(理)如果f(x)在某个区间I内满足:对任意的x1、x2∈I都有［f(x1)+f(x2)］≥f()则称f(x)在I上为下凸函数.已知函数f(x)=-alnx.(1)证明当a＞0时f(x)在(0+∞)上为下凸函数;(2)

Question

(理)如果f(x)在某个区间I内满足:

对任意的x ₁ 、x ₂ ∈I 都有 ［f(x ₁ )+f(x ₂ )］≥f( ) 则称f(x)在I上为下凸函数.

已知函数f(x)= -alnx.

(1)证明当a＞0时 f(x)在(0 +∞)上为下凸函数;

(2)若f′(x)为f(x)的导函数 且x∈［ 2］时 |f′(x)|＜1 求实数a的取值范围.

(文)如果f(x)在某个区间I内满足:

对任意的x ₁ 、x ₂ ∈I 都有 ［f(x ₁ )+f(x ₂ )］≥f( ) 则称f(x)在I上为下凸函数 已知函数f(x)=ax ² +x.

(1)证明当a＞0时 f(x)在R上为下凸函数;

(2)若x∈(0 1)时 |f(x)|≤1 求实数a的取值范围.

Answer 1

(理)(1)证明:任取x ₁ 、x ₂ ∈(0 +∞)

则 ［f(x ₁ )+f(x ₂ )］

= ［ -alnx ₁ + -alnx ₂ ］

= -aln                                                          

                                          

∵x ₁ ² +x ₂ ² ≥2x ₁ x ₂ ∴(x ₁ +x ₂ ) ² ≥4x ₁ x ₂ .

又x ₁ ＞0 x ₂ ＞0 ∴ .                                            

又 ≥ a＞0

∴-aln ≥-aln                                                   

即 ［f(x ₁ )+f(x ₂ )］≥f( ).                                              

∴f(x)为(0 +∞)上的下凸函数.

(2)f′(x)=                                                     

∵|f′(x)|＜1 即| |＜1

∴-(x+ )＜a＜x- .                                                        

∵x∈［ 2］时 |f′(x)|＜1恒成立

∴a∈(-2 ).                                                            

(文)(1)证明：f(x ₁ )+f(x ₂ )-2f( )

=ax ₁ ² +x ₁ +ax ₂ ² +x ₂ -2［a( ) ² - ］

=                                                            

∵a＞0 ∴ ［f(x ₁ )+f(x ₂ )］≥f( ).

∴当a＞0时 f(x)为R上的下凸函数.                                         

(2)∵|f(x)|≤1

∴-1≤ax ² +x≤1

≤a≤ .                                               &

作业帮用户 2017-03-27