已知函数f(x)=2asin(2x-π/6)+b的定义域为[0,π/2],值域为[-5,1],求常数a,b的值x∈[0,π/2]则 2x-π/6 ∈[-π/6,5π/6]所以 sin(x-π/6) ∈[-1/2,1](1) a>0最大值为2a+b=1最小值为-a+b=-5所以 a=2,b=-3(2)a
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![已知函数f(x)=2asin(2x-π/6)+b的定义域为[0,π/2],值域为[-5,1],求常数a,b的值x∈[0,π/2]则 2x-π/6 ∈[-π/6,5π/6]所以 sin(x-π/6) ∈[-1/2,1](1) a>0最大值为2a+b=1最小值为-a+b=-5所以 a=2,b=-3(2)a](/uploads/image/z/1748818-10-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2asin%282x-%CF%80%2F6%29%2Bb%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BA%5B0%2C%CF%80%2F2%5D%2C%E5%80%BC%E5%9F%9F%E4%B8%BA%5B-5%2C1%5D%2C%E6%B1%82%E5%B8%B8%E6%95%B0a%2Cb%E7%9A%84%E5%80%BCx%E2%88%88%5B0%2C%CF%80%2F2%5D%E5%88%99+2x-%CF%80%2F6+%E2%88%88%5B-%CF%80%2F6%2C5%CF%80%2F6%5D%E6%89%80%E4%BB%A5+sin%28x-%CF%80%2F6%29+%E2%88%88%5B-1%2F2%2C1%5D%281%29+a%3E0%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA2a%2Bb%3D1%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA-a%2Bb%3D-5%E6%89%80%E4%BB%A5+a%3D2%2Cb%3D-3%282%29a)
已知函数f(x)=2asin(2x-π/6)+b的定义域为[0,π/2],值域为[-5,1],求常数a,b的值x∈[0,π/2]则 2x-π/6 ∈[-π/6,5π/6]所以 sin(x-π/6) ∈[-1/2,1](1) a>0最大值为2a+b=1最小值为-a+b=-5所以 a=2,b=-3(2)a
已知函数f(x)=2asin(2x-π/6)+b的定义域为[0,π/2],值域为[-5,1],求常数a,b的值
x∈[0,π/2]
则 2x-π/6 ∈[-π/6,5π/6]
所以 sin(x-π/6) ∈[-1/2,1]
(1) a>0
最大值为2a+b=1
最小值为-a+b=-5
所以 a=2,b=-3
(2)a
已知函数f(x)=2asin(2x-π/6)+b的定义域为[0,π/2],值域为[-5,1],求常数a,b的值x∈[0,π/2]则 2x-π/6 ∈[-π/6,5π/6]所以 sin(x-π/6) ∈[-1/2,1](1) a>0最大值为2a+b=1最小值为-a+b=-5所以 a=2,b=-3(2)a
你是不是看错了.sin(2x-π/6) ∈[-1/2,1]画个sin的图像把2x-π/6 ∈[-π/6,5π/6]的最大最小值找到
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