作业帮f(x)=(log2^x)^2-log2^x+2怎么解
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f(x)=(log2^x)^2-log2^x+2怎么解
f(x)=(log2^x)^2-log2^x+2怎么解
f(x)=(log2^x)^2-log2^x+2怎么解
定义域是{x|x>0}
设log2 x=t,(t∈R)
则f(x)=g(t)=t²-t+2=(t-1/2)²+7/4
∴值域是[7/4,+∞)
单调递减区间是(0,√2),单调递增区间是[√2,+∞)
你好
f(x)=(log2^x)^2-log2^x+2
=（log2^x-1/2）^2-1/4+2
=（log2^x-1/2）^2+7/4
当log2^x=1/2，即x=√2/2时，函数有最小值7/4
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