Question

已知函数f(x)=1/3x^3+ax^2+bx (a,b∈R) 在x= -1 时取得极值

(1)试用含a 的代数式表示b
(2)求f(x)的单调区间

Answer 1

(1)f'(x)=x²+ax+b
∴f'(-1)=1-a+b=0
∴b=a-1
(2)由(1)知b=a-1
∴f'(x)=x²+ax+a-1=[x-(a-1)](x-1)
令f'(x)=0,则x=a-1或x=1
①当a-1=1即a=2时
f'(x)=[x-(a-1)](x-1)≥0恒成立
∴f(x)在R上递增
②当a-1＞1即a＞2时
令f'(x)＞0,则x＜1或x＞a-1,f(x)在(-∞,1),(a-1,+∞)上递增
令f'(x)＜0,则1＜x＜a-1,f(x)在(1,a-1)上递减
③当a-1＜1即a＜2时
令f'(x)＞0,则x＜a-1或x＞1,f(x)在(-∞,a-1),(1,+∞)上递增
令f'(x)＜0,则a-1＜x＜1,f(x)在(a-1,1)上减
综上所述,①当a=2时,f(x)在R上递增
②当a＞2时,f(x)在(-∞,1),(a-1,+∞)上递增,在(1,a-1)上递减
③当a＜2时,f(x)在(-∞,a-1),(1,+∞)上递增,在(a-1,1)上减