Question

已知函數(shù)f(x)=ax³+bx²-3x在x=±1處取得極值.

(1)求函數(shù)f(x)的解析式；

(2)若過(guò)A(1,m)(m≠-2)可作曲線y=f(x)的三條切線，求實(shí)數(shù)m的取值范圍.

Answer 1

答案：(1)解：f′(x)=3ax²+2bx-3,依題意，f′(1)=f′(-1)=0,

即得a=1,b=0.

∴f(x)=x³-3x.                                                                 

(2)f′(x)=3x²-3=3(x+1)(x-1).                                                     

∵曲線方程為y=x³-3x且m≠-2,∴點(diǎn)?A(1，m)不在曲線上.

設(shè)切點(diǎn)為M(x₀,y₀)，則y₀=x₀³-3x₀,

∵f′(x₀)=3(x₀²-1),故切線斜率為3(x₀²-1)=,

整理，得：2x₀³-3x₀²+m+3=0.

∵過(guò)A(1，m)可作曲線的三條切線,

∴關(guān)于x₀的方程2x₀³-3x₀²+m+3=0有三個(gè)實(shí)根.                                   

設(shè)g(x₀)=2x₀³-3x₀²+m+3,

則g′(x₀)=6x₀²-6x₀,由g′(x₀)=0得x₀=0或x₀=1,                                   

∴函數(shù)g(x₀)的極值點(diǎn)為0,1,

∴關(guān)于x₀的方程2x₀³-3x₀²+m+3=0有三個(gè)實(shí)根的充要條件是g(0)g(1)＜0,

即(m+3)(m+2)＜0,∴-3＜m＜-2,

故所求m的范圍是-3＜m＜-2.