设函数f(x)=-1/3x³+x²+(m²-1)x (x∈R)其中M>0﹙1)当M=1时曲线y=f(x)在点 [ 1.f(1)]外切线斜率(2) 求函数单调区间与极值
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![设函数f(x)=-1/3x³+x²+(m²-1)x (x∈R)其中M>0﹙1)当M=1时曲线y=f(x)在点 [ 1.f(1)]外切线斜率(2) 求函数单调区间与极值](/uploads/image/z/3738365-53-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D-1%2F3x%26%23179%3B%2Bx%26%23178%3B%2B%28m%26%23178%3B-1%29x+%28x%E2%88%88R%29%E5%85%B6%E4%B8%ADM%EF%BC%9E0%EF%B9%991%EF%BC%89%E5%BD%93M%3D1%E6%97%B6%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%9C%A8%E7%82%B9+%5B+1.f%281%29%5D%E5%A4%96%E5%88%87%E7%BA%BF%E6%96%9C%E7%8E%87%EF%BC%882%EF%BC%89+%E6%B1%82%E5%87%BD%E6%95%B0%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4%E4%B8%8E%E6%9E%81%E5%80%BC)
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