Question

已知函数fx=In(x-1)-k(x-1)+1求单调区间

Answer 1

f'(x)=1/(x-1)-k,(x>1)

当k≤0时,f'(x)>0,f(x)在(1,+∞)单调递增

即f'(x)的增区间为(1,+∞) 无减区间

当k>0时,令f'(x)=0得x=1+1/k

所以增区间为(1,1+1/k) 减区间为(1+1/k,+∞)

(2)f(x)=ln(x-1)-k(x-1)+1=ln(x-1)+1

k>=[ln(x-1)+1]/(x-1)

设g(x)=[ln(x-1)+1]/(x-1)

g'(x)=[1/(x-1)*(x-1)-(ln(x-1)+1)]/(x-1)^2=[1-ln(x-1)-1]/(x-1)^2=-ln(x-1)/(x-1)^2=0

ln(x-1)=0

x=2

在（1,2）上有g'(x)>0,在（2,+无穷）上有g'(x)=1.