...其图象在点(1,f(1))处的切线与直线x-6y-7=0垂直,导函数f
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发布时间：2024-10-04 17:08
共1个回答
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时间：2024-11-04 12:37
∴f（-x）=-f（x），
即-ax3-bx+c=-ax3-bx-c，
∴c=0，
∵f′（x）=3ax2+b的最小值为-12，
∴b=-12，
又直线x-6y-7=0的斜率为16，则切线的斜率为-6，
∴f′（1）=3a+b=-6，
∴a=2，b=-12，c=0；
（2）∵f（x）=2x3-12x，
∴f′（x）=6x2-12=6（x+2）（x-2），列表如下：
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