT)
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