函数的对称性怎么判断?已知定义在R上的增函数y=f(x)满足①f(x)=f(2-x);②x大于或等于0小于或等于1时,f(x)= x平方问(1)求f(5点5)的值(2)证明:x属于R时,f(x+2)=f(x)答案点播时说 因为f(x)=f(x+2)
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![函数的对称性怎么判断?已知定义在R上的增函数y=f(x)满足①f(x)=f(2-x);②x大于或等于0小于或等于1时,f(x)= x平方问(1)求f(5点5)的值(2)证明:x属于R时,f(x+2)=f(x)答案点播时说 因为f(x)=f(x+2)](/uploads/image/z/6605181-45-1.jpg?t=%E5%87%BD%E6%95%B0%E7%9A%84%E5%AF%B9%E7%A7%B0%E6%80%A7%E6%80%8E%E4%B9%88%E5%88%A4%E6%96%AD%3F%E5%B7%B2%E7%9F%A5%E5%AE%9A%E4%B9%89%E5%9C%A8R%E4%B8%8A%E7%9A%84%E5%A2%9E%E5%87%BD%E6%95%B0y%3Df%28x%29%E6%BB%A1%E8%B6%B3%E2%91%A0f%28x%29%3Df%282-x%29%3B%E2%91%A1x%E5%A4%A7%E4%BA%8E%E6%88%96%E7%AD%89%E4%BA%8E0%E5%B0%8F%E4%BA%8E%E6%88%96%E7%AD%89%E4%BA%8E1%E6%97%B6%2Cf%28x%29%3D+x%E5%B9%B3%E6%96%B9%E9%97%AE%EF%BC%881%EF%BC%89%E6%B1%82f%EF%BC%885%E7%82%B95%EF%BC%89%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89%E8%AF%81%E6%98%8E%EF%BC%9Ax%E5%B1%9E%E4%BA%8ER%E6%97%B6%2Cf%28x%2B2%29%3Df%28x%29%E7%AD%94%E6%A1%88%E7%82%B9%E6%92%AD%E6%97%B6%E8%AF%B4+%E5%9B%A0%E4%B8%BAf%EF%BC%88x%29%3Df%28x%2B2%29)
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