Question

数学函数求求求急急急急急在线等

已知函数F(x)=2lnx-x2-ax
(1)当a=0时,求函数f(x)的极值
(2)设f(x)的图像与x轴交于A(x1,0)B(x2,0)(x1<x2)线段AB的重点为C((x1+x2)/2,0),过点C作平行于y轴的直线交f(x)于点D.求证:函数f(x)在点D的切线与y轴不垂直

Answer 1

1) f'(x) = 2/x - 2x
when x = 1, f(1) = -1 which is maximum

2) for f(x) = 2lnx - x^2 -ax
f'(x) = 2/x - 2x - a
for f'(x) = 0, 2/x - 2x - a = 0
2x^2 + ax - 2 = 0
when x = [-a + sqrt(a^2 + 16)] / 4
f'(x) = 0
you can then check the distance of this x, which is not (x1+x2)/2