设函数y=f(x)是定义域在R,并且满足f(x+y)=f(x)+f(y),f(1/3)=1,且当x>0时,f(x)>01.求f(0)值2.判断函数奇偶性3.如果f(x)+f(2+x)<2,求x的取值范围
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![设函数y=f(x)是定义域在R,并且满足f(x+y)=f(x)+f(y),f(1/3)=1,且当x>0时,f(x)>01.求f(0)值2.判断函数奇偶性3.如果f(x)+f(2+x)<2,求x的取值范围](/uploads/image/z/4344243-51-3.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0y%EF%BC%9Df%EF%BC%88x%EF%BC%89%E6%98%AF%E5%AE%9A%E4%B9%89%E5%9F%9F%E5%9C%A8R%2C%E5%B9%B6%E4%B8%94%E6%BB%A1%E8%B6%B3f%EF%BC%88x%EF%BC%8By%EF%BC%89%EF%BC%9Df%EF%BC%88x%EF%BC%89%EF%BC%8Bf%EF%BC%88y%EF%BC%89%2Cf%EF%BC%881%EF%BC%8F3%EF%BC%89%EF%BC%9D1%2C%E4%B8%94%E5%BD%93x%EF%BC%9E0%E6%97%B6%2Cf%EF%BC%88x%EF%BC%89%EF%BC%9E01%EF%BC%8E%E6%B1%82f%EF%BC%880%EF%BC%89%E5%80%BC2%EF%BC%8E%E5%88%A4%E6%96%AD%E5%87%BD%E6%95%B0%E5%A5%87%E5%81%B6%E6%80%A73%EF%BC%8E%E5%A6%82%E6%9E%9Cf%EF%BC%88x%EF%BC%89%EF%BC%8Bf%EF%BC%882%EF%BC%8Bx%EF%BC%89%EF%BC%9C2%2C%E6%B1%82x%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
设函数y=f(x)是定义域在R,并且满足f(x+y)=f(x)+f(y),f(1/3)=1,且当x>0时,f(x)>01.求f(0)值2.判断函数奇偶性3.如果f(x)+f(2+x)<2,求x的取值范围
设函数y=f(x)是定义域在R,并且满足f(x+y)=f(x)+f(y),f(1/3)=1,且当x>0时,f(x)>0
1.求f(0)值
2.判断函数奇偶性
3.如果f(x)+f(2+x)<2,求x的取值范围
设函数y=f(x)是定义域在R,并且满足f(x+y)=f(x)+f(y),f(1/3)=1,且当x>0时,f(x)>01.求f(0)值2.判断函数奇偶性3.如果f(x)+f(2+x)<2,求x的取值范围
1、
f(0 + 0) = f(0) + f(0)
f(0) = 2f(0)
f(0) = 0
2、
f[x + (-x)] = f(x) + f(-x)
f(0) = f(x) + f(-x)
0 = f(x) + f(-x)
f(-x) = -f(x)
根据定义,这个是奇函数
3、
因为f(x)解析式无法求出(是抽象函数),所以先求出f(x) = 2时的x值
f(1/3 + 1/3) = f(1/3) + f(1/3) = 1 + 1 = 2
f(2/3) = 2
f(x) + f(2+x) < 2
f[x + (2 + x)] < f(2/3)
f(2x + 2) < f(2/3)
要解上述不等式,需求出f(x)单调性
因为 当x>0时,f(x)>0,f(x)是奇函数,所以
当x < 0时,f(x)< 0
又假设 a > b > 0,
f(a + b) = f(a) + f(b)
因为 a + b > a ,b > 0,f(b) > 0,
所以f(a + b) > f(a)
所以f(x)在x>0时单调递增
又因为f(x)是奇函数,所以f(x)在R上单调递增
上述不等式f(2x + 2) < f(2/3)
得 2x + 2 < 2/3
2x < -4/3
x < -2/3