已知定义在(-∞,0)U(0,+∞)上的偶函数f(x)满足对任意正数x,y满足f(x×y)=f(x)×f(y),且x>1时,0
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![已知定义在(-∞,0)U(0,+∞)上的偶函数f(x)满足对任意正数x,y满足f(x×y)=f(x)×f(y),且x>1时,0](/uploads/image/z/15104506-58-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9A%E4%B9%89%E5%9C%A8%28-%E2%88%9E%2C0%29U%280%2C%2B%E2%88%9E%29%E4%B8%8A%E7%9A%84%E5%81%B6%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3%E5%AF%B9%E4%BB%BB%E6%84%8F%E6%AD%A3%E6%95%B0x%2Cy%E6%BB%A1%E8%B6%B3f%28x%C3%97y%29%3Df%28x%29%C3%97f%28y%29%2C%E4%B8%94x%3E1%E6%97%B6%2C0)
已知定义在(-∞,0)U(0,+∞)上的偶函数f(x)满足对任意正数x,y满足f(x×y)=f(x)×f(y),且x>1时,0
已知定义在(-∞,0)U(0,+∞)上的偶函数f(x)满足对任意正数x,y满足f(x×y)=f(x)×f(y),且x>1时,0
2.求证f(x)在(0,+∞)上是减函数
3.若f(4)=1/2,解不等式f(x)-4≥0
4.求证恒有f(x)>0
已知定义在(-∞,0)U(0,+∞)上的偶函数f(x)满足对任意正数x,y满足f(x×y)=f(x)×f(y),且x>1时,0
(1)
put y=1
f(x)=f(x)f(1)
=> f(1) = 1
(2)
for y>x and x,y∈(0,+∞)
then y = kx where k > 1
f(y) = f(kx)
= f(k)f(x)
< f(x)
f是减函数
(3)
for |x| >1 then f(x) 1 is not solution of f(x)-4≥0
Consider |x| < 1
f(4) = 1/2
put x = 4 ,y= 1/4
f(1) = f(4)f(1/4)
1 = (1/2) f(1/4)
f(1/4) = 2
put x=y= 1/4
f(1/16) = f(1/4)f(1/4)
f(1/16) = 4
f(x)-4≥0
0 < x ≤ 1/16 or -1/16 ≤ x < 0
(4)
x ∈(0,+∞)
put y=x
f(x^2) = f(x) f(x) > 0
f(x) > 0 for x ∈(0,+∞)
for x∈(-∞,0)
x -x >0
f(x) = f(-x) > 0
恒有f(x)>0
1、f(2*1)=f(2)*f(1);
因为f(2)>0所以f(1)=1;
2、任取x1,x2∈(0,+∞);
x1>x2; x1=(x1/x2)*x2;
则有f(x1)=f((x1/x2)*x2)=f(x1/x2)*f(x2);
所以f(x1)/f(x2)=f(x1/x2);
因为x1/x2>1;所以0
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1、f(2*1)=f(2)*f(1);
因为f(2)>0所以f(1)=1;
2、任取x1,x2∈(0,+∞);
x1>x2; x1=(x1/x2)*x2;
则有f(x1)=f((x1/x2)*x2)=f(x1/x2)*f(x2);
所以f(x1)/f(x2)=f(x1/x2);
因为x1/x2>1;所以0
f(x)在(0,+∞)上是减函数
3、因为f(4*0.25)=f(1)=f(4)*f(0.25)=1;
所以f(0.25)=2;
f(0.625)=f(0.25*0.25)=f(0.25)*f(0.25)=4;
因为f(x)-4≥0;
所以f(x)≥4;
f(x)≥f(0.625);
因为f(x)在(0,+∞)上是减函数;
所以0≤x≤0.625;
又因为
f(x)是定义在(-∞,0)U(0,+∞)上的偶函数;
所以
当-0.625≤x≤0时
也有f(x)-4≥0;
综上得:-0.625≤x≤0.625;
(注:0.625=1/16);
4、任取 x∈(0,1),
则(1/x)>1
f(1)=f(x*(1/x))=f(x)*f(1/x)=1;
因为0
所以f(x)在(0,1)U{1}U(1,+∞)上均大于0;
即f(x)在(0,+∞)上恒有f(x)>0;
因为f(x)是偶函数;
所以恒有f(x)>0;
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