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Fx lnx

(1)f(x)=lnx-ax+b/x 满足fx+f(1/x)=lnx-ax+b/x-lnx-a/x+b=(b-a)(x+1/x)=0, ∴b=a.f(x)=lnx-ax+a/x, f'(x)=1/x-a-a/x^2,f(1)=0,f'(1)=1-2a, ∴y=f(x)的图像在x=1处的切线:y=(1-2a)(x-1),它过点(0,-5), ∴-5=-1+2a,a=-2. (2)0

不祥

(0,√5/2+1/2) 解: 定义域:(0,+∞) f'(x) =[lnx-(x-1)²/2]' =1/x-(x-1) =[1-x(x-1)]/x =-(x²-x-1)/x =-[(x-1/2)²-5/4]/x 令f'(x)>0,解得:0

f(x)=(lnx)'=1/x ∴f'(x)=-1/x²

定义域为x>0, 且x≠1 f'(x)=[(x-1)/x-lnx]/(x-1)²=[x-1-xlnx]/[x(x-1)²] 令g(x)=x-1-xlnx g'(x)=1-lnx-1=-lnx=0得x=1 g(1)=0, 为g(x)的极大值点,因此有g(x)

(1)f'(x)=1/x+(a-2),那么f'(1)=1+a-2=0,∴a=1 f(x)=lnx-x,f'(x)=1/x-1=-(x-1)/x (x>0) 令f'(x)≥0,那么0

X为在范围0 X> 1时 函数f(x)'= -1 / X + 1 / X-2 -1 当X = 0函数f(x )`= 1 + LN(2) 当X = 2 F(X)'= LN(2)3

fx=lnx+2/x 的定义域是 {x| x>0}; 对f(x)求导,有:f'(x)=1/x-2/x^2=(x-2)/x^2 令f'(x)=0,x=2,即fx在x=2处取极值; 故函数单调性讨论如下: 当0

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