设f(x)在[a,b]上可微,且f(a)=f(b)=0,证明在(a,b)内存在一点z,使f'(z)=f(z)
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![设f(x)在[a,b]上可微,且f(a)=f(b)=0,证明在(a,b)内存在一点z,使f'(z)=f(z)](/uploads/image/z/13224287-47-7.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E5%8F%AF%E5%BE%AE%2C%E4%B8%94f%28a%29%3Df%28b%29%3D0%2C%E8%AF%81%E6%98%8E%E5%9C%A8%28a%2Cb%29%E5%86%85%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9z%2C%E4%BD%BFf%27%28z%29%3Df%28z%29)
设f(x)在[a,b]上可微,且f(a)=f(b)=0,证明在(a,b)内存在一点z,使f'(z)=f(z)
设f(x)在[a,b]上可微,且f(a)=f(b)=0,证明在(a,b)内存在一点z,使f'(z)=f(z)
设f(x)在[a,b]上可微,且f(a)=f(b)=0,证明在(a,b)内存在一点z,使f'(z)=f(z)
令g(x)=e^(-x)f(x)
则g(a)=g(b)=0
所以存在z,使得
g'(z)=e^(-z)f'(z)-e^(-z)f(z)=0
即 f'(z)-f(z)=0
f'(z)=f(z)
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