已知函数f(x)=ax^3+3/2(a-1)x^2 - 知识问答

已知函数f(x)=ax^3+3/2(a-1)x^2-3x(a为常数).讨论函数y=f(x)的单调性,并写出相应的单调递增

0为递增区间3ax^2+3(a-1)x-3＞0→ ax^2+(a-1"}}}'>

2个回答

对f(x)求导：f(x)'=3ax^2+3(a-1)x-3 (x≠0)
f(x)'>0为递增区间
3ax^2+3(a-1)x-3＞0
→ ax^2+(a-1)x-1>0,（因式分解得）ax（x+1）-(x+1)>0
→(ax-1)(x+1)>0
→ax-1>0且x+1>0
ⅰ) a>0
x>1/a
ⅱ)-1< a

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