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
【数学辅导团】为您解答,不理解请追问,理解请及时选为满意回答!(*^__^*)谢谢!
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