Question

已知函數(shù)f(x)=x³x²+(k²-k-2)x(k∈R),

(1)若k=0,求f(x)的單調(diào)區(qū)間;

(2)若函數(shù)f(x)同時(shí)滿足以下三個(gè)條件:

①對(duì)任意實(shí)數(shù)x＜0,都有f(x)＜f(0);

②對(duì)任意實(shí)數(shù)x＞2,都有f(x)＞f(2);

③存在實(shí)數(shù)x₁＜1＜x₂,使得f(x₁)＞f(1)＞f(x₂).

求實(shí)數(shù)k的取值范圍.

Answer 1

解:(1)若k=0,f′(x)=7x²-13x-2=(7x+1)(x-2),

f′(x)＞0x＞2或x＜;f′(x)＜0＜x＜2.

所以f(x)的單調(diào)增區(qū)間為(-∞,),(2,+∞);單調(diào)減區(qū)間為(,2).

(2)由題意可得:

由③知f(x)存在減區(qū)間(m,n),且1∈(m,n),

∴f′(1)＜0,解得-2＜k＜4.

由②知f(x)在(2,+∞)上單調(diào)遞增,∴f′(2)≥0,解得k≥3或k≤0;

由①知f(x)在(-∞,0)上單調(diào)遞增,∴f′(0)≥0,解得k≥2或k≤-1.

綜上,k的取值范圍為［3,4)∪(-2,-1］.