Question

如果函數(shù)f(x)的定義域?yàn)镽，對(duì)于任意實(shí)數(shù)a,b滿足f(a+b)=f(a)·f(b)

(1)設(shè)f(1)=k,(k≠0),試求f(n)(n∈N^*);

(2)設(shè)當(dāng)x＜0時(shí)，f(x)＞1，試解不等式f(x+5)＞.

Answer 1

解：(1)∵f(n+1)=f(n)·f(1),∴=f(1)=k≠0 x∈R,f(x)=f(+)

=f²()≥0.

∴｛f(n)｝是以k為首項(xiàng)，k為公比的等比數(shù)列,

∴f(n)=f(1)·［f(1)］^n-1=kⁿ(k∈N^*)

(2)對(duì)于任意的x∈R,f(x)=f(+)=f²()≥0.

假定存在x₀∈R,使f(x₀)=0,則可取x＜0則f(x)=f(x-x₀+x₀)=f(x-x₀)·f(x₀)=0.這與已知矛盾，則f(x₀)≠0,于是，對(duì)于任意x∈R必有f(x)＞0,

∵f(0)=f(0+0)=f²(0)≠0,∴f(0)=1.

設(shè)x₁＜x₂,則x₁-x₂＜0,則f(x₁-x₂)＞1,

又∵(fx₂)＞0,∴f(x₁)=f(x₁-x₂+x₂)=f(x₁-x₂)·f(x₂)＞f(x₂),

∴f(x)為R上的減函數(shù).

對(duì)f(x+5)＞,∵f(x)＞0,∴原不等式等價(jià)于f(x+5)·f(x)＞1,

即f(2x+5)＞f(0),又∵f(x)為R上的減函數(shù)，

∴2x+5＜0，解得x＜-.