1. **Problem:** Determine if $f(x) = x^3 - 5x^2 + 5x + 1$ is continuous at $x = \sqrt{2}$. 2. **Formula and rules:** Polynomial functions are continuous everywhere, so $f(x)$ is continuous at all real numbers including $x = \sqrt{2}$. 3. **Intermediate work:** Since $f$ is a polynomial, no discontinuities exist. 4. **Conclusion:** $f(x)$ is continuous at $x = \sqrt{2}$. 1. **Problem:** Determine if $f(x) = \frac{x^3 + 3x^2 - x - 3}{x+1}$ is continuous at $x = -1$. 2. **Formula and rules:** A rational function is continuous where the denominator is not zero. Check denominator at $x = -1$: $$x+1 = -1 + 1 = 0$$ So the function is undefined at $x = -1$. 3. **Intermediate work:** Factor numerator: $$x^3 + 3x^2 - x - 3 = (x+1)(x^2 + 2x - 3)$$ Factor quadratic: $$x^2 + 2x - 3 = (x+3)(x-1)$$ So, $$f(x) = \frac{(x+1)(x+3)(x-1)}{x+1}$$ Cancel common factor: $$f(x) = \frac{\cancel{(x+1)}(x+3)(x-1)}{\cancel{(x+1)}} = (x+3)(x-1), \quad x \neq -1$$ 4. **Check limit at $x = -1$:** $$\lim_{x \to -1} f(x) = (-1 + 3)(-1 - 1) = 2 \times (-2) = -4$$ 5. **Check function value at $x = -1$:** Undefined. 6. **Conclusion:** Since $f(-1)$ is undefined but the limit exists, $f$ is discontinuous at $x = -1$. 1. **Problem:** Determine if the piecewise function $$f(x) = \begin{cases} 2x + 3, & x \leq 3 \\ x - 5, & x > 3 \end{cases}$$ is continuous at $x = 3$. 2. **Formula and rules:** For continuity at $x = 3$, left-hand limit, right-hand limit, and function value must be equal. 3. **Calculate left-hand limit:** $$\lim_{x \to 3^-} f(x) = 2(3) + 3 = 6 + 3 = 9$$ 4. **Calculate right-hand limit:** $$\lim_{x \to 3^+} f(x) = 3 - 5 = -2$$ 5. **Function value at $x=3$:** $$f(3) = 2(3) + 3 = 9$$ 6. **Conclusion:** Left limit and function value are 9, right limit is -2, so limits are not equal. Therefore, $f$ is discontinuous at $x = 3$. **Final answers:** 1. Continuous at $x = \sqrt{2}$. 2. Discontinuous at $x = -1$. 3. Discontinuous at $x = 3$.