求函数y=x-2+根号下(4-x²)的值域.函数定义域是 -2≤x≤2,可设 x=2sinθ,θ∈[-π/2,π/2],则:根号下(4-x^2)=2cosθ (不带绝对值,因为 θ∈[-π/2,π/2])原函数即为 y=2sinθ -2 + 2cosθ=2(sinθ+cosθ)-2=2√2
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 01:03:07
![求函数y=x-2+根号下(4-x²)的值域.函数定义域是 -2≤x≤2,可设 x=2sinθ,θ∈[-π/2,π/2],则:根号下(4-x^2)=2cosθ (不带绝对值,因为 θ∈[-π/2,π/2])原函数即为 y=2sinθ -2 + 2cosθ=2(sinθ+cosθ)-2=2√2](/uploads/image/z/6897895-7-5.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0y%3Dx-2%2B%E6%A0%B9%E5%8F%B7%E4%B8%8B%284-x%26%23178%3B%29%E7%9A%84%E5%80%BC%E5%9F%9F.%E5%87%BD%E6%95%B0%E5%AE%9A%E4%B9%89%E5%9F%9F%E6%98%AF+-2%E2%89%A4x%E2%89%A42%2C%E5%8F%AF%E8%AE%BE+x%EF%BC%9D2sin%CE%B8%2C%CE%B8%E2%88%88%5B-%CF%80%2F2%2C%CF%80%2F2%5D%2C%E5%88%99%EF%BC%9A%E6%A0%B9%E5%8F%B7%E4%B8%8B%284-x%5E2%29%EF%BC%9D2cos%CE%B8+%28%E4%B8%8D%E5%B8%A6%E7%BB%9D%E5%AF%B9%E5%80%BC%2C%E5%9B%A0%E4%B8%BA+%CE%B8%E2%88%88%5B-%CF%80%2F2%2C%CF%80%2F2%5D%29%E5%8E%9F%E5%87%BD%E6%95%B0%E5%8D%B3%E4%B8%BA+y%EF%BC%9D2sin%CE%B8+-2+%2B+2cos%CE%B8%EF%BC%9D2%28sin%CE%B8%2Bcos%CE%B8%29-2%EF%BC%9D2%E2%88%9A2)
求函数y=x-2+根号下(4-x²)的值域.函数定义域是 -2≤x≤2,可设 x=2sinθ,θ∈[-π/2,π/2],则:根号下(4-x^2)=2cosθ (不带绝对值,因为 θ∈[-π/2,π/2])原函数即为 y=2sinθ -2 + 2cosθ=2(sinθ+cosθ)-2=2√2
求函数y=x-2+根号下(4-x²)的值域.
函数定义域是 -2≤x≤2,可设 x=2sinθ,θ∈[-π/2,π/2],
则:根号下(4-x^2)=2cosθ (不带绝对值,因为 θ∈[-π/2,π/2])
原函数即为 y=2sinθ -2 + 2cosθ=2(sinθ+cosθ)-2=2√2·sin(θ+π/4) - 2 .
因 -π/4≤θ+π/4≤3π/4 ,故 -√2/2 ≤sin(θ+π/4)≤sin(π/2)=1,这一步怎么出来的的.为啥是≤sin(π/2)=1?
于是 y 的值域是 [-4 ,2√2 - 2 ]
其中,√2 表示 根号2 .
求函数y=x-2+根号下(4-x²)的值域.函数定义域是 -2≤x≤2,可设 x=2sinθ,θ∈[-π/2,π/2],则:根号下(4-x^2)=2cosθ (不带绝对值,因为 θ∈[-π/2,π/2])原函数即为 y=2sinθ -2 + 2cosθ=2(sinθ+cosθ)-2=2√2
因为 sinx在x=π/2 时取最大值
不懂再问