设 $f(x)$ 二阶连续可导, $f(0)=f(1)=0$, $dps{max_{0leq xleq 1}f(x)=2}$. 证明: $$ex min_{0leq xleq 1}f''(x)leq -16. eex$$
证明: 设 $$ex xiin (0,1),st f(xi)=max_{0leq xleq 1}f(x)=2 a f'(xi)=0. eex$$ 在 $xi$ 处由 Taylor 展式, $$eex ea 0=f(0)=f(xi)+f'(xi)(-xi)+cfrac{f''(eta)}{2}(-xi)^2,&0<eta<xi,\ 0=f(1)=f(xi)+f'(xi)(1-xi)+cfrac{f''(zeta)}{2}(1-xi)^2,&xi<zeta<1. eea eeex$$ 于是 $$ex f''(eta)=-cfrac{4}{eta^2},quad f''(zeta)=-cfrac{4}{(1-xi)^2}. eex$$ 若 $0<xileq cfrac{1}{2}$, 则 $$ex min_{0leq xleq 1}f''(x)leq f''(eta)leq -16; eex$$ 若 $cfrac{1}{2}<xi<1$, 则 $$ex min_{0leq xleq 1}f''(x)leq f''(zeta)=-16. eex$$