f(x)在[0,1]上有二阶导数 f(0)=f(1)=0 f"(x)的绝对值≤M求证 f'(x)的绝对值≤0.5M
问题描述：
f(x)在[0,1]上有二阶导数 f(0)=f(1)=0 f"(x)的绝对值≤M
求证 f'(x)的绝对值≤0.5M
答
任取x,由泰勒公式：f(0)=f(x)+f'(x)(-x)+f''(a)x^2/2f(1)=f(x)+f'(x)(1-x)+f''(b)(1-x)^2/2x相减得：0=f'(x)+f''(b)(1-x)^2/2-f''(a)x^2/2|f'(x)|=|f''(b)(1-x)^2/2-f''(a)x^2/2|《0.5M((1-x)^2+x^2)现考虑g(x)=((1-...