I have a question about the function $f(x)$ where $ f: [0,\infty)\to\mathbb R$ is given by: $$f(x)=\begin{cases} 5-x, & 0\leq x<4\\ 2+x, & x\geq4\end{cases}$$
I would have thought the minimum would be 1 as that is the limit from the left as x approaches 4, Can a limit not be a minimum?
Any clarification would be appreciated, Thanks
The function $f: [0, \infty) \to \mathbb{R}$ defined by $$f(x) = \begin{cases} 5 - x, & 0 \leq x < 4\\ 2 + x, & x \geq 4 \end{cases} $$ assumes each value in the interval $(1, 5]$ on the interval $[0, 4)$ and assumes each value in the interval $[6, \infty)$ on the interval $[4, \infty)$. Therefore, its range is $(1, 5] \cup [6, \infty)$. At no point in its domain is the value of the function equal to $1$. Therefore, $1$ is not a minimum. It is a lower bound, meaning that for each $x$ in the domain, $f(x) > 1$. In fact, it is the greatest lower bound (or infimum) since $1$ is the largest number that is a lower bound for the range.