Question

函数f（X）=sin2x+2根号2cos（π/4+x）+3的值域

今天就用！！！！！！！！！！！！！！！

Answer 1

f(X)=sin2x+2√2cos（π/4+x）+3
=cos(2x-π/2)+2√2cos（π/4+x）+3
=1-2sin²(x-π/4)-2√2sin（x-π/4）+3
=1-[2sin²(x-π/4)+2√2sin（x-π/4)]+3
=4-[2sin²(x-π/4)+2√2sin（x-π/4)+1]+1
=5-[2sin²(x-π/4)+2√2sin（x-π/4)+1]
=5-[√2sin（x-π/4)+1]^2
-1<=sin（x-π/4)<=1
-√2<=√2sin（x-π/4)<=√2
1-√2<=√2sin（x-π/4)+1<=1+√2
3-2√2<=[√2sin（x-π/4)+1]^2<=3+2√2
-3-2√2<=-[√2sin（x-π/4)+1]^2<=2√2-3
-3-2√2+5<=5-[√2sin（x-π/4)+1]^2<=2√2-3+5
2-2√2<=5-[√2sin（x-π/4)+1]^2<=2√2+2

f(X)值域y∈[2-2√2,2√2+2]

Answer 2

是f（X）=sin2x+2根号【2cos（π/4+x）】+3，还是f（X）=sin2x+2根号【2cos（π/4+x）+3】？
因为-1<sin2x<1，-1<cos（π/4+x）<1
所以-2<2cos（π/4+x）<2，于是
如果是前者：
0<2根号【2cos（π/4+x）】<2根号2,，于是2<f（X）<4+2根号2
如果是后者：
2根号2<2根号【2cos（π/4+x）+3】<2根号5

Answer 3

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