Question

设f(x)=ln|（x-1）（x-2）（x-3）|,则方程f '(x)=0的根的个数为多少

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则f'(x)是单调函数,与x轴只有一个交点,即根的个数为1