1. **Problem Statement:** We are given a piecewise function: $$f(x)=\begin{cases} \frac{2}{x^2} & \text{for } x \leq -1 \\\ \frac{x+3}{\cos(x+1)} & \text{for } -1 < x < \frac{\pi - 2}{2} \end{cases}$$ We need to determine if $f$ is continuous at $x = -1$. 2. **Continuity Definition:** A function $f$ is continuous at $x = a$ if: $$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$$ 3. **Evaluate $f(-1)$:** Since $x = -1$ falls in the first case ($x \leq -1$), $$f(-1) = \frac{2}{(-1)^2} = \frac{2}{1} = 2$$ 4. **Left-hand limit as $x \to -1^-$:** Using the first case, $$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \frac{2}{x^2} = \frac{2}{(-1)^2} = 2$$ 5. **Right-hand limit as $x \to -1^+$:** Using the second case, $$\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \frac{x+3}{\cos(x+1)}$$ Evaluate numerator and denominator at $x = -1$: $$x+3 = -1 + 3 = 2$$ $$\cos(x+1) = \cos(0) = 1$$ So, $$\lim_{x \to -1^+} f(x) = \frac{2}{1} = 2$$ 6. **Conclusion:** Since $$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) = f(-1) = 2,$$ $f$ is continuous at $x = -1$. **Final answer:** A Yes