Question

定義在R上的函數(shù)f(x)滿足:如果對任意x₁、x₂∈R都有f()≤［f(x₁)+f(x₂)］,則稱f(x)為R上的凹函數(shù),已知二次函數(shù)f(x)=ax²+x(a∈R,a≠0).??

(1)求證:當(dāng)a＞0時,函數(shù)f(x)是凹函數(shù);?

(2)若x∈［0,1］時,|f(x)|≤1,試求實數(shù)a的取值范圍.??

Answer 1

(1)證明:對任意x₁、x₂∈R,∵a＞0,?

∴［f(x₁)+f(x₂)］-2f()?

=ax₁²+x₁+ax₂²+x₂-2［a()²+］??

=ax₁²+ax₂²-a(x₁²+x₂²+2x₁x₂)?

=a(x₁-x₂)²≥0.?

∴f()≤［f(x₁)+f(x₂)］.?

∴f(x)為凹函數(shù).?

(2)解析:由|f(x)|≤1-1≤f(x)≤1-1≤ax²+x≤1.(*)?

當(dāng)x=0時,a∈R;?

當(dāng)x∈(0,1］時,(*)即?

恒成立.?

∵x∈(0,1］,

∴≥1.?

∴當(dāng)=1時,-(+)²+取得最大值是-2;?

當(dāng)=1時,(- )²-取得最小值0.?

∴-2≤a≤0,結(jié)合a≠0,得-2≤a＜0.??

綜上,a的范圍是［-2,0).