题目内容

(本题满分15分)

设函数.

(Ⅰ)当时,解不等式:

(Ⅱ)求函数的最小值;

(Ⅲ)求函数的单调递增区间.

 

【答案】

【解析】解:(Ⅰ)…………3分

(Ⅱ)令

(1)当时,上单调递增,故

(2)当时,可证上单调递增,故

(3)当时,

综合得,当时,;当时,…………9分

(Ⅲ),令,可得

时,单调递增区间为

时,由

(2)当时,单调递增区间为

(3)当时,单调递增区间为…………15分

 

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