1. **State the problem:** Determine if the piecewise function $$f(x)=\begin{cases}4x-1 & \text{if } x<2 \\ 4 & \text{if } x=2 \\ 2x & \text{if } x>2\end{cases}$$ is continuous at $$x=2$$. 2. **Recall continuity condition:** A function is continuous at $$x=2$$ if $$\lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x)$$. 3. **Evaluate left-hand limit:** For $$x<2$$, $$f(x) = 4x-1$$. Calculate $$\lim_{x \to 2^-} f(x) = 4(2) - 1 = 8 - 1 = 7$$. 4. **Evaluate right-hand limit:** For $$x>2$$, $$f(x) = 2x$$. Calculate $$\lim_{x \to 2^+} f(x) = 2(2) = 4$$. 5. **Evaluate function value:** $$f(2) = 4$$. 6. **Compare limits and function value:** - Left limit = 7 - Right limit = 4 - $$f(2) = 4$$ Since $$\lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x)$$, the limit $$\lim_{x \to 2} f(x)$$ does not exist. 7. **Conclusion:** The function is **not continuous** at $$x=2$$ because the left-hand and right-hand limits are different, so the limit as $$x \to 2$$ does not equal $$f(2)$$. **Final answer:** $$f(x)$$ is discontinuous at $$x=2$$.