求二次函数f(x)=x²-2(2a-1)x+5a²-4a+2在[0,1]上最小值g(a)的解析式.
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求二次函数f(x)=x²-2(2a-1)x+5a²-4a+2在[0,1]上最小值g(a)的解析式.
求二次函数f(x)=x²-2(2a-1)x+5a²-4a+2在[0,1]上最小值g(a)的解析式.
求二次函数f(x)=x²-2(2a-1)x+5a²-4a+2在[0,1]上最小值g(a)的解析式.
f(x)=x²-2(2a-1)x+5a²-4a+2
=(x-2a+1)²+a²-8a+1
x-2a+1=0,即x=1-2a时f(x)取得最小值,最小值g(a)的解析式为:
g(a)=a²-8a+1