To determine if a function $F(x)$ is continuous at a point $x=c$, three conditions must be met: 1. $F(c)$ must be defined. 2. $\lim_{x \to c} F(x)$ must exist. 3. $\lim_{x \to c} F(x) = F(c)$. 1. Determine whether the function $F(x) = 3x^2 + 4x - 10$ is continuous at $x = 3$ or not. Step 1: Evaluate $F(3)$. $$F(3) = 3(3)^2 + 4(3) - 10$$ $$F(3) = 3(9) + 12 - 10$$ $$F(3) = 27 + 12 - 10$$ $$F(3) = 39 - 10$$ $$F(3) = 29$$ $F(3)$ is defined. Step 2: Evaluate $\lim_{x \to 3} F(x)$. Since $F(x)$ is a polynomial function, the limit as $x$ approaches $3$ can be found by direct substitution. $$\lim_{x \to 3} (3x^2 + 4x - 10) = 3(3)^2 + 4(3) - 10$$ $$\lim_{x \to 3} (3x^2 + 4x - 10) = 27 + 12 - 10$$ $$\lim_{x \to 3} (3x^2 + 4x - 10) = 29$$ The limit exists. Step 3: Compare $F(3)$ and $\lim_{x \to 3} F(x)$. Since $F(3) = 29$ and $\lim_{x \to 3} F(x) = 29$, we have $\lim_{x \to 3} F(x) = F(3)$. Therefore, the function $F(x) = 3x^2 + 4x - 10$ is continuous at $x=3$. $$\boxed{\text{Continuous}}$$ 2. Is the function $F(x) = \frac{2x^2 + 3}{2x + 1}$ continuous at (i) $x = -1$ and (ii) $x = -\frac{1}{2}$? A rational function is continuous everywhere except where its denominator is zero. First, find the values of $x$ for which the denominator is zero: $$2x + 1 = 0$$ $$2x = -1$$ $$x = -\frac{1}{2}$$ The function is undefined at $x = -\frac{1}{2}$. a) Continuity at $x = -1$ Step 1: Evaluate $F(-1)$. $$F(-1) = \frac{2(-1)^2 + 3}{2(-1) + 1}$$ $$F(-1) = \frac{2(1) + 3}{-2 + 1}$$ $$F(-1) = \frac{2 + 3}{-1}$$ $$F(-1) = \frac{5}{-1}$$ $$F(-1) = -5$$ $F(-1)$ is defined. Step 2: Evaluate $\lim_{x \to -1} F(x)$. Since $x=-1$ does not make the denominator zero, the limit can be found by direct substitution. $$\lim_{x \to -1} \frac{2x^2 + 3}{2x + 1} = \frac{2(-1)^2 + 3}{2(-1) + 1}$$ $$\lim_{x \to -1} \frac{2x^2 + 3}{2x + 1} = \frac{5}{-1}$$ $$\lim_{x \to -1} \frac{2x^2 + 3}{2x + 1} = -5$$ The limit exists. Step 3: Compare $F(-1)$ and $\lim_{x \to -1} F(x)$. Since $F(-1) = -5$ and $\lim_{x \to -1} F(x) = -5$, we have $\lim_{x \to -1} F(x) = F(-1)$. Therefore, the function $F(x) = \frac{2x^2 + 3}{2x + 1}$ is continuous at $x=-1$. $$\boxed{\text{Continuous at } x = -1}$$ b) Continuity at $x = -\frac{1}{2}$ Step 1: Evaluate $F(-\frac{1}{2})$. $$F\left(-\frac{1}{2}\right) = \frac{2\left(-\frac{1}{2}\right)^2 + 3}{2\left(-\frac{1}{2}\right) + 1}$$ $$F\left(-\frac{1}{2}\right) = \frac{2\left(\frac{1}{4}\right) + 3}{-1 + 1}$$ $$F\left(-\frac{1}{2}\right) = \frac{\frac{1}{2} + 3}{0}$$ $$F\left(-\frac{1}{2}\right) = \frac{\frac{7}{2}}{0}$$ Since the denominator is zero, $F\left(-\frac{1}{2}\right)$ is undefined. Because $F\left(-\frac{1}{2}\right)$ is undefined, the first condition for continuity is not met. Therefore, the function $F(x) = \frac{2x^2 + 3}{2x + 1}$ is discontinuous at $x=-\frac{1}{2}$. $$\boxed{\text{Discontinuous at } x = -\frac{1}{2}}$$