1. We are asked to find \(\lim_{x \to -1} f(x)\). Looking at the graph, the function on the interval including \(x=-1\) is a straight line segment from (-3, -1) to (-1, 3). 2. To find the limit as \(x\) approaches -1, we evaluate the behavior of the straight line segment near \(x=-1\). Since the segment is continuous and defined at those points, the limit equals the y-value at \(x=-1\). 3. The point on the line segment at \(x=-1\) is clearly \(y=3\). Therefore, $$\lim_{x \to -1} f(x) = 3$$ 4. Next, determine continuity at \(x=1\) by finding the left-hand and right-hand limits. 5. The function is defined as the parabola-like curve near \(x=1\) on \((0, 3]\) and a solid point at \((1, 3)\). 6. The left-hand limit \(\lim_{x \to 1^-} f(x)\) is found by analyzing the parabola-like curve approaching \(1\) from values less than 1. From the graph, the curve peaks near \(y=2\) at \(x=1\). 7. So, $$\lim_{x \to 1^-} f(x) = 2$$ 8. The right-hand limit \(\lim_{x \to 1^+} f(x)\) is from values greater than 1 on the parabola-like curve going down. The curve passes through approximately \( (1,2) \) and decreases. 9. Therefore, $$\lim_{x \to 1^+} f(x) = 2$$ 10. The two one-sided limits at \(x=1\) are equal: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = 2$$ 11. However, the function value at \(x=1\) is given by the solid point at \(y=3\), which is not equal to the limit 2. 12. Since the limit exists but is not equal to the function value at \(x=1\), the function is not continuous at \(x=1\). **Final answers:** - \(\lim_{x \to -1} f(x) = 3\) - The function is **not continuous** at \(x=1\) because \(\lim_{x \to 1} f(x) = 2 \neq f(1) = 3\).