函數的單調區間一

分析以下函數的單調區間：(1) y(x)=3x^4-6x^2+2=3(x^4-2x^2+1)+2-3=3(x^2-1)^2-1x=+-1 => min=y(-1)=-1y(x)>=-1.........ans
(2) y(x)=3*2^(2x)-6*2^x+2=3(2^x-1)^2-1x=0 => min=y(0)=-1y(x)>=-1..........ans
(3) y(x)=3sin^2(x)-6sin(x)+2=3[sin(x)-1]^2-1x=π/2 => y(π/2)=-1y(x)>=-1..........ans


2013-02-11 08:57:20 補充：
(3) 繼續

y(0)=2, y(3π/2)=3+6+2=11

-1<=y(x)<=11......ans

2013-02-12 09:48:59 補充：
(1) x=+-1 => min=y(+-1)=-1

y(0)=2

x=~-1 => 遞減

x=-1~0 => 遞增

x==0~1 => 遞減

x=1~ => 遞增

2013-02-12 14:14:27 補充：
(3) y(x)=3sin^2(x)-6sin(x)+2=3[sin(x)-1]^2-1

x=π/2 => y(π/2)=-1

y(x)>=-1

y(-3π/2)=y(-5π/2)=y(-7π/2)=....=3(-2)^2-1=11

y(-π/2)=y(π/2)=y(3π/2)=....=-1

週期: T=π

..............

x=-5π/2~-3π/2 => 遞減

x=-3π/2~-π/2 => 遞增

x=-π/2~π/2 => 遞減

x=π/2~3π/2 => 遞增

........

2013-02-12 14:15:00 補充：
(1) y(x)=3x^4-6x^2+2

=3(x^4-2x^2+1)+2-3

=3(x^2-1)^2-1

x=+-1 => min=y(+-1)=-1

y(0)=2

x=~-1 => 遞減

x=-1~0 => 遞增

x==0~1 => 遞減

x=1~ => 遞增

2013-02-12 14:30:10 補充：
(2) y(x)=3*2^(2x)-6*2^x+2=3(2^x-1)^2-1

x=0 => min=y(0)=-1

y(x)>=-1

y(-∞)=3(0-1)^2-1=2

x=-∞~0 => 遞減

x=0~∞ => 遞增