我们高二期末考试有一道题目是:已知x= -3 是函数f[x]=aln[1-x]+x^2+10x的一个极值点,【1】求a【2】求函数f[x]的单调区间.我算的a是-16 但是我的同学算的是16
网友回答
f'(x) = -a/(1-x) + 2x + 10
f'(-3) = -a/(1-(-3)) + 2(-3) + 10
= -a/4 + 4
= 0(1) a = 16
(2) a = 16,
f'(x) = -16/(1-x) + 2x + 10 = (-16 + 2x(1-x) + 10(1-x))/(1-x)
= (2x^2 + 8x + 6)/(x-1) = 2(x+1)(x+3)/(x-1)
x0, increasing
-1