设f(x)在[0,1]上可导,f’(x)>0,求φ(x)=∫01|f(x)一f(t)|dt的极值点.

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问题 设f(x)在[0,1]上可导,f’(x)>0,求φ(x)=∫01|f(x)一f(t)|dt的极值点.

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答案由f’(x)>0可知,f(x)单调增加.于是可知, [*] 所以φ(x)=∫0x[f(x)一f(t)]dt+∫x1[f(t)一f(x)]dt =xf(x)一∫0xf(t)dt+∫x1f(t)dt+xf(x)一f(x) =(2x-1)f(x)一∫0xf(t)dt一∫1xf(t)dt, φ’(x)=2f(x)+(2x一1)f’(x)一f(x)一f(x) =(2x一1)f’(x). [*]

解析 先去掉φ(x)的绝对值符号,再求极值.
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