Question

已知函數(shù)f(x)=x-alnx(a∈R).

(1)當(dāng)a=2時,求曲線y=f(x)在點A(1,f(1))處的切線方程.

(2)求函數(shù)f(x)的極值.

Answer 1

【解析】函數(shù)f(x)的定義域為(0,+∞),f′(x)=1-.

(1)當(dāng)a=2時,f(x)=x-2lnx,f′(x)=1-(x>0),所以f(1)=1,f′(1)= -1,

所以y=f(x)在點A(1,f(1))處的切線方程為y-1=-(x-1),即x+y-2=0.

(2)由f′(x)=1-=,x>0可知:

①當(dāng)a≤0時,f′(x)>0,函數(shù)f (x)為(0,+∞)上的增函數(shù),函數(shù)f(x)無極值;

②當(dāng)a>0時,由f′(x)=0,解得x=a;

因為x∈(0, a)時,f′(x)<0,x∈(a,+∞)時,f′(x)>0,

所以f(x)在x=a處取得極小值,且極小值為f(a)=a-alna,無極大值.

綜上:當(dāng)a≤0時,函數(shù)f(x)無極值,

當(dāng)a>0時,函數(shù)f(x)在x=a處取得極小值a-alna,無極大值.