Question

已知兩函數(shù)f(x)=8x²+16x-k,g(x)=2x³+5x²+4x,其中k為實(shí)數(shù).
(1)對(duì)任意x∈[-3,3]都有f(x)≤g(x)成立,求k的取值范圍.
(2)存在x∈[-3,3]使f(x)≤g(x)成立,求k的取值范圍.
(3)對(duì)任意x₁,x₂∈[-3,3]都有f(x₁)≤g(x₂),求k的取值范圍.

Answer 1

(1) k≥45   (2) k≥-7   (3) k≥141

(1)設(shè)h(x)=g(x)-f(x)=2x³-3x²-12x+k,
問(wèn)題轉(zhuǎn)化為x∈[-3,3]時(shí),h(x)≥0恒成立,
即h(x)_min≥0,x∈[-3,3].
令h'(x)=6x²-6x-12=0,得x=2或x=-1.
∵h(yuǎn)(-3)=k-45,h(-1)=k+7,h(2)=k-20,
h(3)=k-9,
∴h(x)_min=k-45≥0,得k≥45.
(2)據(jù)題意:存在x∈[-3,3],使f(x)≤g(x)成立,
即為h(x)=g(x)-f(x)≥0在x∈[-3,3]上能成立,
∴h(x)_max≥0.∴h(x)_max=k+7≥0,得k≥-7.
(3)據(jù)題意:f(x)_max≤g(x)_min,x∈[-3,3],
易得f(x)_max=f(3)=120-k,g(x)_min=g(-3)=-21.
∴120-k≤-21,得k≥141.