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
问题描述:
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
答
f(x) = 6x³ + 3(a + 2)x² + 2axf'(x) = 18x² + 6(a + 2)x + 2af'(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 =...