Show that the minimum value of $f(x)$ is $f(t_{50})$.

Question

Let $t_1<t_2<\cdots<t_{99}$ be real numbers, and consider the function $f: \mathbb R \to \mathbb R$ be given by $f(x)=|x-t_1|+|x-t_2|+\cdots+|x-t_{99}|$. Show that $\min\{f(x): x \in \mathbb R\} = f(t_{50})$.

Answer 1

Hint: if x is less than $t_1$ then f is the sum of 99 linear functions each of which has slope -1. So f has slope -99. As x increases and passes over $t_1$ one of these pieces changes from slope -1 to slope +1. So how does the slope of f change as x passes over $t_1$ ? As x continues to increase, how does the slope of f change as x passes over each $t_n$ ? At what point does the slope of f change from a negative value to a positive value ?