Question

高数，微分中值定理

Answer 1

由带lagrange余项的Taylor展式，存在a<e<c<f<b 使得f(a)-f(c)=f'(c)(a-c)+f''(e)(a-c)^2/2; f(b)-f(c)=f'(c)(b-c)+f''(f)(b-c)^2/2 消去f'(c) 所以[f(a)-f(c)]/(a-c) - [f(b)-f(c)]/(b-c)=f''(e)(a-c)/2 - f''(f)(b-c)/2
设M=maxf''(x) m=minf''(x) 则 (a-b)M/2=<f''(e)(a-c)/2 - f''(f)(b-c)/2<=(a-b)m/2
即 m=<[f''(e)(a-c) - f''(f)(b-c)]/(a-b)<=M 由导函数的介值性（达布定理）知存在ξ属于(a,b)使得
f''(ξ)=[f''(e)(a-c) - f''(f)(b-c)]/(a-b) 所以[f(a)-f(c)]/(a-c) - [f(b)-f(c)]/(b-c)=(a-b)f''(ξ)/2
整理后可得要证的式子

追问

解释下 (a-b)M/2=<f''(e)(a-c)/2 - f''(f)(b-c)/2<=(a-b)m/2，怎么来的。