f(x)=x³/3 + a*x² + bx
f ' (x) = x²+ 2ax + b,代入已知條件 f'(-1)=0,得到：f'(-1)=1 -2a + b = 0,即b = 2a-1
所以：f ' (x) = x²+ (b+1)x + b = (x+b)(x+1)
f(x) 單調(diào)區(qū)間為：
⑴ 當(dāng) -b≥ -1時(shí),即b≤ 1時(shí)x∈[-b,∞]
⑵ 當(dāng) -b＜ -1時(shí),即b＞1時(shí)x∈(-∞,-b]
令a=-1,設(shè)函數(shù)f(x)在x1,x2(x1