Question

已知函数g(x)=a/x,a为常数,且a>0,若f(x)=ln(x-1)+g(x),且a=6,求函数f(x)的单调区间拜

Answer 1

f(x)=a(x平方-x-1)*e^(3-x) f'(x)=a(2x-1)*e^(3-x) a(x-x-1)*e^(3-x)*(-1)=a*e^(3-x)[(2x-1)-(x平方-x-1)] =a*e^(3-x)(-x平方 3x)=-a*e^(3-x)*x(x-3) a>0时 令f'(x)3或x0,x>3或x