已知f(x)在定义域x>0上为增函数,满足f(xy)=f(x)+f(y),且f(2)=1(1)求f(1) (2)求满足f(x)+f(x-3)≤2的x的取值范围
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![已知f(x)在定义域x>0上为增函数,满足f(xy)=f(x)+f(y),且f(2)=1(1)求f(1) (2)求满足f(x)+f(x-3)≤2的x的取值范围](/uploads/image/z/1265830-70-0.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%E5%9C%A8%E5%AE%9A%E4%B9%89%E5%9F%9Fx%3E0%E4%B8%8A%E4%B8%BA%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E6%BB%A1%E8%B6%B3f%28xy%29%3Df%28x%29%2Bf%28y%29%2C%E4%B8%94f%282%29%3D1%281%29%E6%B1%82f%281%29+%282%29%E6%B1%82%E6%BB%A1%E8%B6%B3f%28x%29%2Bf%28x-3%29%E2%89%A42%E7%9A%84x%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
已知f(x)在定义域x>0上为增函数,满足f(xy)=f(x)+f(y),且f(2)=1(1)求f(1) (2)求满足f(x)+f(x-3)≤2的x的取值范围
已知f(x)在定义域x>0上为增函数,满足f(xy)=f(x)+f(y),且f(2)=1
(1)求f(1)
(2)求满足f(x)+f(x-3)≤2的x的取值范围
已知f(x)在定义域x>0上为增函数,满足f(xy)=f(x)+f(y),且f(2)=1(1)求f(1) (2)求满足f(x)+f(x-3)≤2的x的取值范围
1.令X=Y=1
f(xy)=f(x)+f(y),f(1)=2f(1),f(1)=0
2.2=1+1=f(2)+f(2)=f(4)
f(x)+f(x-3)=f(x^2-3x)
(1)
因为f(xy)=f(x)+f(y),f(2)=1
f(4)=f(2*2)=2f(2)=2
f(8)=f(2*4)=f(2)+f(4)=1+2=3
(2)
f(x)+f(x-3)=f(x^2-3x)<=2=f(4),0
x^2-3x<=4
(x-4)(x+1)<=0
-1<=x<=4
3
f(xy)=f(x)+f(y),
令x=y=1,则xy=1
f(1)=f(1)+f(1)
所以f(1)=0
f(x)+f(y)=f(xy)
所以f(x)+f(x-3)=f(x^2-3x)
f(xy)=f(x)+f(y),
令x=y=2,则xy=4
所以f(4)=f(2)+f(2)=2
所以f(x^2-3x)≤f(4)
所以x^2-3x≤4
x^2-3x-4≤0
-1≤x≤4
又定义域
x>0,x-3>0,x>3
所以3
f(1*1)=f(1)+f(1)
=>f(1)=2f(1)
=>f(1)=0;
f(x)+f(x-3)=f(x*(x-3))
f(x)是增函数,f(4)=2f(2)=2
所以,0
3
(1)代入:
f(xy)=f(x)+f(y)
f(1*1)=f(1)+f(1)
f(1)=2f(1)
f(1)=0
(2)运用函数单调性:a>b>0,则f(a)>f(b)
f(x)+f(x-3)=f(x^2-3x)≤2=f(2)+f(2)=f(4)
x^2-3x≤4
(x-4)(x+1)≤0 解得:
-1≤x≤4
又x-3>0,
所以3