f(x)=(ax^2-x)(lnx-1/2ax^2+x)

f(x)=(ax^2-x)(lnx-1/2ax^2+x)

The first is derivation
After deriving, we get f '(x) = (2ax-1) LNX (X & gt; 0)
(1) If a ≤ 0 X & gt; 0, then 2ax-1 & lt; 0 Let f '(x) = (2ax-1) LNX & lt; 0, then when 0 & lt; X & lt; 1, f' (x) & gt; 0; X & gt; 1, f '(x) & lt; 0
So f (x) increases on (0,1) and decreases on (1, + ∞)
(2) 0 & lt; a & lt; 1 / 2, Let f '(x) = (2ax-1) LNX = 0, then x = 1 / (2a) or x = 1, when 0 & lt; a & lt; 1 / 2, 1 / (2a) & gt; 1,
So when x belongs to (0,1), f '(x) & gt; 0, f (x) monotonically increases in (0,1); when x belongs to (1,1 / 2a), f' (x) & lt; 0, f (x) monotonically decreases in (1,1 / 2a); when X & gt; 1 / 2a, f '(x) & gt; 0, f (x) monotonically increases in (1 / 2a, + ∞)
(3) If f '(x) = (2ax-1) LNX = 0, then x = 1 / (2a) or x = 1, when a & gt; 1 / 2, 1 / (2a) & lt; 1,
So when x belongs to (0,1 / 2a), f '(x) & gt; 0, f (x) monotonically increases in (0,1 / 2a); when x belongs to (1 / 2a, 1), f' (x) & lt; 0, f (x) monotonically decreases in (1 / 2a, 1); when X & gt; 1, f '(x) & gt; 0, f (x) monotonically increases in (1, + ∞)
The discussion is over. When we discuss the classification, we should clarify our thinking. When we discuss a, we should first divide the interval of a, and then discuss it one by one, so that there will be no confusion