The function $f(x)=|x-1|$ is strictly increasing in the interval .

(0, 1) (-∞, 0) (-∞, -1) (1, ∞)

(1, ∞)

The correct answer is Option (4) → (1, ∞) $f(x)=|x-1|=\left\{\begin{matrix}x-1,&x≥1\\1-x,&x<1\end{matrix}\right.$ $f'(x)=\left\{\begin{matrix}1,&x≥1\\-1,&x<1\end{matrix}\right.$ f(x) is strictly increasing in (1, ∞)