设函数f(x)=(a+1)lnx+ax^2+1,a

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设函数f(x)=(a+1)lnx+ax^2+1,a

数学人气:701 ℃时间:2020-10-01 05:30:21
优质解答
f'(x)=(a+1)/x+2ax=(a+1+2ax^2)/x,对任意x1,x2∈(0,∞),都有|f(x1)-f(x2)|≥4|x1-x2|,∴|[f(x1)-f(x2)]/(x1-x2)|>=4,∴|f'(x)|>=4,x>0,∴|a+1+2ax^2|>=4x,∴a+1+2ax^2>=4x,或a+1+2ax^2=(4x-1)/(2x^2+1),或a0,g(x)...
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f'(x)=(a+1)/x+2ax=(a+1+2ax^2)/x,对任意x1,x2∈(0,∞),都有|f(x1)-f(x2)|≥4|x1-x2|,∴|[f(x1)-f(x2)]/(x1-x2)|>=4,∴|f'(x)|>=4,x>0,∴|a+1+2ax^2|>=4x,∴a+1+2ax^2>=4x,或a+1+2ax^2=(4x-1)/(2x^2+1),或a0,g(x)...