已知函数f(x)=x³-3ax²-3(2a+1)x-3,x∈R,a是常熟.(1)若a=1/2,求函数y=f(x)在区间[-3,3]上零点的求助!
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![已知函数f(x)=x³-3ax²-3(2a+1)x-3,x∈R,a是常熟.(1)若a=1/2,求函数y=f(x)在区间[-3,3]上零点的求助!](/uploads/image/z/5427822-30-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dx%26%23179%3B-3ax%26%23178%3B-3%282a%2B1%29x-3%2Cx%E2%88%88R%2Ca%E6%98%AF%E5%B8%B8%E7%86%9F.%281%29%E8%8B%A5a%3D1%2F2%2C%E6%B1%82%E5%87%BD%E6%95%B0y%3Df%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5B-3%2C3%5D%E4%B8%8A%E9%9B%B6%E7%82%B9%E7%9A%84%E6%B1%82%E5%8A%A9%21)
已知函数f(x)=x³-3ax²-3(2a+1)x-3,x∈R,a是常熟.(1)若a=1/2,求函数y=f(x)在区间[-3,3]上零点的求助!
已知函数f(x)=x³-3ax²-3(2a+1)x-3,x∈R,a是常熟.(1)若a=1/2,求函数y=f(x)在区间[-3,3]上零点的
求助!
已知函数f(x)=x³-3ax²-3(2a+1)x-3,x∈R,a是常熟.(1)若a=1/2,求函数y=f(x)在区间[-3,3]上零点的求助!
题目不完整啊 这题可以用导数来求.利用单调性
SB
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