设函数f(x)=2sin(ωx+π/6)(ω>0)对于x∈R有f(x1)≤f(x)≤f(x2)且点A(x1,f(x1))与点B(x2,f(x2))之间距离为√20,则ω最小值为
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![设函数f(x)=2sin(ωx+π/6)(ω>0)对于x∈R有f(x1)≤f(x)≤f(x2)且点A(x1,f(x1))与点B(x2,f(x2))之间距离为√20,则ω最小值为](/uploads/image/z/60977-65-7.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D2sin%28%CF%89x%2B%CF%80%2F6%29%28%CF%89%3E0%29%E5%AF%B9%E4%BA%8Ex%E2%88%88R%E6%9C%89f%28x1%29%E2%89%A4f%28x%29%E2%89%A4f%28x2%29%E4%B8%94%E7%82%B9A%28x1%2Cf%28x1%29%29%E4%B8%8E%E7%82%B9B%28x2%2Cf%28x2%29%29%E4%B9%8B%E9%97%B4%E8%B7%9D%E7%A6%BB%E4%B8%BA%E2%88%9A20%2C%E5%88%99%CF%89%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA)
设函数f(x)=2sin(ωx+π/6)(ω>0)对于x∈R有f(x1)≤f(x)≤f(x2)且点A(x1,f(x1))与点B(x2,f(x2))之间距离为√20,则ω最小值为
设函数f(x)=2sin(ωx+π/6)(ω>0)对于x∈R有f(x1)≤f(x)≤f(x2)且点A(x1,f(x1))与点B(x2,f(x2))之间距离为√20,则ω最小值为
设函数f(x)=2sin(ωx+π/6)(ω>0)对于x∈R有f(x1)≤f(x)≤f(x2)且点A(x1,f(x1))与点B(x2,f(x2))之间距离为√20,则ω最小值为
我们知道,如果要满足对于x∈R有f(x1)≤f(x)≤f(x2),则f(x1)和f(x2)必须是f(x)的最小和最大值,所以f(x1)=-2,f(x2)=2
A(x1,f(x1))与点B(x2,f(x2))之间距离为√20,即A 和B的距离的平方等于20
就是(x2-x1)²+(f(x2)-f(x1))²=20
f(x2)-f(x1)=4
所以(x2-x1)²+16=20
所以x2-x1=2
因为这连个点分别是最大和最小值点,所以他们之间的距离的最小值就是半个周期
2是半个周期,那么一个周期就是4
所以w的最大值当半个周期取最小的时候的值
根据T=2π/w得w=2/π 也就是ω最小值为2/π
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