Question

函数f(x)=2x-根号4x-x^2的值域是多少

Answer 1

f(x)=2x-√(4x-x^2)
=2x-√(4-(x-2)^2)
又4x-x^2>=0
=>x(x-4)<=0
=>0<=x<=4
令x=2*sinA+2 （-90<=A<=90)

则f(x)=2x-√(4-(x-2)^2)
=4+4sinA-√(4-4sinA^2)
=4+4sinA-2cosA (i)

设cosB=2/√5,sinB=1/√5,0<B<30
4sinA-2cosA
=2√5(2/√5*sinA-1/√5*cosA)
=2√5*(sinA*cosB-cosA*sinB)
=2√5*sin(A-B) (ii)

因为-90<=A<=90
=>-90-B<=A-B<=90-B
sin(A-B)>=-1
sin(A-B)<=sin(90-B)=cosB=2/√5

=>-2√5<=2√5*sin(A-B)<=2/√5*2√5=4 (iii)
(i)(ii)(iii)
=>4-2√5<=f(x)<=4+4=8
因此有
f(x)∈[4-2√5,8]

Answer 2

看这位的解法， 可以参考。
http://zhidao.baidu.com/question/181084139.html

但这里纠正和补充一点
2x-y=sqrt(4x-x^2)>=0.
:. y<=2x
又4x-x^2>=0, x ∈[0,4]
y<=2*4=8.

y ∈[4-2√5, 8] .这个是正确结果。

Answer 3

[0,4]