Introduction to Analysis: Bounded and Defined This question has been rattling my brain for a while. If $f(x)$ is bounded and defined on the interval $[a,b]$ does this imply $(f(x))^2$ is bounded and defined on $[a,b]$? I would say so. Just not sure if it is a given fact or not.
Yes, $f(x)^2$ is defined and is also bounded.

To see that it is defined, note that $f(x)^2 = f(x) \cdot f(x)$. Since $f(x)$ is defined for all $x \in [a,b]$ and since multiplication of two real numbers is defined, we have $f(x)^2$ is defined.

To see that it is bounded, since $\vert f(x) \vert \leq M$ for some $M \in \mathbb{R}^+$, we have $f(x)^2 = \vert f(x) \vert^2 \leq M^2$.
Remember that the product of two functions is continuous. Now use the extreme value theorem. This is another way to think about it.