Question

已知定義在R上的函數(shù)f(x)對(duì)任意的實(shí)數(shù)x₁、x₂滿足關(guān)系f(x₁+x₂)?=f(x₁)+f(x₂)+2.

(1)證明f(x)的圖象關(guān)于點(diǎn)(0,-2)成中心對(duì)稱圖形;

(2)若x＞0,則有f(x)＞-2,求證:f(x)在(-∞,+∞)上是增函數(shù).

Answer 1

剖析:對(duì)于(1),只要證明=-2即可;對(duì)于(2),注意到f(x)是抽象函數(shù),欲證單調(diào)性,需對(duì)f(x)進(jìn)行適當(dāng)?shù)淖冃?

證明:(1)令x₁=x₂=0,則f(0+0)=f(0)+f(0)+2,

    所以f(0)=-2.

    對(duì)任意實(shí)數(shù)x,令x₁=x,x₂=-x,有f(x-x)=f(x)+f(-x)+2,

    即f(0)-2=f(x)+f(-x),得=-2.

    又=0,

    這表明點(diǎn)M(x,f(x))與點(diǎn)N(-x,f(-x))的中點(diǎn)是(0,-2),即點(diǎn)M₁N關(guān)于點(diǎn)(0,-2)成中心對(duì)稱.

    由點(diǎn)M的任意性知:函數(shù)f(x)的圖象關(guān)于點(diǎn)(0,-2)成中心對(duì)稱.

    (2)對(duì)任意實(shí)數(shù)x₁、x₂,且x₁＜x₂.

    由x₂-x₁＞0,有f(x₂-x₁)＞-2.

    于是f(x₂)=f［(x₂-x₁)+x₁］=f(x₂-x₁)+f(x₁)+2.

    所以f(x₂)-f(x₁)=f(x₂-x₁)+2＞-2+2=0,

    即f(x₂)＞f(x₁).

    所以f(x)在(-∞,+∞)上是增函數(shù).

講評(píng):對(duì)于(1),求出f(0)=-2是解題的關(guān)鍵;對(duì)于(2),變形f(x₂)=f［(x₂-x₁)+x₁］=f(x₂-x₁)+f(x₁)+2是解題的關(guān)鍵.