f(x)在[0,1]上有二阶导数 f(0)=f(1)=0 f"(x)的绝对值≤M
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/10 08:11:25
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
求证 f'(x)的绝对值≤0.5M
任取x,由泰勒公式:
f(0)=f(x)+f'(x)(-x)+f''(a)x^2/2
f(1)=f(x)+f'(x)(1-x)+f''(b)(1-x)^2/2
x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-x)^2+x^2),g'(x)=2(x-1)+2x=4x-2 x=1/2为极值点
比较g(0)=1,g(1)=1,g(1/2)=1/2知:g(x)《1
所以:|f'(x)|=《0.5M
f(0)=f(x)+f'(x)(-x)+f''(a)x^2/2
f(1)=f(x)+f'(x)(1-x)+f''(b)(1-x)^2/2
x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-x)^2+x^2),g'(x)=2(x-1)+2x=4x-2 x=1/2为极值点
比较g(0)=1,g(1)=1,g(1/2)=1/2知:g(x)《1
所以:|f'(x)|=《0.5M
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