f（x）=ln(x-1)(x-2)(x-3),f(x)拐点的个数是0,为什么（二阶导肿么就恒小于0了）

Question

f（x）=ln【(x-1)(x-2)(x-3)】 ,f(x)拐点的个数是0,为什么
（二阶导肿么就恒小于0了）

Answer 1

f(x)=ln [(x-1)(x-2)(x-3)]
= ln(x-1)+ln(x-2)+ln(x-3)
f'(x)=1/(x-1)+1/(x-2)+1/(x-3)
f''(x)=- [1/(x-1)^2+1/(x-2)^2+1/(x-3)^2] < 0
根据拐点的条件,函数 f(x)=ln [(x-1)(x-2)(x-3)] 没有拐点.即拐点的个数为零.
//:拐点条件：(1) f''(x0)=0 (2)在x0两侧f''(x)变号.