Question

已知定義R⁺的函數(shù)f(x)滿足:對任意的x、y∈R⁺,恒有f(xy)=f(x)+f(y).

(1)求證:f()=-f(x)(x∈R⁺);

(2)若x＞1時,恒有f(x)＜0,求證:f(x)必有反函數(shù);

(3)設(shè)f(x)的反函數(shù)是f^-1(x),求證:f^-1(x)對于定義域內(nèi)任意的x₁、x₂恒有f^-1 (x₁+x₂)=f^-1(x₁)·f^-1(x₂).

Answer 1

證明:(1)對任意的x、y∈R⁺,

f(xy)=f(x)+f(y)恒成立,

令x=y=1,得f(1)=2f(1),可得f(1)=0.

而f(x)+f()=f(x·)=f(1),

即f(x)+f()=0.

由此可得f()=-f(x).

(2)任取x₁、x₂∈R⁺,且x₁＜x₂,則＞1.

∵x＞1時,f(x)＜0恒成立,可知f()＜0.

又f()=f(x₂·)=f(x₂)+f()=f(x₂)-f(x₁),

∴f(x₂)-f(x₁)＜0,即f(x₁)＞f(x₂).

由此可知f(x)是R⁺上的減函數(shù),故f(x)必有反函數(shù).

(3)設(shè)f^-1(x₁)=u₁,f^-1(x₂)=u₂,

則x₁=f(u₁),x₂=f(u₂).

∵f(u₁·u₂)=f(u₁)+f(u₂)=x₁+x₂,

∴f^-1(x₁+x₂)=f^-1［f(u₁·u₂)］

=u₁·u₂=f^-1(x₁)·f^-1(x₂).