作业帮设f(x)=[2(logx(x))^2]+2alog2(1/x)+b,已知x=1/2时,f(x)有最小值-8,求a,b设f(x)=[2(logx(x))^2]+2alog2(1/x)+b,已知x=1/2时,f(x)有最小值-8.(1)求a,b(2)求满足f(x)>0的x的集合A设f(x)=[2(log2(x))^2]+2alog2(1/x)+b,已知x=1/2时，f(x)有
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![设f(x)=[2(logx(x))^2]+2alog2(1/x)+b,已知x=1/2时,f(x)有最小值-8,求a,b设f(x)=[2(logx(x))^2]+2alog2(1/x)+b,已知x=1/2时,f(x)有最小值-8.(1)求a,b(2)求满足f(x)>0的x的集合A设f(x)=[2(log2(x))^2]+2alog2(1/x)+b,已知x=1/2时，f(x)有](/uploads/image/z/5498247-39-7.jpg?t=%E8%AE%BEf%28x%29%3D%5B2%28logx%28x%29%29%5E2%5D%2B2alog2%281%2Fx%29%2Bb%2C%E5%B7%B2%E7%9F%A5x%3D1%2F2%E6%97%B6%2Cf%28x%29%E6%9C%89%E6%9C%80%E5%B0%8F%E5%80%BC-8%2C%E6%B1%82a%2Cb%E8%AE%BEf%28x%29%3D%5B2%28logx%28x%29%29%5E2%5D%2B2alog2%281%2Fx%29%2Bb%2C%E5%B7%B2%E7%9F%A5x%3D1%2F2%E6%97%B6%2Cf%28x%29%E6%9C%89%E6%9C%80%E5%B0%8F%E5%80%BC-8.%281%29%E6%B1%82a%2Cb%282%29%E6%B1%82%E6%BB%A1%E8%B6%B3f%28x%29%3E0%E7%9A%84x%E7%9A%84%E9%9B%86%E5%90%88A%E8%AE%BEf%28x%29%3D%5B2%28log2%28x%29%29%5E2%5D%2B2alog2%281%2Fx%29%2Bb%2C%E5%B7%B2%E7%9F%A5x%3D1%2F2%E6%97%B6%EF%BC%8Cf%28x%29%E6%9C%89)
设f(x)=[2(logx(x))^2]+2alog2(1/x)+b,已知x=1/2时,f(x)有最小值-8,求a,b设f(x)=[2(logx(x))^2]+2alog2(1/x)+b,已知x=1/2时,f(x)有最小值-8.(1)求a,b(2)求满足f(x)>0的x的集合A设f(x)=[2(log2(x))^2]+2alog2(1/x)+b,已知x=1/2时，f(x)有
设f(x)=[2(logx(x))^2]+2alog2(1/x)+b,已知x=1/2时,f(x)有最小值-8,求a,b
设f(x)=[2(logx(x))^2]+2alog2(1/x)+b,已知x=1/2时,f(x)有最小值-8.(1)求a,b(2)求满足f(x)>0的x的集合A
设f(x)=[2(log2(x))^2]+2alog2(1/x)+b,已知x=1/2时，f(x)有最小值-8.(1)求a,b(2)求满足f(x)>0的x的集合A 刚才写错了。
设f(x)=[2(logx(x))^2]+2alog2(1/x)+b,已知x=1/2时,f(x)有最小值-8,求a,b设f(x)=[2(logx(x))^2]+2alog2(1/x)+b,已知x=1/2时,f(x)有最小值-8.(1)求a,b(2)求满足f(x)>0的x的集合A设f(x)=[2(log2(x))^2]+2alog2(1/x)+b,已知x=1/2时，f(x)有
f(x)=[2(log2(x))^2]+2alog2(1/x)+b
=[2(log2(x))^2]-2alog2(x)+b
令log2(x)=t
则可换为：
f(t)=2t^2-2at+b
而抛物线在对称轴处取得最值,而次抛物线开口向上,则取得最小值
即：f(t)=2(t-a/2)^2+b-a^2/2
而此时t=a/2时,取得最小值
由题意,当x=1/2时,f(x)有最小值-8
则：x=1/2,则t=-1
a/2=-1
b-a^2/2=-8
则,a=-2,b=-6
则：
f(t)=2t^2+4t-6=2(t+3)(t-1)
f(x)>0
则t>1,或者t<-3
则可知
log2(x)>1,或者log2(x)<-3
则,0
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