已知函数f(x)定义在[-1,1],且满足① f(1)=1;②f(-x)=-f(x);③m,n∈[-1,1],m+n不等于0,有[f(m)+f(n)]/(m+n)>0. 请解不等式f(x+0.5)
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![已知函数f(x)定义在[-1,1],且满足① f(1)=1;②f(-x)=-f(x);③m,n∈[-1,1],m+n不等于0,有[f(m)+f(n)]/(m+n)>0. 请解不等式f(x+0.5)](/uploads/image/z/8643952-64-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%E5%AE%9A%E4%B9%89%E5%9C%A8%5B-1%2C1%5D%2C%E4%B8%94%E6%BB%A1%E8%B6%B3%E2%91%A0+f%281%29%3D1%3B%E2%91%A1f%28-x%29%3D-f%28x%29%3B%E2%91%A2m%2Cn%E2%88%88%5B-1%2C1%5D%2Cm%2Bn%E4%B8%8D%E7%AD%89%E4%BA%8E0%2C%E6%9C%89%5Bf%EF%BC%88m%EF%BC%89%2Bf%EF%BC%88n%EF%BC%89%5D%2F%EF%BC%88m%2Bn%EF%BC%89%3E0.+++++++%E8%AF%B7%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%EF%BC%88x%2B0.5%EF%BC%89)
已知函数f(x)定义在[-1,1],且满足① f(1)=1;②f(-x)=-f(x);③m,n∈[-1,1],m+n不等于0,有[f(m)+f(n)]/(m+n)>0. 请解不等式f(x+0.5)
已知函数f(x)定义在[-1,1],且满足① f(1)=1;②f(-x)=-f(x);③m,n∈[-1,1],m+n不等于0,有[f(m)+f(n)]/(m+n)>0. 请解不等式f(x+0.5)
已知函数f(x)定义在[-1,1],且满足① f(1)=1;②f(-x)=-f(x);③m,n∈[-1,1],m+n不等于0,有[f(m)+f(n)]/(m+n)>0. 请解不等式f(x+0.5)
由函数的定义域[-1,1],②f(-x)=-f(x)可判断f(x)在[-1,1]上为奇函数.
因为奇函数关于原点对称,故f(-1)=-f(1)=-1.
又此题f(x)的定义域为[-1,1],没有除去0点,故函数图像过原点,所以f(0)=0.
由条件③有f(x+0.5)+f[-1/(x+1)]/[x+0.5-1/(x+1)]>0…⑴
由f(x+0.5)<f[1/(x+1)]有:f(x+0.5)-f[1/(x+1)]<0
又奇函数f(x),从而:f(x+0.5)+f[-1/(x+1)]<0…⑵
故要满足⑴,且f(x+0.5)+f[-1/(x+1)]<0,故x+0.5-1/(x+1)<0…⑶
解这个不等式就要注意了!
由函数f(x)定义为[-1,1],得函数f(x+0.5)定义域:-1≤x+0.5≤1,即-1.5≤x≤0.5…⑷
函数f[1/(x+1)]的定义域:-1≤1/(x+1)≤1,即:|1/(x+1)|≤1,解得:x≤-2或x≥0…⑸
由⑷⑸得0≤x≤0.5,从而不等式⑶可化为(x+0.5)(x+1)<1,解得:
(-3-√17)/4<x<(-3+√17)/4,再综合0≤x≤0.5
得:0≤x<(-3+√17)/4.