1. **Problem:** Determine the type of discontinuity for the piecewise function $$f(x) = \begin{cases} 2x^3 + x, & x < -2 \\ x^4 - x^2, & x \geq -2 \end{cases}$$ at $x = -2$. 2. **Step 1: Find $f(-2)$** Since $x = -2$ falls in the second piece ($x \geq -2$), $$f(-2) = (-2)^4 - (-2)^2 = 16 - 4 = 12$$ 3. **Step 2: Find the left-hand limit $\lim_{x \to -2^-} f(x)$** Use the first piece: $$\lim_{x \to -2^-} (2x^3 + x) = 2(-2)^3 + (-2) = 2(-8) - 2 = -16 - 2 = -18$$ 4. **Step 3: Find the right-hand limit $\lim_{x \to -2^+} f(x)$** Use the second piece: $$\lim_{x \to -2^+} (x^4 - x^2) = (-2)^4 - (-2)^2 = 16 - 4 = 12$$ 5. **Step 4: Find the overall limit $\lim_{x \to -2} f(x)$** Since left and right limits differ, $$\lim_{x \to -2} f(x) \text{ does not exist (DNE)}$$ 6. **Step 5: Compare $f(-2)$ and the limit** $f(-2) = 12$, but $\lim_{x \to -2} f(x)$ DNE. 7. **Conclusion:** Because the left and right limits are not equal, the function has a **jump discontinuity** at $x = -2$. --- **Summary:** - $f(-2) = 12$ - $\lim_{x \to -2^-} f(x) = -18$ - $\lim_{x \to -2^+} f(x) = 12$ - $\lim_{x \to -2} f(x)$ does not exist - Discontinuity type: **Jump Discontinuity**