I attempted to find the extrema of the following piecewise function $f$ on the closed interval [3,5]: $$f(x) =\left\{ \begin{array}{c}\frac{2}{x-5}, x \neq 5\\ 2, x=5 \end{array} \right.$$ I came out with the critical numbers $3$ and $5$, the endpoints, and they yielded a maximum of $(5,2)$ and a minimum of $(3,-1)$.
My teacher says that the correct solution is a maximum of $(5,2)$ but no minimum. Why? Shouldn't there be a minimum if there is a maximum? And why is $(5,2)$ a maximum? It doesn't appear as if there's either a maximum or minimum on the graph. This is not homework help, I'm just really confused.
Note that $$\lim_{x\to 5^{-}} \dfrac{2}{x-5} = -\infty$$ then $f$ hasn't a minimum in a neighborhood of $x=5$ (in $[3,5]$). Besides, we have $$f'(x)=-\dfrac{2}{(x-5)^2} < 0 \; \forall x\in \mathbb{R}$$ then $f$ is decreasing in $[3,5[$.
So, since $f(3) = -1$ and $f$ is decreasing, we have that $x=5$ is a maximum point of $f$ (because $f(5)=2$).