设函数y=f(x)(x∈R,且x≠0)对任意非零实数x1,x2,满足f(x1)+f(x2)=f(x1x2)
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设函数y=f(x)(x∈R,且x≠0)对任意非零实数x1,x2,满足f(x1)+f(x2)=f(x1x2)
1.求f(1)+f(—1)的值
2.判断函数y=f(x)的奇偶性.
1.求f(1)+f(—1)的值
2.判断函数y=f(x)的奇偶性.
![设函数y=f(x)(x∈R,且x≠0)对任意非零实数x1,x2,满足f(x1)+f(x2)=f(x1x2)](/uploads/image/z/8268236-44-6.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%B0x1%2Cx2%2C%E6%BB%A1%E8%B6%B3f%28x1%29%2Bf%28x2%29%3Df%28x1x2%29)
1.f(x1)+f(x2)=f(x1x2),即f(x1x2)=f(x1)+f(x2),
令x1=x2=1,代入得f(1*1)=f(1)+f(1),所以f(1)=0;
令x1=x2=-1,代入得f(1)=f[(-1)*(-1)]=f(-1)+f(-1)=2f(-1)=0,所以f(-1)=0;
2.令x1=-1,代入得f(-x2)=f(-1)+f(x2)=0+f(x2)=f(x2),所以f(x)是偶函数.
令x1=x2=1,代入得f(1*1)=f(1)+f(1),所以f(1)=0;
令x1=x2=-1,代入得f(1)=f[(-1)*(-1)]=f(-1)+f(-1)=2f(-1)=0,所以f(-1)=0;
2.令x1=-1,代入得f(-x2)=f(-1)+f(x2)=0+f(x2)=f(x2),所以f(x)是偶函数.
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