f(x)=2sin(π/2x+π/5)若对任意的x∈R,都有f(x1)≤f(x)≤f(x2)成立,则|x1-x2|最小值为——要过
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![f(x)=2sin(π/2x+π/5)若对任意的x∈R,都有f(x1)≤f(x)≤f(x2)成立,则|x1-x2|最小值为——要过](/uploads/image/z/5243271-15-1.jpg?t=f%EF%BC%88x%EF%BC%89%3D2sin%EF%BC%88%CF%80%2F2x%2B%CF%80%2F5%EF%BC%89%E8%8B%A5%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84x%E2%88%88R%2C%E9%83%BD%E6%9C%89f%28x1%29%E2%89%A4f%EF%BC%88x%EF%BC%89%E2%89%A4f%EF%BC%88x2%EF%BC%89%E6%88%90%E7%AB%8B%2C%E5%88%99%7Cx1-x2%7C%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA%E2%80%94%E2%80%94%E8%A6%81%E8%BF%87)
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f(x)=2sin(π/2x+π/5)若对任意的x∈R,都有f(x1)≤f(x)≤f(x2)成立,则|x1-x2|最小值为——要过
f(x)=2sin(π/2x+π/5)若对任意的x∈R,都有f(x1)≤f(x)≤f(x2)成立,则|x1-x2|最小值为——要过
f(x)=2sin(π/2x+π/5)若对任意的x∈R,都有f(x1)≤f(x)≤f(x2)成立,则|x1-x2|最小值为——要过
设函数F(x)=3cos(πx/2+π/3),若对任意x∈R都有f(x1)≤f(x)≤f(x2)成立,则|x1-x2|的最小值为?
函数f(x)周期为4,|f(x)|≤3.
要使f(x1)≤f(x)≤f(x2)恒成立,必须有f(x2)取最大值3,f(x1)取最小值-3,
也就是x1与x2相隔半个周期的奇数倍,
故│x1-x2│最小值为半个周期 2.