An image that is bounded - attains maximum but does not attain a minimum

Question

I'm trying to think of a function that is bounded above and attains a maximum and is bounded below but does not attain a minimum. I have an exam coming up and I'm trying to think of possible scenarios for true or false type questions. I thought this may be a good one but I can't think of an example. The function will be discontinuous but that's as much as I can get. I've been pondering over this for quite a while. Any ideas?

Edit: $\ f:(a,b)\longrightarrow \mathbb{R}$.

Answer 1

$$f(x) = e^{-\left(x-\tfrac{a+b}{2}\right)^2}$$

has a maximum of $1$ at $x=\frac{a+b}{2} \in (a, b)$, but no minimum is attained, though it is bounded below by $0$.

More importantly, this function is not discontinuous. It is in fact very not discontinuous (it is infinitely differentiable).

Answer 2

$$ x\mapsto \frac 1 {1+x^2} $$ This has a maximum value at $x=0.$ Its greatest lower bound is $0,$ but it is everywhere positive.

Answer 3

$f(x) = \frac{x(1-x)}{1+\frac{1}{x}}$ on $(0,1)$