Question

решите неравенство f'(x)<0, если: f(x)=x/(x-3)

Answer 1

$mathtt{f(x)=frac{x}{x-3};~f'(x)=frac{x'(x-3)-x(x-3)'}{(x-3)^2}=-frac{3}{(x-3)^2};}\mathtt{f'(x) textless 0~to~-frac{3}{(x-3)^2} textless 0~to~frac{1}{(x-3)^2} textgreater 0~to~xin(infty;3)U(3;+infty)}$