作业帮maths question!help me!已知函数f(x)=lg(2+x)+lg(2-x) ,若不等式f（x）>m有解,求实数m的取值范围.
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maths question!help me!已知函数f(x)=lg(2+x)+lg(2-x) ,若不等式f（x）>m有解,求实数m的取值范围.
maths question!help me!
已知函数f(x)=lg(2+x)+lg(2-x) ,若不等式f（x）>m有解,求实数m的取值范围.
maths question!help me!已知函数f(x)=lg(2+x)+lg(2-x) ,若不等式f（x）>m有解,求实数m的取值范围.
f(x)=lg(2+x)+lg(2-x)>=m
定义域为2+x>0,2-x>0,即 -2
要使f(x)>m有解，必须让f(x)的最大值大于m，对f(x)求导进行单调性分析可知，x=0时有最大值2lg2，所以m∈(-∞,2lg2)
对函数求导，可得其上升区间为(-2,0],下降区间为[0,2);
所以在x=0处有最大值2lg2
即f(x)≤2lg2, 若m大于此值，则显然无解，
所以有m<2lg2
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