数学函数对称及周期问题3道2.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (x)是奇函数,则f (x)的周期是 .A.2 B.4 C.6 D.83.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (x)是偶函数,则f
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![数学函数对称及周期问题3道2.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (x)是奇函数,则f (x)的周期是 .A.2 B.4 C.6 D.83.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (x)是偶函数,则f](/uploads/image/z/4949884-28-4.jpg?t=%E6%95%B0%E5%AD%A6%E5%87%BD%E6%95%B0%E5%AF%B9%E7%A7%B0%E5%8F%8A%E5%91%A8%E6%9C%9F%E9%97%AE%E9%A2%983%E9%81%932%EF%BC%8E%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f+%28x%29%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E6%98%AFR%2C%E4%B8%94f+%282+-+x%29+%3D+-+f+%28x+%2B+2%29%2C%E8%8B%A5f+%28x%29%E6%98%AF%E5%A5%87%E5%87%BD%E6%95%B0%2C%E5%88%99f+%28x%29%E7%9A%84%E5%91%A8%E6%9C%9F%E6%98%AF+.A.2+B.4+C.6+D.83%EF%BC%8E%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f+%28x%29%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E6%98%AFR%2C%E4%B8%94f+%282+-+x%29+%3D+-+f+%28x+%2B+2%29%2C%E8%8B%A5f+%28x%29%E6%98%AF%E5%81%B6%E5%87%BD%E6%95%B0%2C%E5%88%99f)
数学函数对称及周期问题3道2.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (x)是奇函数,则f (x)的周期是 .A.2 B.4 C.6 D.83.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (x)是偶函数,则f
数学函数对称及周期问题3道
2.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (x)是奇函数,则f (x)的周
期是 .
A.2 B.4 C.6 D.8
3.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (x)是偶函数,则f (x)的周
期是 .
A.2 B.4 C.6 D.8
4.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (1+ x) = f (1- x),则f (x)
的周期是 .
A.2 B.4 C.6 D.8
数学函数对称及周期问题3道2.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (x)是奇函数,则f (x)的周期是 .A.2 B.4 C.6 D.83.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (x)是偶函数,则f
2.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (x)是奇函数,则f (x)的周期是 4 .
分析:∵f (2 - x) = - f (x + 2),
∴f(x+4)=f[(x+2)+2]=-f[2-(x+2)]=-f(-x),
又f (x)是奇函数,得f(-x)= - f(x),
∴f(x+4)= -f(-x)=f(x),T=4.
3.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (x)是偶函数,则f (x)的周期是 8 .
分析:∵f (2 - x) = - f (x + 2),
∴f(x+4)=f[(x+2)+2]=-f[2-(x+2)]=-f(-x),
又f (x)是偶函数,得f(-x)= f(x),
∴f(x+4)= -f(-x)=-f(x),
f(x+8)=f[(x+4)+4]=-f(x+4)=f(x),T=8.
4.已知函数f (x)的定义域是R,且f (2 - x) = - f (x + 2),若f (1+ x) = f (1- x),则f (x)的周期是 4 .
分析:∵f (1+ x) = f (1- x),
∴f(2-x)=f[1+(1-x)]=f[1-(1-x)]=f(x),
又f (2 - x) = - f (x + 2),
∴f(x+2)=-f(x),
f(x+4)=f[(x+2)+2]=-f(x+2)=f(x),T=4.