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函数y=2^（x+2）-3X4^x,若x^2+x
题目详情
函数y=2^（x+2）-3X4^x,若x^2+x<=0,则函数的最大值和最小值是?
▼优质解答
答案和解析
a=2^x
x(x+1)<=0
-1<=x<=0
则1/2<=a<=1
y=a*2²-3a²
=-3a²+4a
=-3(a-2/3)²+4/3
1/2<=a<=1
所以
a=2/3,最大=4/3
a=1,最小=1
x(x+1)<=0
-1<=x<=0
则1/2<=a<=1
y=a*2²-3a²
=-3a²+4a
=-3(a-2/3)²+4/3
1/2<=a<=1
所以
a=2/3,最大=4/3
a=1,最小=1
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