Question

函数f（x）＝x除以x2+1怎么求值域

Answer 1

求函数f(x)=x/(x²+1)的值域

解：令f'(x)=(x²+1-2x²)/(x²+1)²=(1-x²)/(x²+1)²=-(x²-1)/(x²+1)²=-(x-1)(x+1)/(x²+1)²=0

得驻点 x₁=-1，x₂=1；  当x<-1时f'(x)<0；当-1<x<1时f'(x)>0，∴x₁=-1是极小点，

极小值f(x)=f(-1)=-1/2；

当x>1时f'(x)<0；故x₂=1是极大点，极大值f(x)=f(1)=1/2；

又x→∞limf(x)=x→∞lim[x/(x²+1)]=0;

∴该函数的值域是[-1/2，1/2]；其图像如下：

Answer 2

令y=f(x)=x/(x²+1)，那么y(x²+1)=x，∴yx²-x+y=0
当y=0时，x=0，符合要求；当y≠0时，那么△=1-4y²≥0，∴-1/2≤y≤1/2，且y≠0
综上,-1/2≤y≤1/2，即值域为[-1/2,1/2]

Answer 3

f(x) = x/(x^2+1)
f'(x)
= (x^2+1 - 2x^2 )/(x^2+1)^2
= (-x^2+1 )/(x^2+1)^2
f'(x) =0 => x=1 or -1
f'(x) | x=1+ <0 , f'(x) | x=1- >0 => x=1 (max)
f'(x) | x=-1+ >0 , f'(x) | x=-1- <0 => x=-1 (min)
max f(x) = f(1) =1/(1+1)=1/2
min f(x) = f(-1) =-1/(1+1)=-1/2
x->∞, f(x) ->0
x->-∞, f(x) ->0
值域 = [ -1/2, 1/2]