Question

已知函數(shù)f(x)=x³-x²-3x+,直線l:9x+2y+c=0.

(1)求證:直線l與函數(shù)y=f(x)的圖象不相切;

(2)若當(dāng)x∈［-2,2］時(shí),函數(shù)y=f(x)的圖象在直線l的下方,求c的范圍.

Answer 1

(1)證法一:f′(x)=x²-2x-3=(x-1)²-4≥-4.                                                            ?

根據(jù)導(dǎo)數(shù)的幾何意義知,函數(shù)y=f(x)的圖象上任意一點(diǎn)處切線的斜率均不小于-4.       ?

而直線l:9x+2y+c=0的斜率為-＜-4,?

所以直線l與函數(shù)y=f(x)的圖象不相切.                                                                    ?

證法二:f′(x)=x²-2x-3.                                                                                         ?

假設(shè)直線l:9x+2y+c=0與函數(shù)y=f(x)的圖象相切,?

則x²-2x-3=-有實(shí)數(shù)解,即x²-2x+=0有實(shí)數(shù)解.                                             ?

因?yàn)棣?-2＜0,方程x²-2x+=0無(wú)實(shí)數(shù)解,?

所以直線l與函數(shù)y=f(x)的圖象不相切.                                                                    ?

(2)解:當(dāng)x∈［-2,2］時(shí),函數(shù)y=f(x)的圖象在直線l的下方,即x³-x²-3x+-(-x-)＜0對(duì)一切x∈［-2,2］都成立,                                                                               9分?

即c＜-x³+2x²-3x-對(duì)一切x∈［-2,2］都成立.                                                   

令g(x)=-x³+2x²-3x-.?

因?yàn)?I >g′(x)=-2x²-4x-3=-2(x-1)²-1＜0,?

所以g(x)在［-2,2］上單調(diào)遞減.                                                                       ?

所以當(dāng)x∈［-2,2］時(shí),［g(x)］_Min=g(2)=-×2³+2×2²-3×2-=-6.               ?

所以c＜-6,所以c的范圍是(-∞,-6).