1. **Problem Statement:** We need to explain why each function is continuous at every number in its domain using Theorems 4, 5, 7, and 9, and state the domain for each function. 2. **Recall Theorems:** - Theorem 4: Polynomials are continuous everywhere. - Theorem 5: The sum, difference, product, and quotient (where denominator is not zero) of continuous functions are continuous. - Theorem 7: The composition of continuous functions is continuous. - Theorem 9: Root functions are continuous on their domain. --- ### 27. \(f(x) = \frac{x^2}{\sqrt{x^4 + 2}}\) 3. **Domain:** The denominator \(\sqrt{x^4 + 2}\) is defined for all real \(x\) because \(x^4 + 2 > 0\) for all real \(x\). So domain is \(\mathbb{R}\). 4. **Continuity:** - \(x^2\) is a polynomial, continuous everywhere (Theorem 4). - \(x^4 + 2\) is polynomial, continuous everywhere (Theorem 4). - \(\sqrt{\cdot}\) is continuous on its domain (Theorem 9). - Composition \(\sqrt{x^4 + 2}\) is continuous (Theorem 7). - Denominator never zero, so quotient is continuous (Theorem 5). --- ### 28. \(g(v) = \frac{3v - 1}{v^2 + 2v - 15}\) 5. **Domain:** Denominator \(v^2 + 2v - 15 = (v+5)(v-3)\) is zero at \(v = -5, 3\), so domain is \(\mathbb{R} \setminus \{-5, 3\}\). 6. **Continuity:** - Numerator \(3v - 1\) is polynomial, continuous everywhere (Theorem 4). - Denominator is polynomial, continuous everywhere (Theorem 4). - Quotient is continuous where denominator \(\neq 0\) (Theorem 5). --- ### 29. \(h(t) = \frac{\cos(t^2)}{1 - e^t}\) 7. **Domain:** Denominator \(1 - e^t = 0\) when \(e^t = 1\) \(\Rightarrow t=0\). So domain is \(\mathbb{R} \setminus \{0\}\). 8. **Continuity:** - \(t^2\) is polynomial, continuous everywhere (Theorem 4). - \(\cos(x)\) is continuous everywhere. - Composition \(\cos(t^2)\) is continuous (Theorem 7). - \(e^t\) is continuous everywhere. - Denominator \(1 - e^t\) continuous everywhere. - Quotient continuous where denominator \(\neq 0\) (Theorem 5). --- **Final answers:** - \(f(x)\) continuous on \(\mathbb{R}\). - \(g(v)\) continuous on \(\mathbb{R} \setminus \{-5,3\}\). - \(h(t)\) continuous on \(\mathbb{R} \setminus \{0\}\).