y=3x-4x³
y'=3-12x²=0
(x+1/2)(x-1/2)=0
x=1/2 (因为x∈【0,1】)
f(0)=0,f(1/2)=1,f(1)=3-4=-1
所以
最大值=f(1/2)=1
最小值=f(1)=-1
函数y=3x-4x的立方在区间【0,1】的最大值和最小值是多少
函数y=3x-4x的立方在区间【0,1】的最大值和最小值是多少
数学人气:637 ℃时间:2019-08-21 20:51:41
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