Question

(理)已知函數(shù)f(x)=()²(x＞0).

(1)判斷f(x)的單調(diào)性,并證明你的結(jié)論;

(2)求f(x)的反函數(shù)f^-1(x);

(3)若關(guān)于x的不等式(x-1)f^-1(x)＞a(a-)在x∈［3,+∞)上恒成立,求實(shí)數(shù)a的取值范圍.

Answer 1

答案：(理)解:(1)∵f(x)=(1+)²=1++,f′(x)==-2(+),

當(dāng)x＞0時(shí),f′(x)＜0,∴f(x)在(0,+∞)上是減函數(shù).

(2)令y=()²,∵x＞0,＞0,∴==1+.

∴=,x=.∴f^-1(x)=(x＞1).

(3)由(x-1)＞a()在［3,+∞)上恒成立,即,也即＜0要在x∈［3,+∞)上恒成立.

(a+1)(a--1)＜0在［3,+∞)上恒成立,

亦即-1＜a＜+1在x∈［3,+∞)上恒成立.

由于+1在［3,+∞)上是單調(diào)遞增函數(shù).∴+1在［3,+∞)上有最小值+1.

從而實(shí)數(shù)a的取值范圍是-1＜a＜+1.