Question

函數(shù)f(x)的定義域?yàn)镽，并滿足以下條件：

①對(duì)任意x∈R，有f(x)＞0;

②對(duì)任意x、y∈R，有f(xy)=［f(x)］^y；

③f()＞1.

(1)求f(0)的值;

(2)求證:f(x)在R上是單調(diào)增函數(shù);

(3)若a＞b＞c＞0,且b²=ac,求證:f(a)+f(c)＞2f(b).

Answer 1

解法一:（1）解：令x=0,y=2,得f(0)=[f(0)]².

∵f(0)＞0,∴f(0)=1.

(2)證明：任取x₁、x₂∈(-∞,+∞),且x₁＜x₂.

設(shè)x₁=p₁,x₂=p₂,則p₁＜p₂.

f(x₁)-f(x₂)=f(p₁)-f(p₂)=[f()]-[f()].

∵f()＞1,p₁＜p₂,∴f(x₁)＜f(x₂).∴f(x)在R上是單調(diào)增函數(shù).

(3)證明：由(1)(2)知f(b)＞f(0)=1,∴f(b)＞1.

∵f(a)=f(b·)=,f(c)=f(b·)=,

∴f(a)+f(c)=+[f(b)]＞2.而a+c＞2=2=2b,

∴2＞2=2f(b).∴f(a)+f(c)＞2f(b).

解法二:(1)解：∵對(duì)任意x、y∈R,有f(xy)=[f(x)]^y,

∴f(x)=f(x·1)=[f(1)]^x.∴當(dāng)x=0時(shí)f(0)=[f(1)]⁰.

∵任意x∈R,f(x)＞0,∴f(0)=1.

(2)證明：∵f()＞1, ∴f(1)=f(3×)=[f()]³＞1.∴f(x)=[f(1)]^x是R上的單調(diào)增函數(shù)，

即f(x)是R上的單調(diào)增函數(shù).

(3)證明：f(a)+f(c)=[f(1)]^a+[f(1)]^c＞2.而a+c＞2=2=2b,

∴2＞2=2f(b).∴f(a)+f(c)＞2f(b).