已知:函数f(x)=
(1)求函数y=f(x)的单调递增区间; (2)若函数y=f(x)在x=2取极值,求函数y=f(x)在区间[e-2,e2]上的最大值. |
(1)函数f(x)定义域为x>0,
f′(x)=x-a+
=a-1 x
=x2-ax+a-1 x
.[x-(a-1)](x-1) x
由f'(x)>0且x>0
得
即x>0 x2-ax+a-1>0 x>0 [x-(a+1)](x-1)>0
(i)当a-1=1即a=2时,f(x)在(0,+∞)上为增函数;
(ii)当a>2时,x>a-1或0<x<1,∴f(x)在(a-1,+∞),(0,1)上为增函数;
(iii)当1<a<2时,0<x<a-1或x>1,∴f(x)在(0,a-1),(1,+∞)上为增函数.
综上可知:f(x)的单调区间为:当a=2时,(0,+∞)
当a>2时,(a-1,+∞),(0,1)
当1<a<2时,(0,a-1),(1,+∞).
(2)x=2是f(x)极值点,∴f'(2)=0,即2-a+
=0,解得a=3.a-1 2
∴f(x)=
x2-3x+2lnx(x>0),f′(x)=1 2
.(x-1)(x-2) x
∵
<1<2<e2,且当2<x<e2时,f′(x)>0;当1<x<2时,f′(x)<0;当1 e2
<x<1时,f′(x)>0.1 e2
∴函数f(x)在区间[
,1)及(2,e2]上单调递增,在区间(1,2)上单调递减.1 e2
∴f(x)在[
,e2]最大值应在x=1和x=e2处取得1 e2
又f(1)=-
,f(e2)=5 2
-3e2+4=e4 2
>-(e2-2)(e2-4) 2
,5 2
∴f(x)max=
.(e2-2)(e2-4) 2