设f(x)在[a,b]上二阶可导且f'(a)=f'(b)=0,试证:存在c属于(a,b),使得If''(c)I>=4/(b-a)²*If(b)-f(a)I设f(x)在[0,1]上三阶连续可导,f(0)=1,f(1)=2,f'(1/2)=0,证明:至少存在一点c属于(0,1),使得If
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![设f(x)在[a,b]上二阶可导且f'(a)=f'(b)=0,试证:存在c属于(a,b),使得If''(c)I>=4/(b-a)²*If(b)-f(a)I设f(x)在[0,1]上三阶连续可导,f(0)=1,f(1)=2,f'(1/2)=0,证明:至少存在一点c属于(0,1),使得If](/uploads/image/z/2534652-36-2.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E4%BA%8C%E9%98%B6%E5%8F%AF%E5%AF%BC%E4%B8%94f%26%2339%3B%28a%29%3Df%26%2339%3B%28b%29%3D0%2C%E8%AF%95%E8%AF%81%EF%BC%9A%E5%AD%98%E5%9C%A8c%E5%B1%9E%E4%BA%8E%28a%2Cb%29%2C%E4%BD%BF%E5%BE%97If%26%2339%3B%26%2339%3B%28c%29I%26gt%3B%3D4%2F%28b-a%29%26%23178%3B%2AIf%28b%29-f%28a%29I%E8%AE%BEf%28x%29%E5%9C%A8%5B0%2C1%5D%E4%B8%8A%E4%B8%89%E9%98%B6%E8%BF%9E%E7%BB%AD%E5%8F%AF%E5%AF%BC%2Cf%280%29%3D1%2Cf%281%29%3D2%2Cf%26%2339%3B%281%2F2%29%3D0%2C%E8%AF%81%E6%98%8E%EF%BC%9A%E8%87%B3%E5%B0%91%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9c%E5%B1%9E%E4%BA%8E%280%2C1%29%2C%E4%BD%BF%E5%BE%97If%26%23)
设f(x)在[a,b]上二阶可导且f'(a)=f'(b)=0,试证:存在c属于(a,b),使得If''(c)I>=4/(b-a)²*If(b)-f(a)I设f(x)在[0,1]上三阶连续可导,f(0)=1,f(1)=2,f'(1/2)=0,证明:至少存在一点c属于(0,1),使得If
设f(x)在[a,b]上二阶可导且f'(a)=f'(b)=0,试证:存在c属于(a,b),使得If''(c)I>=4/(b-a)²*If(b)-f(a)I
设f(x)在[0,1]上三阶连续可导,f(0)=1,f(1)=2,f'(1/2)=0,证明:至少存在一点c属于(0,1),使得If'''(c)I>=24
求下列曲线的曲率半径:(1)r=aθ;(2)r=ae^(mθ)
设函数f(x),g(x)在[a,b]上连续,在(a,b)内具有二阶导数且存在相等的最大值,f(a)=g(a),f(b)=g(b),证明:存在c属于(a,b),使得f''(c)=g"(c)
已知函数f(x)在[0,1]上连续,在(0,1)内可导且f(0)=0,f(1)=1,证明:存在两个不同的点η,ζ属于(0,1),使得f'(η)f'(ζ)=1
假设函数f(x)和g(x)在[a,b]上存在二阶导数,并且g"(x)!=0,f(a)=f(b)=g(a)=g(b)=0,试证:在(a,b)内至少存在一点c,使f(c)/g(c)=f"(c)/g"(c)
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设f(x)在[a,b]上二阶可导且f'(a)=f'(b)=0,试证:存在c属于(a,b),使得If''(c)I>=4/(b-a)²*If(b)-f(a)I设f(x)在[0,1]上三阶连续可导,f(0)=1,f(1)=2,f'(1/2)=0,证明:至少存在一点c属于(0,1),使得If
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因为f'(a)=f'(b)=0 根据泰勒公式 f(x)=f(a)+[f''(ξ1)/2](x-a)^2 f(x)=f(b)+[f''(ξ2)/2](x-b)^2 取x= (b+a)/2得: 全部展开 因为f'(a)=f'(b)=0 根据泰勒公式 f(x)=f(a)+[f''(ξ1)/2](x-a)^2 f(x)=f(b)+[f''(ξ2)/2](x-b)^2 取x= (b+a)/2得: f((b+a)/2)=f(a)+ [f''(ξ1)/2]((b-a)/2)^2 f((b+a)/2)=f(b)+ [f''(ξ2)/2]((b-a)/2)^2 两式相减,得: f(b)-f(a)={[f''(ξ1)/2]-[f''(ξ2)/2]}((b-a)/2)^2 4 | f(b)-f(a)|/(b-a)^2=(1/2)|f''(ξ1)-f''(ξ2)|≤(1/2)[|f''(ξ1|+|f''(ξ2)|]≤max{|f''(ξ1)|,|f''(ξ2)|} 令c满足|f''(c)|=max{|f''(ξ1)|,|f''(ξ2)|} 则:|f''(c)|≥ 4 | f(b)-f(a)|/(b-a)^2
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