1. **State the problem:** Determine if the piecewise function $$g(x) = \begin{cases} \frac{1}{2}x + 2, & x \leq 2 \\ 4x - 5, & x > 2 \end{cases}$$ is continuous at $x=2$. 2. **Recall the continuity condition:** A function $g(x)$ is continuous at $x=a$ if $$\lim_{x \to a^-} g(x) = g(a) = \lim_{x \to a^+} g(x).$$ 3. **Calculate $g(2)$:** Since $2 \leq 2$, use the first piece: $$g(2) = \frac{1}{2} \times 2 + 2 = 1 + 2 = 3.$$ 4. **Calculate the left-hand limit as $x \to 2^-$:** Use the first piece: $$\lim_{x \to 2^-} g(x) = \lim_{x \to 2^-} \left( \frac{1}{2}x + 2 \right) = \frac{1}{2} \times 2 + 2 = 3.$$ 5. **Calculate the right-hand limit as $x \to 2^+$:** Use the second piece: $$\lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} (4x - 5) = 4 \times 2 - 5 = 8 - 5 = 3.$$ 6. **Compare values:** $$\lim_{x \to 2^-} g(x) = 3, \quad g(2) = 3, \quad \lim_{x \to 2^+} g(x) = 3.$$ All three are equal. 7. **Conclusion:** Since the left limit, right limit, and function value at $x=2$ are equal, the function $g(x)$ is continuous at $x=2$. **Final answer:** Yes, the function $g(x)$ is continuous at $x=2$.