已知a∈R,函数f(x)=2x³-3(a+1)x²+6ax.若|a|>1,求f(x)在闭区间[0,|2a|]上的最小值.f'(x)=6x²-6(a+1)x+6a=6[x²-(a+1)x+a]=6(x-1)(x-a)令f'(x)=0得x₁=a,x₂=1接下来又不知怎
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![已知a∈R,函数f(x)=2x³-3(a+1)x²+6ax.若|a|>1,求f(x)在闭区间[0,|2a|]上的最小值.f'(x)=6x²-6(a+1)x+6a=6[x²-(a+1)x+a]=6(x-1)(x-a)令f'(x)=0得x₁=a,x₂=1接下来又不知怎](/uploads/image/z/3941518-22-8.jpg?t=%E5%B7%B2%E7%9F%A5a%E2%88%88R%2C%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D2x%26%23179%3B-3%EF%BC%88a%2B1%EF%BC%89x%26%23178%3B%2B6ax.%E8%8B%A5%7Ca%7C%3E1%2C%E6%B1%82f%28x%29%E5%9C%A8%E9%97%AD%E5%8C%BA%E9%97%B4%5B0%2C%7C2a%7C%5D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC.f%27%EF%BC%88x%EF%BC%89%3D6x%26%23178%3B-6%EF%BC%88a%EF%BC%8B1%EF%BC%89x%EF%BC%8B6a%3D6%5Bx%26%23178%3B-%EF%BC%88a%2B1%EF%BC%89x%2Ba%5D%3D6%EF%BC%88x-1%EF%BC%89%EF%BC%88x-a%EF%BC%89%E4%BB%A4f%27%EF%BC%88x%EF%BC%89%3D0%E5%BE%97x%26%238321%3B%3Da%2Cx%26%238322%3B%3D1%E6%8E%A5%E4%B8%8B%E6%9D%A5%E5%8F%88%E4%B8%8D%E7%9F%A5%E6%80%8E)
已知a∈R,函数f(x)=2x³-3(a+1)x²+6ax.若|a|>1,求f(x)在闭区间[0,|2a|]上的最小值.f'(x)=6x²-6(a+1)x+6a=6[x²-(a+1)x+a]=6(x-1)(x-a)令f'(x)=0得x₁=a,x₂=1接下来又不知怎
已知a∈R,函数f(x)=2x³-3(a+1)x²+6ax.若|a|>1,求f(x)在闭区间[0,|2a|]上的最小值.
f'(x)=6x²-6(a+1)x+6a=6[x²-(a+1)x+a]=6(x-1)(x-a)令f'(x)=0得x₁=a,x₂=1
接下来又不知怎么下手了~麻烦老师解析下吧
已知a∈R,函数f(x)=2x³-3(a+1)x²+6ax.若|a|>1,求f(x)在闭区间[0,|2a|]上的最小值.f'(x)=6x²-6(a+1)x+6a=6[x²-(a+1)x+a]=6(x-1)(x-a)令f'(x)=0得x₁=a,x₂=1接下来又不知怎
分类讨论
(1)a1时,f'(x)>0
00, f(x)递增
10 f(x)递增
则 最小值是f(0)和f(a)中的小的那一个
f(0)=0,f(a)=2a³-3a³-3a²+6a²=3a²-a³=a²(3-a)
① a>3, a²(3-a)