Let f(x) = |x – 2| + |x – 3| + |x – 4|, then the minimum value of f(x + 1) occur at x equal to

2

$f(x + 1) = |x – 1| + |x – 2| + |x – 3|$

$f(x + 1) =\left\{\begin{matrix}6-3x&x<1\\4-x&1≤x<2\\x&2≤x<3\\3x-6&x≥3\end{matrix}\right.$

$f'(x) =\left\{\begin{matrix}-3&x<1\\-1&1≤x<2\\1&2≤x<3\\3&x≥3\end{matrix}\right.$

$f'(x) <0$ for $x<2$

$f'(x) >0$ for $x≥2$

⇒ 2 → point of minima

$f(2+1)=2$ → min. value