设函数y=f(x)(x∈R),对任意非零实数x1,x2满足f(x1x2)=f(x1)+f(x2),又f(x)在(0,+∞
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设函数y=f(x)(x∈R),对任意非零实数x1,x2满足f(x1x2)=f(x1)+f(x2),又f(x)在(0,+∞)是增函数,则不等式f(x)+f(x-1/2)≤0的解集为
![设函数y=f(x)(x∈R),对任意非零实数x1,x2满足f(x1x2)=f(x1)+f(x2),又f(x)在(0,+∞](/uploads/image/z/15287360-32-0.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0y%3Df%28x%29%28x%E2%88%88R%29%2C%E5%AF%B9%E4%BB%BB%E6%84%8F%E9%9D%9E%E9%9B%B6%E5%AE%9E%E6%95%B0x1%2Cx2%E6%BB%A1%E8%B6%B3f%EF%BC%88x1x2%EF%BC%89%3Df%EF%BC%88x1%EF%BC%89%2Bf%EF%BC%88x2%EF%BC%89%2C%E5%8F%88f%EF%BC%88x%EF%BC%89%E5%9C%A8%EF%BC%880%2C%2B%E2%88%9E)
/>∵f(1)=f(1)+f(1),∴f(1)=0
又f(1)=f(-1)+f(-1),∴f(-1)=0
∵f(-x)=f(x)+f(-1)=f(x),∴f(x)是偶函数
∵f(x)在(0,+∞)增,那么x在[-1,1]上小于等于0
f(x)+f(x-1/2)=f(x^2-0.5x)≤0
-1≤x^2-0.5x≤1
解得1/4 (1-√(17))
再问: 不对啊老师说是错
再答: 绝对没有问题,除非是你题目打错了
又f(1)=f(-1)+f(-1),∴f(-1)=0
∵f(-x)=f(x)+f(-1)=f(x),∴f(x)是偶函数
∵f(x)在(0,+∞)增,那么x在[-1,1]上小于等于0
f(x)+f(x-1/2)=f(x^2-0.5x)≤0
-1≤x^2-0.5x≤1
解得1/4 (1-√(17))
再问: 不对啊老师说是错
再答: 绝对没有问题,除非是你题目打错了
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