设f(x)在[a,b]上可微,f'(x)不等于0,0

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设f(x)在[a,b]上可微,f'(x)不等于0,0
设f(x)在[a,b]上可微,f'(x)不等于0,0

设f(x)在[a,b]上可微,f'(x)不等于0,0
对f(x)和x^2用Cauchy中值定理有,存在d,使得【f(b)--f(a)】/【b^2--a^2】=f'(d)/2d.
对f(x)用Lagrange中值定理有,存在c,使得【f(b)--f(a)】/(b--a)=f'(c).
两式比较（相除）即得结论.

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