设f(x)在[0,1]上有二阶连续导数,且满足f(1)=f(0)及|f''(x)|

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设f(x)在[0,1]上有二阶连续导数,且满足f(1)=f(0)及|f''''(x)|设f(x)在[0,1]上有二阶连续导数,且满足f(1)=f(0)及|f''''(x)|设f(x)在[0,1]上有二阶连续导

设f(x)在[0,1]上有二阶连续导数,且满足f(1)=f(0)及|f''(x)|
设f(x)在[0,1]上有二阶连续导数,且满足f(1)=f(0)及|f''(x)|<=M(x∈[[0,1],证明一切x∈[0,1],|f'(x)<=(M/2)

设f(x)在[0,1]上有二阶连续导数,且满足f(1)=f(0)及|f''(x)|
Taylor展式:对任意的x,
f(0)=f(x)+f'(x)(0-x)+f''(c1)(0-x)^2/2,
f(1)=f(x)+f'(x)(1-x)+f''(c2)(1-x)^2/2.
两式相减,得
f'(x)=f''(c1)x^2/2-f''(c2)(1-x)^2/2,
取绝对值并利用条件得
|f'(x)|<=M/2(x^2+(1-x)^2)<=M/2.
最后的不等式是因为x^2+(1-x)^2在[0,1]
上的最大值是1.

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