Question

已知函數(shù)f(x)=(x²+bx+c)e^x,其中b,c∈R為常數(shù).

(1)若b²＞4(c-1),討論函數(shù)f(x)的單調(diào)性;

(2)若b²≤4(c-1),且=4,試證:-6≤b≤2.

Answer 1

(1)解:求導(dǎo)得f′(x)=［x²+(b+2)x+b+c］e^x.

因b²＞4(c-1),故方程f′(x)=0,即x²+(b+2)x+b+c=0有兩根;

x₁=＜x₂=.

令f′(x)＞0,解得x＜x₁或x＞x₂;

又令f′(x)＜0,解得x₁＜x＜x₂.

故當(dāng)x∈(-∞,x₁)時(shí),f(x)是增函數(shù);

當(dāng)x∈(x₂,+∞)時(shí),f(x)也是增函數(shù);

但當(dāng)x∈(x₁,x₂)時(shí),f(x)是減函數(shù).

(2)證明:易知f(0)=c,f′(0)=b+c,因此

=f′(0)=b+c.

所以由已知條件得

因此b²+4b-12≤0.

解得-6≤b≤2.