Question

已知f(x)=x²-x+c定义在区间［0,1］上,x₁、x₂∈［0,1］且x₁≠x₂,证明

(1)f(0)=f(1);

(2)|f(x₂)-f(x₁)|＜|x₁-x₂|;

(3)|f(x₁)-f(x₂)|＜.

Answer 1

证明：(1)f(0)=c,f(1)=c,

    故f(0)=f(1).

    (2)|f(x₂)-f(x₁)|=|x₂²-x₂+c-x₁²+x₁-c|

    =|x₂-x₁||x₂+x₁-1|.

    ∴0≤x₁≤1,0≤x₂≤1,0＜x₁+x₂＜2(x₁≠x₂).

    ∴-1＜x₁+x₂-1＜1.

    ∴|f(x₂)-f(x₁)|＜|x₂-x₁|.

    (3)不妨设x₂＞x₁.

    由(2)知|f(x₂)-f(x₁)|＜x₂-x₁.                    ①

    又∵|f(x₂)-f(x₁)|=|f(x₂)-f(1)+f(0)-f(x₁)|

    ≤|f(x₂)-f(1)|+|f(0)-f(x₁)|

    ＜|1-x₂|+|x₁|=1-x₂+x₁.                        ②

    ①+②得2|f(x₂)-f(x₁)|＜1,即|f(x₂)-f(x₁)|＜.