已知m∈R,函數(shù)f(x)=(x2+mx+m)•ex.
(1)若函數(shù)f(x)沒有零點(diǎn),求實(shí)數(shù)m的取值范圍;
(2)當(dāng)m>2時(shí),求函數(shù)f(x)的極大值.
解:(1)令f(x)=(x2+mx+m)•ex=0.
∵ex>0,∴x2+mx+m=0.
∵函數(shù)f(x)沒有零點(diǎn),∴方程x2+mx+m=0無實(shí)根.
則△=m2-4m<0,解得:0<m<4.
所以函數(shù)f(x)沒有零點(diǎn)的實(shí)數(shù)m的取值范圍是(0,4);
(2)由f(x)=(x2+mx+m)•ex.
得:f′(x)=(2x+m)ex+(x2+mx+m)ex
=(x2+2x+mx+2m)ex=(x+2)(x+m)ex.
令f′(x)=0,得:x=-2或x=-m.
當(dāng)m>2時(shí),-m<-2.
所以,當(dāng)x∈(-∞,-m)時(shí),f′(x)>0,函數(shù)f(x)為增函數(shù);
當(dāng)x∈(-m,-2)時(shí),f′(x)<0,函數(shù)f(x)為減函數(shù);
當(dāng)x∈(2,+∞)時(shí),f′(x)>0,f(x)為增函數(shù);
所以,當(dāng)x=-m時(shí),f(x)取得極大值,極大值為f(-m)=[(-m)2+m•(-m)+m]e-m=me-m.
分析:(1)給出的函數(shù)是一個(gè)二次三項(xiàng)式和一個(gè)指數(shù)式的乘積,指數(shù)式恒大與0,要使原函數(shù)沒有零點(diǎn),只需要二次三項(xiàng)式對應(yīng)的二次方程的判別式小于0即可;
(2)求出函數(shù)的導(dǎo)函數(shù),由m>2,得-m<-2,由導(dǎo)函數(shù)的兩個(gè)零點(diǎn)-m,-2把函數(shù)的定義域分段,借助于二次函數(shù)判斷導(dǎo)函數(shù)在各區(qū)間段內(nèi)的符號,從而得到原函數(shù)在各區(qū)間段內(nèi)的增減性,得到極大值點(diǎn),把極大值點(diǎn)的橫坐標(biāo)代入原函數(shù)求得函數(shù)的極大值.
點(diǎn)評:本題考查了函數(shù)零點(diǎn)的判斷,考查了利用函數(shù)導(dǎo)函數(shù)研究函數(shù)的單調(diào)性與極值,連續(xù)函數(shù)在定義域內(nèi)的某點(diǎn)處,左右兩側(cè)的單調(diào)性不同,則該點(diǎn)為函數(shù)的極值點(diǎn),先增后減為極大值點(diǎn),先減后增為極小值點(diǎn).此題是中檔題.