已知函数f(x)=x^3+ax^2+bx,若y=f(x)的导数f'(x)对x∈[-1,1]都有f'(x)≤2,求b/(a-1)的范围 请用线性规划回答,答案是b/(a-1)∈(-∞,-2)∪[-1,+∞),还有-1的范围是怎么判断的?怎么想到那个点的?
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![已知函数f(x)=x^3+ax^2+bx,若y=f(x)的导数f'(x)对x∈[-1,1]都有f'(x)≤2,求b/(a-1)的范围 请用线性规划回答,答案是b/(a-1)∈(-∞,-2)∪[-1,+∞),还有-1的范围是怎么判断的?怎么想到那个点的?](/uploads/image/z/5348689-25-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dx%5E3%2Bax%5E2%2Bbx%2C%E8%8B%A5y%3Df%28x%29%E7%9A%84%E5%AF%BC%E6%95%B0f%27%28x%29%E5%AF%B9x%E2%88%88%5B-1%2C1%5D%E9%83%BD%E6%9C%89f%27%28x%29%E2%89%A42%2C%E6%B1%82b%2F%28a-1%29%E7%9A%84%E8%8C%83%E5%9B%B4+%E8%AF%B7%E7%94%A8%E7%BA%BF%E6%80%A7%E8%A7%84%E5%88%92%E5%9B%9E%E7%AD%94%2C%E7%AD%94%E6%A1%88%E6%98%AFb%2F%28a-1%29%E2%88%88%28-%E2%88%9E%2C-2%29%E2%88%AA%5B-1%2C%2B%E2%88%9E%29%2C%E8%BF%98%E6%9C%89-1%E7%9A%84%E8%8C%83%E5%9B%B4%E6%98%AF%E6%80%8E%E4%B9%88%E5%88%A4%E6%96%AD%E7%9A%84%EF%BC%9F%E6%80%8E%E4%B9%88%E6%83%B3%E5%88%B0%E9%82%A3%E4%B8%AA%E7%82%B9%E7%9A%84%EF%BC%9F)
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