1. **Problem statement:** We have a piecewise function defined as: $$f(x) = \begin{cases} \frac{1}{x} + 4 & \text{if } x < 2 \\ 16 & \text{if } x = 2 \\ -2x^2 + 2x - 8 & \text{if } x > 2 \end{cases}$$ We need to determine if $f$ is differentiable at $x=2$ and find the derivative at that point if it exists. 2. **Check continuity at $x=2$:** For differentiability, $f$ must be continuous at $x=2$. - Left limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \left( \frac{1}{x} + 4 \right) = \frac{1}{2} + 4 = 4.5$ - Right limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (-2x^2 + 2x - 8) = -2(2)^2 + 2(2) - 8 = -8 + 4 - 8 = -12$ - Value at $x=2$: $f(2) = 16$ Since left limit $4.5 \neq 16$ and right limit $-12 \neq 16$, $f$ is not continuous at $x=2$. 3. **Conclusion:** Since $f$ is not continuous at $x=2$, it cannot be differentiable there. **Final answer:** The function $f$ is not differentiable at point $x=2$.