f(x)在a,b二阶可导,f'(a)=f'(b)=0,证明存在m属于(a,b),使|f"(m)|>=4÷(b-a)^2×|f(b)-f(a)

Question

f(x)在【a,b】二阶可导,f'(a)=f'(b)=0,证明存在m属于(a,b),使|f"(m)|
>=4÷(b-a)^2×|f(b)-f(a)

Answer 1

f(x)在a,b两点泰勒展开
f(x)=f(a)+f`(a)(x-a)+f``(ξ1)(x-a)^2/2=f(a)+f``(ξ1)(x-a)^2/2
f(x)=f(b)+f`(b)(x-b)+f``(ξ2)(x-b)^2/2=f(b)+f``(ξ2)(x-b)^2/2
取x=a+b/2 代入后两式相减得
8(f(b)-f(a))/(b-a)^2=f``(ξ1)-f``(ξ2)
所以|8(f(b)-f(a))/(b-a)^2|=|f``(ξ1)-f``(ξ2)|