Question

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

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

(2)|f(x₂)-f(x₁)|＜|x₁-x₂|;

(3)|f(x₁)-f(x₂)|＜ ;

(4)|f(x₁)-f(x₂)|≤.

Answer 1

证明:(1)f(0)=c,f(1)=c,

∴f(0)=f(1).

(2)|f(x₂)-f(x₁)|=|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

而由f(0)=f(1),从而|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)|＜.

(4)|f(x₂)-f(x₁)|≤f_max-f_min=f(0)-f()=.