设函数y=f(x)(x∈R,且x≠0)对任意非零实数x,y都有f(xy)=f(x)+f(y)成立 (1)求证f(1)=f(-1)=0,且f(1/x)=-f(x)(x≠0)(2) 判断f(x)的奇偶性(3)若f(x)在(0,正无穷)上单调递增,解不等式f(1/x)-f(2x-1
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![设函数y=f(x)(x∈R,且x≠0)对任意非零实数x,y都有f(xy)=f(x)+f(y)成立 (1)求证f(1)=f(-1)=0,且f(1/x)=-f(x)(x≠0)(2) 判断f(x)的奇偶性(3)若f(x)在(0,正无穷)上单调递增,解不等式f(1/x)-f(2x-1](/uploads/image/z/3648104-8-4.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0y%3Df%EF%BC%88x%EF%BC%89%EF%BC%88x%E2%88%88R%2C%E4%B8%94x%E2%89%A00%EF%BC%89%E5%AF%B9%E4%BB%BB%E6%84%8F%E9%9D%9E%E9%9B%B6%E5%AE%9E%E6%95%B0x%2Cy%E9%83%BD%E6%9C%89f%EF%BC%88xy%EF%BC%89%3Df%EF%BC%88x%EF%BC%89%2Bf%EF%BC%88y%EF%BC%89%E6%88%90%E7%AB%8B+%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81f%281%29%3Df%28-1%29%3D0%2C%E4%B8%94f%EF%BC%881%2Fx%29%3D-f%28x%29%28x%E2%89%A00%29%282%29+%E5%88%A4%E6%96%ADf%28x%29%E7%9A%84%E5%A5%87%E5%81%B6%E6%80%A7%283%29%E8%8B%A5f%28x%29%E5%9C%A8%280%2C%E6%AD%A3%E6%97%A0%E7%A9%B7%EF%BC%89%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%2C%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%EF%BC%881%2Fx%29-f%282x-1)
设函数y=f(x)(x∈R,且x≠0)对任意非零实数x,y都有f(xy)=f(x)+f(y)成立 (1)求证f(1)=f(-1)=0,且f(1/x)=-f(x)(x≠0)(2) 判断f(x)的奇偶性(3)若f(x)在(0,正无穷)上单调递增,解不等式f(1/x)-f(2x-1
设函数y=f(x)(x∈R,且x≠0)对任意非零实数x,y都有f(xy)=f(x)+f(y)成立 (1)求证f(1)=f(-1)=0,且f(1/x)=-f(x)(x≠0)
(2) 判断f(x)的奇偶性
(3)若f(x)在(0,正无穷)上单调递增,解不等式f(1/x)-f(2x-1)≥0.
设函数y=f(x)(x∈R,且x≠0)对任意非零实数x,y都有f(xy)=f(x)+f(y)成立 (1)求证f(1)=f(-1)=0,且f(1/x)=-f(x)(x≠0)(2) 判断f(x)的奇偶性(3)若f(x)在(0,正无穷)上单调递增,解不等式f(1/x)-f(2x-1
(1)另y=1,则有f(x)=f(x)+f(1),所以f(1)=0,另y=1/x,则有f(1)=f(x)+f(1/x)=0,所以f(1/x)=-f(x).另x=y=-1,则f(1)=2f(-1)=0,所以f(1)=f(-1)=0;
(2)另y=-1,则f(-x)=f(x)+f(-1),由于f(-1)=0,所以f(x)=f(-x)为偶函数
(3)f(1/x)-f(2x-1)≥0,由于在(0,正无穷)单调递增,所以1/x>=2X-1>=0,解得-0.5<=X<1,并且X>0,2X-1>0,得到0.5<=X<=1,
由于f(x)为偶函数,则在(负无穷,0)单调递减,当f(1/x)-f(2x-1)≥0,则有1/x<=2x-1<=0,解得-0.5<=X<1,并且X<0,2X-1<0,得到-0.5<=X<=0,
综上所述 -0.5<=X<=0或0.5<=X<=1
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