已知函数f(x)=x^3+ax^2+bx+c,曲线在点x=1处的切线为3x-y+1=0,若x=2/3时,y=f(x )有极值.求y=f(x)在[-3,1]上的最大值和最小值.
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![已知函数f(x)=x^3+ax^2+bx+c,曲线在点x=1处的切线为3x-y+1=0,若x=2/3时,y=f(x )有极值.求y=f(x)在[-3,1]上的最大值和最小值.](/uploads/image/z/1168777-1-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dx%5E3%2Bax%5E2%2Bbx%2Bc%2C%E6%9B%B2%E7%BA%BF%E5%9C%A8%E7%82%B9x%3D1%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E4%B8%BA3x-y%2B1%3D0%2C%E8%8B%A5x%3D2%2F3%E6%97%B6%2Cy%3Df%28x+%29%E6%9C%89%E6%9E%81%E5%80%BC.%E6%B1%82y%3Df%28x%29%E5%9C%A8%5B-3%2C1%5D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC.)
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