二次函数f(x)=ax^2+bx+c满足①f(-1)=0 ②x≤f(x)≤(x^2+1)/2恒成立,求f(x)的表达式越详细越好,本人理解能力较为薄弱.
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![二次函数f(x)=ax^2+bx+c满足①f(-1)=0 ②x≤f(x)≤(x^2+1)/2恒成立,求f(x)的表达式越详细越好,本人理解能力较为薄弱.](/uploads/image/z/5506825-49-5.jpg?t=%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%28x%29%3Dax%5E2%2Bbx%2Bc%E6%BB%A1%E8%B6%B3%E2%91%A0f%28-1%29%3D0+%E2%91%A1x%E2%89%A4f%28x%29%E2%89%A4%28x%5E2%2B1%29%2F2%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82f%28x%29%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F%E8%B6%8A%E8%AF%A6%E7%BB%86%E8%B6%8A%E5%A5%BD%2C%E6%9C%AC%E4%BA%BA%E7%90%86%E8%A7%A3%E8%83%BD%E5%8A%9B%E8%BE%83%E4%B8%BA%E8%96%84%E5%BC%B1.)
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