Question

数学数学数学数学数学

Answer 1

(1)f'(x)=1/x-ax+b
f'(1)=1/1-a+b=0
b=a-1
(2)f'(x)=1/x-ax+b,把 b=a-1代入得
f'(x)=1/x-ax+a-1
=[-ax^2+(a-1)x+1]/x
=(ax+1)(-x+1)/x
推出
x∈(-∞,-1/a)∪(0,1)时，f'(x)＜0则f(x)是单调递减函数
x∈(-1/a,0)∪(1,+∞)时，f'(x)＞0则f(x)是单调递增函数
(3)当c≥1/2,f(x)单调递增
f(x)max=f(c+1/2)=ln(c+1/2)-(1/4)(c+1/2)^2+(c+1/2)
f(x)min=f(c)=lnc-(1/4)c^2+c
当0＜c＜1/2，f(x)单调递减
f(x)min=f(c+1/2)=ln(c+1/2)-(1/4)(c+1/2)^2+(c+1/2)
f(x)max=f(c)=lnc-(1/4)c^2+c

下面自己写吧！