f(0)=f(c)-f'(c)*c+f''(m)*c^2/2f(1)=f(c)+f'(c)*(1-c)+f''(n)*(1-c)^2/2两式相减,得f'(c)=f(1)-f(0)-f''(m)*c^2/2+f''(n)*(1-c)^2/2所以|f'(c)|<|f(1)|+|f(0)|+|f''(m)|*c^2/2+|f''(n)|*(1-c)^2/2<2a+(b/2)
