Question

已知a∈R，討論函數(shù)f(x)=ln(x-1)-ax的單調(diào)性并求相對(duì)應(yīng)的單調(diào)區(qū)間.

Answer 1

解：函數(shù)f(x)=ln(x-1)-ax的定義域?yàn)?1,+∞),

其導(dǎo)數(shù)f′(x)=-a 

(Ⅰ)當(dāng)a=0時(shí),f′(x)=  ∵x＞1  ∴f′(x)＞o,所以,函數(shù)f(x)在區(qū)間(1,+∞)為增函數(shù),單調(diào)區(qū)間為(1,+∞)   

(Ⅱ)當(dāng)a≠0時(shí),f′(x)=-a=

①當(dāng)a＞0時(shí),令f′(x)＞0,解得1＜x＜令f′(x)＜0,解得x＞

所以,當(dāng)a＞0時(shí),函數(shù)f(x)在區(qū)間(1,)為增函數(shù),在區(qū)間(,+∞)為減函數(shù) 

②當(dāng)a＜0時(shí),令f′(x)＞0,解得  x＞1所以,當(dāng)a＜0時(shí),函數(shù)f(x)在區(qū)間(1,+∞)為增函數(shù)