已知函数f(x)=x²-4ax+2a+6(a∈R) 1.若函数的值域为[0,+∞),求a的值
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1.f(x)=(x-2a)^2-4a^2+2a+6
x=2a,fmin=-4a^2+2a+6=0
a=3/2ora=-1
2.f(x)>=0
fmin>=0
fmin=-4a^2+2a+6>=0
2a^2-a-3<=0
(2a-3)(a+1)<=0
-1<=a<=3/2
g(a)=2-a/a+3/
g(a)=2-a(a+3)=2-a^2-3a=-a^2-3a+2=-(a^2+3a-2)=-((a+3/2)^2-9/4-2)
=-((a+3/2)^2-17/4)=-(a+3/2)^2+17/4
a=-1,gmax=-1+3+2=6
a=3/2,gmin=-9/4-9/2+2=-19/4
g(a):[-19/4,6]
x=2a,fmin=-4a^2+2a+6=0
a=3/2ora=-1
2.f(x)>=0
fmin>=0
fmin=-4a^2+2a+6>=0
2a^2-a-3<=0
(2a-3)(a+1)<=0
-1<=a<=3/2
g(a)=2-a/a+3/
g(a)=2-a(a+3)=2-a^2-3a=-a^2-3a+2=-(a^2+3a-2)=-((a+3/2)^2-9/4-2)
=-((a+3/2)^2-17/4)=-(a+3/2)^2+17/4
a=-1,gmax=-1+3+2=6
a=3/2,gmin=-9/4-9/2+2=-19/4
g(a):[-19/4,6]
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