Question

The minimum value of $f\left(x\right)=|x-1|+|x-2|+|x-3|$ is

• A. $1$
• B. $2$
• C. $6$
• D. $3$

Answer 1

The correct option is B $2$

$f\left(x\right)=|x-1|+|x-2|+|x-3|$

$f\left(x\right)=\{\begin{array}{cc}6-3x,& x<1\\ 4-x,& 1\le x<2\\ x,& 2\le x<3\\ 3x-6,& x\ge 3\end{array}$

For a modulus function, the minimum value exists at a critical point.
Critical points of $f$ are $1,2,3$
$f\left(1\right)=3,f\left(2\right)=2,f\left(3\right)=3$

$\therefore f_{min}=2$ at $x=2$