非零实数集的函数f(xy)=f(x)+f(y),且f(x)在(0,+00)上是增函数,解不等式f(2)+f(x-1/2)小于等于0
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![非零实数集的函数f(xy)=f(x)+f(y),且f(x)在(0,+00)上是增函数,解不等式f(2)+f(x-1/2)小于等于0](/uploads/image/z/1647307-19-7.jpg?t=%E9%9D%9E%E9%9B%B6%E5%AE%9E%E6%95%B0%E9%9B%86%E7%9A%84%E5%87%BD%E6%95%B0f%28xy%29%3Df%28x%29%2Bf%28y%29%2C%E4%B8%94f%28x%29%E5%9C%A8%280%2C%2B00%29%E4%B8%8A%E6%98%AF%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%282%29%2Bf%28x-1%2F2%29%E5%B0%8F%E4%BA%8E%E7%AD%89%E4%BA%8E0)
非零实数集的函数f(xy)=f(x)+f(y),且f(x)在(0,+00)上是增函数,解不等式f(2)+f(x-1/2)小于等于0
非零实数集的函数f(xy)=f(x)+f(y),且f(x)在(0,+00)上是增函数,解不等式f(2)+f(x-1/2)小于等于0
非零实数集的函数f(xy)=f(x)+f(y),且f(x)在(0,+00)上是增函数,解不等式f(2)+f(x-1/2)小于等于0
由题意得 :f(1) = f(1*1) = f(1) + f(1)
所以 f(1) = 0
因为 f(x)在(0,+00)上是增函数
所以 当 x ∈(0,1] 时 f(x) ≤ 0
当 x ∈(1,+∞) 时 f(x) > 0
f(2)+f(x-1/2) = f(2x - 1) ≤ 0
即 0 < 2x - 1 ≤ 1
解得 1/2 < x ≤ 1