I am stuck in understanding the solution of this exercise. We have $f(x) = \begin{cases} 10 & x = 4 \\ x & 4 < x \leq 5 \end{cases}$ where also $f: [4, 5] \to \mathbb{R}$.The request asks to show the function is convex in $[4, 5]$ and in $(4, 5]$.
In the solution there is written that the use of derivative is useless, and then it says: since $f'(x) = 1$ in $(4, 5)$ this means the function is strictly increasing. Since we have no minimum, but the inf exists and it's $4$, and since $f(x) > 0$ for every $x$ in the domain, we can conclude that $f$ is convex.
- I am not sure I got this. It's ok the fact that $f$ is strictly increasing in that interval, but what about $f$ at $x = 4$? I am not sure the function is convex in both $[4, 5]$ and $(4, 5]$.Also I am not sure about how the inf helps us.
Can you please elaborate a little bit more to help me understanding the solution? Thank you!