高中数学函数的周期、对称性若函数满足下列条件则有结论:f(x+2)=f(-x)f(x+2)=-f(x)f(x+2)=-f(-x)f(x+1)=±[1/f(x)]f(x+3)=f(-x+5)f(x+3)=f(x+5)
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 02:42:04
![高中数学函数的周期、对称性若函数满足下列条件则有结论:f(x+2)=f(-x)f(x+2)=-f(x)f(x+2)=-f(-x)f(x+1)=±[1/f(x)]f(x+3)=f(-x+5)f(x+3)=f(x+5)](/uploads/image/z/6893676-36-6.jpg?t=%E9%AB%98%E4%B8%AD%E6%95%B0%E5%AD%A6%E5%87%BD%E6%95%B0%E7%9A%84%E5%91%A8%E6%9C%9F%E3%80%81%E5%AF%B9%E7%A7%B0%E6%80%A7%E8%8B%A5%E5%87%BD%E6%95%B0%E6%BB%A1%E8%B6%B3%E4%B8%8B%E5%88%97%E6%9D%A1%E4%BB%B6%E5%88%99%E6%9C%89%E7%BB%93%E8%AE%BA%EF%BC%9Af%28x%2B2%29%3Df%28-x%29f%28x%2B2%29%3D-f%28x%29f%28x%2B2%29%3D-f%28-x%29f%28x%2B1%29%3D%C2%B1%5B1%2Ff%28x%29%5Df%28x%2B3%29%3Df%28-x%2B5%29f%28x%2B3%29%3Df%28x%2B5%29)
高中数学函数的周期、对称性若函数满足下列条件则有结论:f(x+2)=f(-x)f(x+2)=-f(x)f(x+2)=-f(-x)f(x+1)=±[1/f(x)]f(x+3)=f(-x+5)f(x+3)=f(x+5)
高中数学函数的周期、对称性
若函数满足下列条件则有结论:
f(x+2)=f(-x)
f(x+2)=-f(x)
f(x+2)=-f(-x)
f(x+1)=±[1/f(x)]
f(x+3)=f(-x+5)
f(x+3)=f(x+5)
高中数学函数的周期、对称性若函数满足下列条件则有结论:f(x+2)=f(-x)f(x+2)=-f(x)f(x+2)=-f(-x)f(x+1)=±[1/f(x)]f(x+3)=f(-x+5)f(x+3)=f(x+5)
结论是,布置作业的老师昨晚被wife揍了10000吨。
f(x+2)=f(-x)
令x=x-2代入得f(x)=f(-x+2)
∴f(x)关于直线x=1对称;
f(x+2)=-f(x)
令x=x+2代入得f(x+4)=-f(x+2)=f(x)
∴f(x)是以4为最小正周期的周期函数;
f(x+2)=-f(-x)
f(x+1)=±[1/f(x)]
令x=x+1代入得f(x+2)= ±1/f(x...
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f(x+2)=f(-x)
令x=x-2代入得f(x)=f(-x+2)
∴f(x)关于直线x=1对称;
f(x+2)=-f(x)
令x=x+2代入得f(x+4)=-f(x+2)=f(x)
∴f(x)是以4为最小正周期的周期函数;
f(x+2)=-f(-x)
f(x+1)=±[1/f(x)]
令x=x+1代入得f(x+2)= ±1/f(x+1)=f(x)
∴f(x)是以2为最小正周期的周期函数;
f(x+3)=f(-x+5)
令x=x-3代入得f(x)=f(-x+8)
∴f(x)关于直线x=4对称;
f(x+3)=f(x+5)
令x=x-3代入得f(x)=f(x+2)
∴f(x)是以2为最小正周期的周期函数;
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