設(shè)函數(shù)f(x)=x2+x-l,g(x)=ebx,其中P為自然對(duì)數(shù)的底.
(1)當(dāng)b=-1時(shí),求函數(shù)F(x)=f(x)•g(x)的極大、極小值;
(2)當(dāng)b=-1時(shí),求證:函數(shù)G(x)=f(x)+g(x)有且只有一個(gè)零點(diǎn);
(3)若不等式g(x)≥ex對(duì)?x>0恒成立,求實(shí)數(shù)b的取值范圍.
分析:(1)當(dāng)b=-1時(shí),•F(x)=(x
2+x-1)e
-x,求出函數(shù)的導(dǎo)數(shù),畫出表格,判斷函數(shù)的單調(diào)性,求出函數(shù)的極值
(2)當(dāng)b=-1時(shí),G(x)=x2+x-l+e-x,當(dāng)b=-1時(shí),G(x)=x2+x-l+e-x,求出函數(shù)的導(dǎo)數(shù),判斷函數(shù)的單調(diào)性,判斷函數(shù)的零點(diǎn)
(3)g(x)=e
bx≥ex,等價(jià)于bx≥ln(ex)=1+lnx對(duì)?x>0恒成立,即
b≥對(duì)?x>0恒成立,利用恒成立問題,求b的取值范圍
解答:(1)解:當(dāng)b=-1時(shí),•F(x)=(x
2+x-1)e
-x,
則F'(x)=(2x+1)e
-x+(x
2+x-1)•(-e
-x)=-(x
2-x-2)e
-x=-(x+1)(x-2)e
-x(2分)
令F'(x)=O,得x
1=-1,x
2=2.
當(dāng)x變化時(shí),F(xiàn)'(0)、F(x)的變化情況如下表:
(4分)
∴當(dāng)x=-1時(shí),F(xiàn)(x)
極小值=-e:當(dāng)x=2時(shí),F(xiàn)(x)
極大值=5e
-2(6分)
(2)證:當(dāng)b=-1時(shí),G(x)=x2+x-l+e-x,顯然G(O)=O,當(dāng)b=-1時(shí),G(x)=x2+x-l+e-x.(7分)
∵G’(x)=2x+l-e
-x,則G”(x)=2+e
-x>O,(8分)
∴G’(x)在R上是增函數(shù),
∴當(dāng)x<0時(shí),G'(x)<G'(O)=O,G(x)單調(diào)遞減,G(x)>G(0)=0;
當(dāng)x>0時(shí),G'(x)>G'(0)=O,G(x)單調(diào)遞增,G(x)>G(0)=0.
故函數(shù)G(x)有且只有一個(gè)零點(diǎn)x=0.(注:或說明G(x)
min=G(0)=O)(l0分)
(3)解:g(x)=e
bx≥ex,等價(jià)于bx≥ln(ex)=1+lnx對(duì)?x>0恒成立,即
b≥對(duì)?x>0恒成立,
設(shè)
h(x)=(x>0),則b≥h(x)
max(l2分)
而
h′(x)=,令h'(x)=O,得x=1.
∵x∈(O,1)時(shí),h'(x)>0:當(dāng)x∈(1,+∞)時(shí),h'(x)<O,
∴h(x)
max=h(1)=l,
∴b≥l為所求.(14分)
點(diǎn)評(píng):該題考查函數(shù)導(dǎo)數(shù)的求法,以及利用導(dǎo)數(shù)判斷函數(shù)的單調(diào)性,切記要畫圖表,要會(huì)用最值求未知量的取值范圍