According to Weierstrass theorem, a continuous function on a closed bounded interval is bounded.

Question

What if the function is discontinuous at just one point? Is it bounded anyway?

Why is my example different from the one you proposed?

Answer 1

If the function is discontinuous only at $c \in [a,b]$, and depending on the type of discontinuity, then you can say $f$ is bounded in $[a,c), (c, b]$ (but depending on the type of discontinuity!).

You're probably thinking on a removable discontinuity, but take $\tan x$ on $[0, \pi]$ (which is only discontinuous at $\frac{\pi}{2}$, but not bounded!)

Answer 2

It is not necessarily bounded anymore. For instance, take the function $$f:[-1,1]\to\mathbb R,~f(x)=\begin{cases}0&x\leq0\\\frac{1}{x}&x>0\end{cases}$$ It is defined on a closed, bounded interval (this is important: $\frac1x$ or $\tan$ are not counterexamples, since they are continuous, they are just not defined everywhere), has one discontinuity, and it is not bounded.