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题目已知函数f(x)=ax2-2x+lnx.(1)若f(x)无极值点,但其导函数f’(x)有零点,求a的值;(略)(2)若f(x)有两个极值点,求a的取值范围,并证明f(x)的极小值小于-3/2.
(1) If f (x) has no extremum point but its derivative function f ’(x) has zero point, find the value of a; (2) If f (x) has two extremum points, find the range of a and prove that the minimum value of f (x) is less than -3/2.