Given a function $f(x)=\left\{\begin{matrix} \frac{1}{2\sqrt{x}} & 0 \leq x \leq 4\\ \\ 0 & \text{otherwise} \end{matrix}\right.$ is it a bounded function?
I first checked the continuity of $f(x)$ at points $0$,$4$. It was discontinuous. However this function can be bound by another function $\frac{1}{2x}$ when $x>1$ and $0$ otherwise. But am confused whether this argument be used to prove that the function is bounded.
Can you please let me know if the approach to the above formulation is correct?