.函数f(x)=x3+x2-x在区间[-2,1]上的最大值和最小值分别是 A.1,

.函数f(x)=x3x2x在区间[-2,1]上的最大值和最小值分别是

A.1,-                                                    B.1,-2

C.2,-                                                    D.2,-2

答案

B


解析:

本题考查闭区间上连续函数的最值求解的基本方法.它的求解过程可分两步:第一步,求(a,b)内的极值;第二步,比较各极值与端点值的大小,求得最值.

f(x)=x3x2x,∴f′(x)=3x2+2x-1.

令3x2+2x-1=0,得x1=-1,x2=.

f(-2)=(-2)3+(-2)2-(-2)=-2,f(-1)=1,f()=-,f(1)=1,

f(x)max=1,f(x)min=-2.

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