Question:

Step 1: Simplify the definition of $H(x)$. For $0 \leq x \leq \frac{\pi}{2}$, the function $f(t) = \cos t$ is a decreasing function. Therefore, the minimum value of $f(t)$ for $0 \leq t \leq x$ occurs at $t=x$. Thus, $\min(f(t) : 0 \leq t \leq x) = \cos x$. The function $H(x)$ can be expressed as: $$H(x) = \begin{cases} \cos x & , \text{for } 0 \leq x \leq \frac{\pi}{2} \\ -x & , \text{for } \frac{\pi}{2} < x \leq 3 \end{cases}$$ Step 2: Check continuity at $x = \frac{\pi}{2}$. For $H(x)$ to be continuous at $x = \frac{\pi}{2}$, the left-hand limit, the right-hand limit, and the function value at $x = \frac{\pi}{2}$ must all be equal. The function value at $x = \frac{\pi}{2}$ is determined by the first case: $$H\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0$$ The left-hand limit at $x = \frac{\pi}{2}$ is: $$\lim_{x \to \frac{\pi}{2}^-} H(x) = \lim_{x \to \frac{\pi}{2}^-} \cos x = \cos\left(\frac{\pi}{2}\right) = 0$$ The right-hand limit at $x = \frac{\pi}{2}$ is: $$\lim_{x \to \frac{\pi}{2}^+} H(x) = \lim_{x \to \frac{\pi}{2}^+} (-x) = -\frac{\pi}{2}$$ Comparing these values, we observe that: $$\lim_{x \to \frac{\pi}{2}^-} H(x) = 0$$ $$\lim_{x \to \frac{\pi}{2}^+} H(x) = -\frac{\pi}{2}$$ Since the left-hand limit is not equal to the right-hand limit ($0 \neq -\frac{\pi}{2}$), the function $H(x)$ is discontinuous at $x = \frac{\pi}{2}$.