设函数f(x)定义域(0,+∞),且f(4)=1,对任意正实数x1x2,有f(x1x2)=f(x1)+f(x2),且当x1≠x2时有f(x2)-f(x1)/x2-x1>0(1)判断函数f(x)在(0,+∞)上的单调性(2)求f(1)的值(3)若f(x+6)+f(x)>2,求x的取值范围
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![设函数f(x)定义域(0,+∞),且f(4)=1,对任意正实数x1x2,有f(x1x2)=f(x1)+f(x2),且当x1≠x2时有f(x2)-f(x1)/x2-x1>0(1)判断函数f(x)在(0,+∞)上的单调性(2)求f(1)的值(3)若f(x+6)+f(x)>2,求x的取值范围](/uploads/image/z/3939349-13-9.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%E5%AE%9A%E4%B9%89%E5%9F%9F%280%2C%2B%E2%88%9E%29%2C%E4%B8%94f%284%29%3D1%2C%E5%AF%B9%E4%BB%BB%E6%84%8F%E6%AD%A3%E5%AE%9E%E6%95%B0x1x2%2C%E6%9C%89f%28x1x2%29%3Df%28x1%29%2Bf%28x2%29%2C%E4%B8%94%E5%BD%93x1%E2%89%A0x2%E6%97%B6%E6%9C%89f%28x2%29-f%28x1%29%2Fx2-x1%3E0%281%29%E5%88%A4%E6%96%AD%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%EF%BC%880%2C%2B%E2%88%9E%EF%BC%89%E4%B8%8A%E7%9A%84%E5%8D%95%E8%B0%83%E6%80%A7%EF%BC%882%EF%BC%89%E6%B1%82f%281%29%E7%9A%84%E5%80%BC%EF%BC%883%EF%BC%89%E8%8B%A5f%28x%2B6%29%2Bf%28x%29%3E2%2C%E6%B1%82x%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
设函数f(x)定义域(0,+∞),且f(4)=1,对任意正实数x1x2,有f(x1x2)=f(x1)+f(x2),且当x1≠x2时有f(x2)-f(x1)/x2-x1>0(1)判断函数f(x)在(0,+∞)上的单调性(2)求f(1)的值(3)若f(x+6)+f(x)>2,求x的取值范围
设函数f(x)定义域(0,+∞),且f(4)=1,对任意正实数x1x2,有f(x1x2)=f(x1)+f(x2),且当x1≠x2时
有f(x2)-f(x1)/x2-x1>0(1)判断函数f(x)在(0,+∞)上的单调性(2)求f(1)的值(3)若f(x+6)+f(x)>2,求x的取值范围
设函数f(x)定义域(0,+∞),且f(4)=1,对任意正实数x1x2,有f(x1x2)=f(x1)+f(x2),且当x1≠x2时有f(x2)-f(x1)/x2-x1>0(1)判断函数f(x)在(0,+∞)上的单调性(2)求f(1)的值(3)若f(x+6)+f(x)>2,求x的取值范围
(1)因为f(x2)-f(x1)/x2-x1>0,所以分两类:a.分子分母都大于0.b.分子分母都小于0.然后运用单调性的定义,不管哪种情况,函数都是增函数.
(2)令x1=x2=1,代入f(x1x2)=f(x1)+f(x2),的f(1)=0.
(3) 根据f(x1x2)=f(x1)+f(x2),则f(x+6)+f(x)=f[(x+6)*x],又因为f(4)=1,所以f(4*4)=f(4)+f(4)=4,即由f(16)=2.所以不等式转换为:f[(x+6)*x]>f(16).因为函数是增函数,所以[(x+6)*x>16,解这个一元二次不等式,并注意x>0这个条件即可.最后结果为x>2.
作为小题用特殊函数法吧,设f(x)=logax(以a为底x的对数)f(4)=1,所以a=4