已知函数f(x)=x∧2+ax+3-a,若X∈[-2,2]时,f(x)≥0恒成立,求a的取值范围做的时候为什么写f(-2)≥0和f(2)≥0有问题答案是-7≤a≤-4,用图象法要考虑△吗?为什么?暂不用最值法解题。写f(-2
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![已知函数f(x)=x∧2+ax+3-a,若X∈[-2,2]时,f(x)≥0恒成立,求a的取值范围做的时候为什么写f(-2)≥0和f(2)≥0有问题答案是-7≤a≤-4,用图象法要考虑△吗?为什么?暂不用最值法解题。写f(-2](/uploads/image/z/3018455-71-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dx%E2%88%A72%2Bax%2B3-a%2C%E8%8B%A5X%E2%88%88%5B-2%2C2%5D%E6%97%B6%2Cf%28x%29%E2%89%A50%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E5%81%9A%E7%9A%84%E6%97%B6%E5%80%99%E4%B8%BA%E4%BB%80%E4%B9%88%E5%86%99f%EF%BC%88-2%EF%BC%89%E2%89%A50%E5%92%8Cf%EF%BC%882%EF%BC%89%E2%89%A50%E6%9C%89%E9%97%AE%E9%A2%98%E7%AD%94%E6%A1%88%E6%98%AF-7%E2%89%A4a%E2%89%A4-4%EF%BC%8C%E7%94%A8%E5%9B%BE%E8%B1%A1%E6%B3%95%E8%A6%81%E8%80%83%E8%99%91%E2%96%B3%E5%90%97%EF%BC%9F%E4%B8%BA%E4%BB%80%E4%B9%88%EF%BC%9F%E6%9A%82%E4%B8%8D%E7%94%A8%E6%9C%80%E5%80%BC%E6%B3%95%E8%A7%A3%E9%A2%98%E3%80%82%E5%86%99f%EF%BC%88-2)
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