设函数y=f(x)对任意非零实数x1,x2满足f(x1x2)=f(x1)+f(x2)
(1)证明f(1)=f(-1)=0
(2)证明f(x)是偶函数
(3)已知f(x)为(0,+无穷)上的增函数,且满足f(x)+f(x-1\2)≤0,求x
人气:182 ℃ 时间:2019-08-18 22:15:17
解答
(1)f(1*1)=f(1)+f(1) 则 f(1)=0f(-1*-1)=f(-1)+f(-1)=0 则 f(-1)=0所以 f(1)=f(-1)=0(2)f(-1*x)=f(x)+f(-1) 即 f(-x)=f(x)所以 f(x)是偶函数(3)f[x(x-1/2)] ≤ 0 因为 f(x)为(0,+无穷)上的增函数当x(x-1/2)>...
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