已知定义域为R的函数 f(x)=(-2的x次幂+b)/[2的(x+1)次幂+a]是奇函数,(1)求
求a,b的值(2)若对任意t∈R不等式f(x平方-2t)+f(2t平方-k)
人气:278 ℃ 时间:2019-11-22 17:30:01
解答
1、既然这个奇函数的定义域是R,则f(0)=0、f(-1)=-f(1),联立这两个方程,解得:a=2,b=1;
2、f(x)=(1/2)[(1-2^x)/(1+2^x)]=(1/2){-1+[2/(1+2^x)]}
则可以证明函数f(x)是递减的
又:f(x²-2t)+f(2t²-k)
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