Question

函数f(x)=(x^3-x)/(x^2+1)^2的值域是

Answer 1

f '(x)=[(3x^2-1)(x^2+1)^2-(x^3-x)*2*(x^2+1)*2x]/(x^2+1)^4
let f '=0 -> (3x^2-1)(x^2+1)^2-(x^3-x)*2*(x^2+1)*2x = 0
(3x^2-1)(x^2+1) - 4x^2(x^2-1) = 0
3x^4+3x^2-x^2-1-4x^4+4x^2=0
x^4 - 6x^2 + 1 = 0
令 y=x^2 -> y^2-6y+1=0 y1,2=(6土√32)/2=3土2√2
x = 土√(3土2√2)
X1=√(3+2√2) X2=√(3-2√2) X3=-√(3+2√2) X4=-√(3-2√2)
x1=1+√2 x2=√2 -1 x3=-1-√2 x4=1-√2
f(x1)=0.25 f(x3)=-0.25 f(x2)=-0.25 f(x4)=0.25
可见
f(x) = (x^3-x)/(x^2+1)^2
的值域为： f ∈ [-0.25，0.25] //: 没去考虑是否有简便方法。

Answer 2

f(x)=(x^3-x)/(x^4+2x^2+1)
若x=0,则f(x)=0
若x≠0,分子分母同时处以x^2,则f(x)=(x-1/x)/(x^2+1/x^2+2)=(x-1/x)/((x-1/x)^2+4)
令t=x-1/x∈(-∞,+∞),则f(x)=1/(t+4/t)∈[-1/4,0)∪(0,1/4]
综上,f(x)∈[-1/4,1/4]

Answer 3

分解因式，极限法，画图法。

Answer 4

xfbf