log2x=2-x
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/06 14:06:38
x>0,f(x)=log2(x)x
解由-3≤log1/2x≤-1/2,得-3≤-log2x≤-1/2,即1/2≤log2x≤3令t=log2(x),则t属于[1/2,3]故f(x)=log2x/4·log2x/2=(log2(x)+l
最后的函数能表达得更清楚点吗
f(x)=(log2x)^2-4log2x+1=(log2x-2)²-3,∵x∈[2,16],∴log2x∈[1,4],∴log2x=2时,f(x)有最小值-3,当log2x=4时,f(x)
log1/2(x)=log[(2)^(-1)]x=-log2(x)再问:第一部的负一的负号是怎么变到前面的再答:公式log[m^(n)](x)=1/nlogm(x)再问:3Q≡ω≡
f(x)=(log2x/3)(log2x/4)(2≤x≤8),f(x)=(log2x)²/12x=8时最大f(x)=(log28)²/12=9/12=3/4x=2时最小f(x)=(
Y=log2(x^2+1)-log2x零和负数无对数:x>0Y=log2(x^2+1)-log2x=log2{(x^2+1)/x}=log2{x+1/x}x>0,x+1/x≥2,log2{x+1/x}
log4[log3(log2x)]=0=log4(1)所以log3(log2x)=1=log3(3)所以log2(x)=3x=2³=8x^(-1/2)=1/√8=√2/4
令t=log2x(t∈R)所以y=t²-4t=(t-2)²-4则由图象得,当t∈(-∞,2)时,y单调递减;t∈(2,+∞)时,y单调递增故当x∈(0,4)时,单调递减;x∈(4,
原式可化为y=1+log2x+1/log2xx>=1y>=1+2根(log2x*1/log2x)=3x
由不等式可得-3≤log1/2x≤-1/2,也就得2^1/2≤x≤2^3即√2≤x≤8f(x)=(log2x/4)(log2x/2)=log2(x/4+x/2)=log2(3x/4)接下来就知道怎样做
解题思路:本题考查函数的性质和应用,解题时要认真审题,仔细解答,注意公式的灵活运用.解题过程:
f(x)=2^|log2(x)|-|x-1/x|定义域{x|x>0}f(x)=2^|log2(x)|-|x-1/x|当x≥1时f(x)=x-(x-1/x)=1/x当0
-2-√30所以f(4)=log2(4)=22>0f(2)=log2(2)=1所以原式=f[f(4)]=f(2)=1
mn=1,根据均值定理m/2+n/2>1所以2f(m/2+n/2)=2log2(m/2+n/2)=log2(m/2+n/2)^2=log2n那么(m/2+n/2)^2=n所以m^2+n^2+2m
log3(log2x)=1log2x=3^1=3x=2^3=8x^(1/2)=√8=2√2
g(x)=(log2x)^2+4log2x+2,x属于[1,2]log2x属于[0,1]配方g(x)=(t+2)^2-2最大值为14,最小值为7
因为2≤x≤8,则:1≤log(2)[x]≤3,则:f(x)=【log(2)[x]-3】×【log(2)[x]-2】【设:log(2)[x]=t,则:1≤t≤3】=(t-3)(t-2)=[t-(5/2