已知函数f(x)=|log2(x+1)|,实数m,n在其定义域内,且m0;f(m2)<f(m+n)<f(n2)

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已知函数f(x)=|log2(x+1)|,实数m,n在其定义域内,且m0;f(m2)<f(m+n)<f(n2)

已知函数f(x)=|log2(x+1)|,实数m,n在其定义域内,且m0;f(m2)<f(m+n)<f(n2)
已知函数f(x)=|log2(x+1)|,实数m,n在其定义域内,且m0;f(m2)<f(m+n)<f(n2)

已知函数f(x)=|log2(x+1)|,实数m,n在其定义域内,且m0;f(m2)<f(m+n)<f(n2)
m-log2(m+1)=log2(n+1)
所以,-10,mn<0,且:log2(n+1)+log2(m+1)=0
log2(n+1)(m+1)=0
(n+1)(m+1)=1
1+mn+m+n=1
m+n=-mn>0
f(m+n)-f(m^2)
=log2(m+n+1)-log2(m^2+1)
=log2(m+n+1)/(m^2+1)
=log2(1-mn)/(m^2+1)
>log2(1-mn)/(|m||n|+1)
=log2(1-mn)/(1-mn)
=log2 1
=0
所以,f(m^2)f(m+n)-f(n^2)
=log2(m+n+1)-log2(n^2+1)
=log2(m+n+1)/(n^2+1)
=log2(1-mn)/(n^2+1)
=log2(1-mn)/(1-mn)
=log2 1
=0
所以,f(m+n)所以,f(m2)<f(m+n)<f(n2)