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幼苗
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f(x)=log1/3(x^2-4x+31),
令,Y=x^2-4x+31,有X属于R时,Y>0恒成立,
Y的对称轴X=2.在对称轴的左边递减,对称轴的右边递增,
而log1/3(y)为单调减函数,f(x)在区间(-无穷,2]上,单调递增,此时X=2时,f(x)有最大值,f(x)=log1/3(2^2-4*2+31)=log1/3(27)=-3.
f(x)在区间[2,+无穷)上单调减,
则f(x)的值域为(-无穷,-3].
1年前
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