设函数f(x)=6x3+3(a+2)x2+2ax. 若f(x)的两个极值点为x1,x2,且x1x2=1,求实数a的值f′(x)=18x2+6(a+2)x+2a由已知有f′(x1)=f′(x2)=0,从而x1x2=2a 18 =1,所以a=9;我想知道x1x2=2a 18 =1是怎么来的?
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![设函数f(x)=6x3+3(a+2)x2+2ax. 若f(x)的两个极值点为x1,x2,且x1x2=1,求实数a的值f′(x)=18x2+6(a+2)x+2a由已知有f′(x1)=f′(x2)=0,从而x1x2=2a 18 =1,所以a=9;我想知道x1x2=2a 18 =1是怎么来的?](/uploads/image/z/2690015-23-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D6x3%2B3%EF%BC%88a%2B2%EF%BC%89x2%2B2ax%EF%BC%8E+%E8%8B%A5f%EF%BC%88x%EF%BC%89%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%9E%81%E5%80%BC%E7%82%B9%E4%B8%BAx1%2Cx2%2C%E4%B8%94x1x2%3D1%2C%E6%B1%82%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%80%BCf%E2%80%B2%EF%BC%88x%EF%BC%89%3D18x2%2B6%EF%BC%88a%2B2%EF%BC%89x%2B2a%E7%94%B1%E5%B7%B2%E7%9F%A5%E6%9C%89f%E2%80%B2%EF%BC%88x1%EF%BC%89%3Df%E2%80%B2%EF%BC%88x2%EF%BC%89%3D0%2C%E4%BB%8E%E8%80%8Cx1x2%3D2a+18+%3D1%2C%E6%89%80%E4%BB%A5a%3D9%EF%BC%9B%E6%88%91%E6%83%B3%E7%9F%A5%E9%81%93x1x2%3D2a+18+%3D1%E6%98%AF%E6%80%8E%E4%B9%88%E6%9D%A5%E7%9A%84%3F)
设函数f(x)=6x3+3(a+2)x2+2ax. 若f(x)的两个极值点为x1,x2,且x1x2=1,求实数a的值f′(x)=18x2+6(a+2)x+2a由已知有f′(x1)=f′(x2)=0,从而x1x2=2a 18 =1,所以a=9;我想知道x1x2=2a 18 =1是怎么来的?
设函数f(x)=6x3+3(a+2)x2+2ax. 若f(x)的两个极值点为x1,x2,且x1x2=1,求实数a的值
f′(x)=18x2+6(a+2)x+2a
由已知有f′(x1)=f′(x2)=0,
从而x1x2=2a 18 =1,
所以a=9;
我想知道x1x2=2a 18 =1是怎么来的?不懂啊
设函数f(x)=6x3+3(a+2)x2+2ax. 若f(x)的两个极值点为x1,x2,且x1x2=1,求实数a的值f′(x)=18x2+6(a+2)x+2a由已知有f′(x1)=f′(x2)=0,从而x1x2=2a 18 =1,所以a=9;我想知道x1x2=2a 18 =1是怎么来的?
f(x) = 6x³ + 3(a + 2)x² + 2ax
f'(x) = 18x² + 6(a + 2)x + 2a
f'(x1) = f'(x2) = 0,x1和x2都是f'(x)的根
根据韦达定理,两根之积x1 * x2 = (常数项)/(x²的系数) = (2a)/(18)
所以x1 * x2 = 2a/18 = 1