17、∴h′(x)<0,從而h(x)在(0,1]上單調(diào)遞減,
∴h(x)≥h(1)=0,∴g(x)在(0,1]上單調(diào)遞增,
∴g(x)≤g(1)=2ln2,∴g(x)的最大值為2ln2.
16.[2018·天津卷]設(shè)函數(shù)f(x)=(x-t1)(x-t2)(x-t3),其中t1,t2,t3∈R,且t1,t2,t3是公差為d的等差數(shù)列.
(1)若t2=0,d=1,求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程;
(2)若d=3,求f(x)的極值;
(3)若曲線y=f(x)與直線y=-(x-t2)-6有三個(gè)互異的公共點(diǎn),求d的取值范圍.
解析:(1)解:由已知,可得f(x)=x(x-1)
18、(x+1)=x3-x,故f′(x)=3x2-1.因此f(0)=0,f′(0)=-1.
又因?yàn)榍€y=f(x)在點(diǎn)(0,f(0))處的切線方程為y-f(0)=f′(0)(x-0),故所求切線方程為x+y=0.
(2)解:由已知可得
f(x)=(x-t2+3)(x-t2)(x-t2-3)=(x-t2)3-9(x-t2)=x3-3t2x2+(3t2-9)x-t2+9t2.
故f′(x)=3x2-6t2x+3t2-9.
令f′(x)=0,解得x=t2-或x=t2+.
當(dāng)x變化時(shí),f′(x),f(x)的變化情況如下表:
x
(-∞,t2-)
t2-
(t2-,t2+)
t2+
(
19、t2+,+∞)
f′(x)
+
0
-
0
+
f(x)
極大值
極小值
所以函數(shù)f(x)的極大值為f(t2-)=(-)3-9×(-)=6,
函數(shù)f(x)的極小值為f(t2+)=()3-9×=-6.
(3)解:曲線y=f(x)與直線y=-(x-t2)-6有三個(gè)互異的公共點(diǎn)等價(jià)于關(guān)于x的方程(x-t2+d)(x-t2)(x-t2-d)+(x-t2)+6=0有三個(gè)互異的實(shí)數(shù)解.
令u=x-t2,可得u3+(1-d2)u+6=0.
設(shè)函數(shù)g(x)=x3+(1-d2)x+6,則曲線y=f(x)與直線y=-(x-t2)-6有三個(gè)互異的公共點(diǎn)等價(jià)于函數(shù)y=g(x
20、)有三個(gè)零點(diǎn).
g′(x)=3x2+(1-d2).
當(dāng)d2≤1時(shí),g′(x)≥0,這時(shí)g(x)在R上單調(diào)遞增,不合題意.
當(dāng)d2>1時(shí),
令g′(x)=0,解得x1=-,x2=.
易得,g(x)在(-∞,x1)上單調(diào)遞增,在[x1,x2]上單調(diào)遞減,在(x2,+∞)上單調(diào)遞增.
所以g(x)的極大值為
g(x1)=g=+6>0.
g(x)的極小值為
g(x2)=g=-+6.
若g(x2)≥0,則由g(x)的單調(diào)性可知函數(shù)y=g(x)至多有兩個(gè)零點(diǎn),不合題意.
若g(x2)<0,即(d2-1)>27,也就是|d|>,此時(shí)|d|>x2,g(|d|)=|d|+6>0,且-2|d|