1, discuss the monotonicity of functions,
2. When a =-1/8,0
(1) analysis: ∫ function f (x) = ax 2-lnx
When a=0, f(x)=-lnx, which obviously decreases monotonously in the definition domain;
When a<0, f' (x) = 2ax- 1/x
When a>0, let f' (x) = 2ax-1/x = 0 = = >; x= 1/√(2a)
f''(x)=2a+ 1/x^2>； 0, f(x) takes the minimum value at 1/√(2a),
0<x< when 1/√(2a), f(x) decreases monotonically; X> when = 1/√(2a), f(x) increases monotonically;
(2) Prove that ∵ function f (x) =- 1/8x 2-lnx, f' (x) =-1/4x-1/x.
Let t ∈ (0 0,2) and f (t) =- 1/8t 2-lnt.
The tangent equation at point P(t, f(t) is y =-(1/4t+1/t) (x-t)-(1/8t2+LNT) =-(1/t).
Let g (x) =-1/8x2-lnx-[-(1/4t+1/t) x+1/8t2-lnt+1].
=- 1/8x^2+( 1/4t+ 1/t)x-lnx- 1/8t^2+lnt- 1
g '(x)=- 1/4x+( 1/4t+ 1/t)- 1/x = = & gt； g''(x)=- 1/4+ 1/x^2
Let g'' (x) =-1/4+1/x2 = 0 = = x1= 2, x2=-2 (s).
∴x=2 is the inflection point of the function g(x)
When 0
That is to say, the image of the function g(x) is an S-shape with monotonic decreasing, and the point (2, g(2)) is the turning point.
∴ When t ∈ (0,2), the tangent at point P(t, f(t) is far from the origin, and there must be a second intersection with curve y=f(x).
∴ The curve y=f(x) and its tangent at point P(t, f(t) have at least two different points in common.