Question

已知函數(shù)f(x)=-x³+ax²-4(a∈R).

(1)若函數(shù)y=f(x)的圖象在點P(1,f(1))處的切線的傾斜角為,求a;

(2)設(shè)f(x)的導(dǎo)函數(shù)是f′(x).在(1)的條件下,若m,n∈［-1,1］,求f(m)+f′(n)的最小值;

(3)若存在x₀∈(0,+∞),使f(x₀)＞0,求a的取值范圍.

Answer 1

解:(1)f′(x)=-3x²+2ax.

據(jù)題意,f′(1)=tan=1,∴-3+2a=1,即a=2.

(2)由(1)知f(x)=-x³+2x²-4,

則f′(x)=-3x²+4x.

x	-1	(-1,0)	0	(0,1)	1
f′(x)	-7	-	0	+	1
f(x)	-1	↘	-4	↗	-3

∴對于m∈［-1,1］,f(m)的最小值為f(0)=-4.

∵f′(x)=-3x²+4x的對稱軸為x=,且拋物線開口向下,

∴x∈［-1,1］時,f′(x)最小值為f′(-1)與f′(1)中較小的.

∵f′(1)=1,f′(-1)=-7,

∴當(dāng)x∈［-1,1］時,f′(x)的最小值為-7.

∴當(dāng)n∈［-1,1］時,f′(n)的最小值為-7.

∴f(m)+f′(n)的最小值為-11.8分(3)∵f′(x)=-3x(x),

①若a≤0,當(dāng)x＞0時,f′(x)＜0,∴f(x)在［0,+∞)上單調(diào)遞減.

又f(0)=-4,則當(dāng)x＞0時,f(x)＜-4.∴當(dāng)a≤0時,不存在x₀＞0,使f(x₀)＞0.

②若a＞0,則當(dāng)0＜x＜時,f′(x)＞0,當(dāng)x＞時,f′(x)＜0.

從而f(x)在(0,］上單調(diào)遞增,在［,+∞)上單調(diào)遞減.

∴當(dāng)x∈(0,+∞)時,f(x)_max=f()=+-4=-4.

據(jù)題意,-4＞0,即a³＞27.∴a＞3.

綜上,a的取值范圍是(3,+∞).