Question

設(shè)函數(shù)f(x)=2x³-3(a+1)x²+6ax,a∈R.

(1)當(dāng)a=1時(shí),求證:f(x)為單調(diào)增函數(shù);

(2)當(dāng)x∈［1,3］時(shí),f(x)的最小值為4,求a的值.

Answer 1

(1)證明：當(dāng)a=1時(shí),f(x)=2x³-6x²+6x,f′(x)=6x²-12x+6,

∴f′(x)=6(x-1)²≥0.

故f(x)為單調(diào)增函數(shù).

(用定義法證明單調(diào)性參照給分)

(2)解：f′(x)=6(x-1)(x-a).

①當(dāng)a≤1時(shí),f(x)在區(qū)間［1,3］上是單調(diào)增函數(shù),最小值為f(1).

由于f(1)=4,即2-3(a+1)+6a=4.解得a=＞1(舍去).

②當(dāng)1＜a＜3時(shí),f(x)在區(qū)間(1,a)上是減函數(shù),在區(qū)間(a,3)上是增函數(shù),故f(a)為最小值.

f(a)=4,即a³-3a²+4=0.

解得a=-1(舍去),a=2.

③當(dāng)a≥3時(shí),f(x)在區(qū)間(1,a)上是減函數(shù),f(3)為最小值.

f(3)=4,即54-27(a+1)+18a=4.解得a=＜3(舍去).

綜上所述,a=2.