Key Concepts and Formulas

• Continuity at a point $x=c$: A function $g(x)$ is continuous at $x=c$ if $\lim_{x \to c^-} g(x) = \lim_{x \to c^+} g(x) = g(c)$.
• Differentiability at a point $x=c$: A function $g(x)$ is differentiable at $x=c$ if the left-hand derivative and the right-hand derivative at $x=c$ are equal. That is, $\lim_{h \to 0^-} \frac{g(c+h)-g(c)}{h} = \lim_{h \to 0^+} \frac{g(c+h)-g(c)}{h}$.
• Monotonicity of a function: A function $f(x)$ is decreasing if $f'(x) < 0$ in the given interval.
• Minimum of a function: For a decreasing function $f(t)$ on an interval $(0, x]$, the minimum value is $f(x)$.

Step-by-Step Solution

Step 1: Analyze the function $f(x)$ and its derivative. The function $f(x)$ is given by $f(x) = \frac{x}{2} + \frac{2}{x}$ for $x \in (0, 2)$. We find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}\left(\frac{x}{2} + \frac{2}{x}\right) = \frac{1}{2} - \frac{2}{x^2}$. To determine the monotonicity of $f(x)$, we examine the sign of $f'(x)$ in the interval $(0, 2)$. $f'(x) < 0 \implies \frac{1}{2} - \frac{2}{x^2} < 0 \implies \frac{1}{2} < \frac{2}{x^2} \implies x^2 < 4$. Since $x \in (0, 2)$, we have $0 < x < 2$, which implies $0 < x^2 < 4$. Thus, $f'(x) < 0$ for all $x \in (0, 2)$. This means $f(x)$ is strictly decreasing on the interval $(0, 2)$.

Step 2: Determine the expression for $g(x)$ for $0 < x \leq 1$. The function $g(x)$ is defined as $g(x) = \min \{f(t) \mid 0 < t \leq x\}$ for $0 < x \leq 1$. Since $f(t)$ is a decreasing function on $(0, 2)$, for any $x$ in $(0, 1]$, the minimum value of $f(t)$ for $t \in (0, x]$ occurs at $t=x$. Therefore, for $0 < x \leq 1$, $g(x) = f(x) = \frac{x}{2} + \frac{2}{x}$.

Step 3: Define the function $g(x)$ piecewise. Combining the definition from the problem statement and our finding in Step 2, we have: $g(x) = \begin{cases} \frac{x}{2} + \frac{2}{x}, & 0 < x \leq 1 \\ \frac{3}{2} + x, & 1 < x < 2 \end{cases}$

Step 4: Check for continuity of $g(x)$ at $x=1$. To check for continuity at $x=1$, we need to evaluate the left-hand limit, the right-hand limit, and the function value at $x=1$. Left-hand limit: $\lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} \left(\frac{x}{2} + \frac{2}{x}\right) = \frac{1}{2} + \frac{2}{1} = \frac{1}{2} + 2 = \frac{5}{2}$. Right-hand limit: $\lim_{x \to 1^+} g(x) = \lim_{x \to 1^+} \left(\frac{3}{2} + x\right) = \frac{3}{2} + 1 = \frac{5}{2}$. Function value at $x=1$: $g(1) = \frac{1}{2} + \frac{2}{1} = \frac{5}{2}$. Since $\lim_{x \to 1^-} g(x) = \lim_{x \to 1^+} g(x) = g(1) = \frac{5}{2}$, the function $g(x)$ is continuous at $x=1$.

Step 5: Check for differentiability of $g(x)$ at $x=1$. To check for differentiability at $x=1$, we need to find the left-hand derivative and the right-hand derivative at $x=1$. Left-hand derivative: The derivative of the first piece is $\frac{d}{dx}\left(\frac{x}{2} + \frac{2}{x}\right) = \frac{1}{2} - \frac{2}{x^2}$. So, the left-hand derivative at $x=1$ is: $g'_{-}(1) = \lim_{x \to 1^-} \left(\frac{1}{2} - \frac{2}{x^2}\right) = \frac{1}{2} - \frac{2}{1^2} = \frac{1}{2} - 2 = -\frac{3}{2}$. Right-hand derivative: The derivative of the second piece is $\frac{d}{dx}\left(\frac{3}{2} + x\right) = 1$. So, the right-hand derivative at $x=1$ is: $g'_{+}(1) = \lim_{x \to 1^+} 1 = 1$. Since the left-hand derivative ($-\frac{3}{2}$) is not equal to the right-hand derivative ($1$), the function $g(x)$ is not differentiable at $x=1$.

Step 6: Analyze the continuity and differentiability for $x \in (0, 1)$ and $x \in (1, 2)$. For $0 < x < 1$, $g(x) = \frac{x}{2} + \frac{2}{x}$. This is a rational function, and its denominator is non-zero in this interval. Thus, $g(x)$ is continuous and differentiable for all $x \in (0, 1)$. For $1 < x < 2$, $g(x) = \frac{3}{2} + x$. This is a linear function, which is continuous and differentiable for all $x \in (1, 2)$.

Step 7: Conclude about the properties of $g(x)$ at $x=1$. From Step 4, $g(x)$ is continuous at $x=1$. From Step 5, $g(x)$ is not differentiable at $x=1$.

Therefore, $g$ is continuous but not differentiable at $x=1$.

Common Mistakes & Tips

• Incorrectly evaluating the minimum: Remember that for a decreasing function, the minimum on an interval $(0, x]$ occurs at the right endpoint $x$.
• Forgetting to check both left and right derivatives: Differentiability at a point requires the equality of both the left-hand and right-hand derivatives.
• Assuming differentiability implies continuity: While differentiability implies continuity, the converse is not always true. A function can be continuous at a point but not differentiable there (e.g., sharp corners or cusps).

Summary

We first analyzed the function $f(x)$ and found it to be strictly decreasing on $(0, 2)$. This allowed us to determine that for $0 < x \leq 1$, $g(x) = f(x) = \frac{x}{2} + \frac{2}{x}$. The function $g(x)$ was then defined piecewise. We checked the continuity of $g(x)$ at $x=1$ by comparing the left-hand limit, right-hand limit, and the function value, all of which were equal to $\frac{5}{2}$. Next, we checked the differentiability at $x=1$ by calculating the left-hand derivative ($-\frac{3}{2}$) and the right-hand derivative ($1$). Since these were not equal, $g(x)$ is not differentiable at $x=1$. Thus, $g$ is continuous but not differentiable at $x=1$.

The final answer is \boxed{A}.