设函数f(x)在（0,1）上有三阶导数,f(0)=0,f(1)=0.5,f'(0.5)=0,证明存在0
问题描述：
设函数f(x)在（0,1）上有三阶导数,f(0)=0,f(1)=0.5,f'(0.5)=0,证明存在0=12
答
f(x)在x=0.5处泰勒展开：
f(0)=f(0.5)-f'(0.5)*0.5 + f''(0.5)/2 * 0.5² - f'''(ξ1)/3!*0.5³(0f(1)=f(0.5)+f'(0.5)*0.5 + f''(0.5)/2 * 0.5² + f'''(ξ2)/3!*0.5³(0.5两式相减得
f(1)-f(0)=2f'(0.5)*0.5+[f'''(ξ2)-f'''(ξ1)]/48
即f'''(ξ2)+f'''(ξ1)=24
所以max{f'''(ξ1),f'''(ξ2)}>=12
若f'''(ξ1)>f'''(ξ2)
则取a=ξ1,否则取a=ξ2
则|f'''(a)|>=12