1. Problem: Determine if the function $$f(x) = \frac{x^2 - x - 2}{x + 1}$$ is continuous at $$x=2$$ and $$x=-1$$. 2. Formula and rules: A function is continuous at a point if the limit as $$x$$ approaches that point equals the function's value there. 3. Simplify $$f(x)$$: $$x^2 - x - 2 = (x-2)(x+1)$$ So, $$f(x) = \frac{(x-2)(x+1)}{x+1}$$ Cancel common factor: $$f(x) = \frac{(x-2)\cancel{(x+1)}}{\cancel{x+1}} = x-2, \quad x \neq -1$$ 4. Check continuity at $$x=2$$: $$f(2) = 2 - 2 = 0$$ Since $$f(x) = x-2$$ for $$x \neq -1$$, limit as $$x \to 2$$ is: $$\lim_{x \to 2} f(x) = 2 - 2 = 0$$ Function is defined and limit equals value, so continuous at $$x=2$$. 5. Check continuity at $$x=-1$$: Function is undefined at $$x=-1$$ (denominator zero), so not continuous at $$x=-1$$. 6. Problem: Is $$g(x) = \begin{cases} x^2 + 3x, & x \leq 0 \\ x, & 0 < x \leq 2 \\ 3x^2, & x > 2 \end{cases}$$ continuous at $$x=0$$ and $$x=2$$? 7. Check continuity at $$x=0$$: Left limit: $$\lim_{x \to 0^-} g(x) = 0^2 + 3\cdot0 = 0$$ Right limit: $$\lim_{x \to 0^+} g(x) = 0$$ Value: $$g(0) = 0^2 + 3\cdot0 = 0$$ Continuous at $$x=0$$. 8. Check continuity at $$x=2$$: Left limit: $$\lim_{x \to 2^-} g(x) = 2$$ Right limit: $$\lim_{x \to 2^+} g(x) = 3 \cdot 2^2 = 12$$ Value: $$g(2) = 2$$ Limits not equal, so not continuous at $$x=2$$. 9. Problem: Find constants $$a$$ and $$b$$ so that $$h(x) = \begin{cases} ax + 5, & x \leq 1 \\ -2x^2 - 9, & -1 < x \leq 5 \\ 3x + b, & x > 5 \end{cases}$$ is continuous everywhere. 10. Continuity at $$x=1$$: From left: $$h(1) = a \cdot 1 + 5 = a + 5$$ From right: $$h(1) = -2(1)^2 - 9 = -2 - 9 = -11$$ Set equal: $$a + 5 = -11 \Rightarrow a = -16$$ 11. Continuity at $$x=5$$: From left: $$h(5) = -2(5)^2 - 9 = -2 \cdot 25 - 9 = -50 - 9 = -59$$ From right: $$h(5) = 3 \cdot 5 + b = 15 + b$$ Set equal: $$15 + b = -59 \Rightarrow b = -74$$ 12. Problem: Determine if $$f(x) = x^3 + x^2 - 2$$ is continuous at $$x=1$$. 13. Polynomial functions are continuous everywhere, so $$f(x)$$ is continuous at $$x=1$$. 14. Problem: Determine if $$f(x) = \frac{x^2 - x - 2}{x - 2}$$ is continuous at $$x=0$$. 15. Simplify numerator: $$x^2 - x - 2 = (x-2)(x+1)$$ So, $$f(x) = \frac{(x-2)(x+1)}{x-2}$$ Cancel common factor: $$f(x) = \cancel{\frac{(x-2)}{(x-2)}} (x+1) = x+1, \quad x \neq 2$$ 16. At $$x=0$$: $$f(0) = 0 + 1 = 1$$ Limit as $$x \to 0$$: $$\lim_{x \to 0} f(x) = 0 + 1 = 1$$ Function is continuous at $$x=0$$. 17. Problem: Find interval of length 1 containing a root of $$f(x) = x^3 - 3x + 5$$ using Intermediate Value Theorem (IVT). 18. Evaluate $$f(x)$$ at integer points: $$f(-3) = (-3)^3 - 3(-3) + 5 = -27 + 9 + 5 = -13$$ $$f(-2) = (-2)^3 - 3(-2) + 5 = -8 + 6 + 5 = 3$$ 19. Since $$f(-3) < 0$$ and $$f(-2) > 0$$, by IVT there is a root in $$(-3, -2)$$. Final answers: - $$f(x)$$ continuous at $$x=2$$ but not at $$x=-1$$. - $$g(x)$$ continuous at $$x=0$$ but not at $$x=2$$. - $$a = -16$$ and $$b = -74$$ for $$h(x)$$ continuous everywhere. - $$f(x) = x^3 + x^2 - 2$$ continuous at $$x=1$$. - $$f(x) = \frac{x^2 - x - 2}{x-2}$$ continuous at $$x=0$$. - Root of $$f(x) = x^3 - 3x + 5$$ lies in interval $$(-3, -2)$$.