2.2　最大值、最小值问题

第1课时　函数的最值

1.函数y=f(x)=1/2x-sin x在区间[0,π]上的最大值和最小值分别是(　　)

A.π/2-1,0 B.π/4-1,π/6-√3/2

C.π/2,π/6-√3/2 D.π/2+1,0

解析:因为y'=1/2-cos x,当x∈(π/3 "," π)时,y'>0,则函数在区间(π/3 "," π)上是增加的;当x∈(0"," π/3)时,y'<0,则函数在(0"," π/3)上是减少的,所以x=π/3是极小值点.又因为f(0)=0,f(π/3)=π/6-√3/2,f(π)=π/2,所以函数的最小值为π/6-√3/2,最大值为π/2,故选C.

答案:C

2.函数f(x)=excos x在区间[0"," π/2]上的值域为(　　)

A.[0"," √2/2 e^(π/4) ] B.[0,1]

C.[0"," √2/2 e^(π/4) ) D.[1"," √2/2 e^(π/4) ]

解析:f'(x)=excos x-exsin x=ex(cos x-sin x),

　　∵x∈(0"," π/4),∴f'(x)>0;x∈(π/4 "," π/2),f'(x)<0,

　　∴x=π/4是f(x)的极大值点.

　　又f(0)=1,f(π/4)=√2/2 e^(π/4),f(π/2)=0,

　　∴f(x)min=0,f(x)max=√2/2 e^(π/4).

答案:A

3.函数f(x)=x2·ex+1在区间[-2,1]上的最大值为(　　)

A.4e-1 B.1 C.e2 D.3e2

解析:f'(x)=2xex+1+x2·ex+1=(x2+2x)ex+1.

　　由f'(x)<0,得-20,得x<-2或x>0.

　　结合题意知f(x)在(-2,0)上递减,在(0,1)上递增.

　　∵f(-2)=4/e,f(1)=e2,f(1)>f(-2),

　　∴函数f(x)在区间[-2,1]上的最大值为e2.

答案:C

4.函数y=xe-x在区间[0,4]上的最大值是(　　)