1. Problem 4(a): Compare the natural domains of \(f(x) = \frac{x^2 + x}{x + 1}\) and \(g(x) = x\). - The domain of \(f\) excludes values that make the denominator zero. - Set denominator \(x + 1 = 0 \Rightarrow x = -1\). - So, domain of \(f\) is all real numbers except \(-1\). - The function \(g(x) = x\) is defined for all real numbers. - Therefore, domain of \(g\) is all real numbers. 2. Problem 4(b): Compare the natural domains of \(f(x) = \frac{x\sqrt{x} + \sqrt{x}}{x + 1}\) and \(g(x) = \sqrt{x}\). - For \(f\), \(\sqrt{x}\) requires \(x \geq 0\). - Denominator \(x + 1 \neq 0 \Rightarrow x \neq -1\), but since \(x \geq 0\), \(-1\) is excluded naturally. - So domain of \(f\) is \([0, \infty)\). - For \(g(x) = \sqrt{x}\), domain is also \([0, \infty)\). 3. Problem 7(a): Find \(f(0), f(2), f(-2), f(3), f(\sqrt{2}), f(3t)\) for \(f(x) = 3x^2 - 2\). - \(f(0) = 3(0)^2 - 2 = -2\) - \(f(2) = 3(2)^2 - 2 = 3(4) - 2 = 12 - 2 = 10\) - \(f(-2) = 3(-2)^2 - 2 = 3(4) - 2 = 10\) - \(f(3) = 3(3)^2 - 2 = 3(9) - 2 = 27 - 2 = 25\) - \(f(\sqrt{2}) = 3(\sqrt{2})^2 - 2 = 3(2) - 2 = 6 - 2 = 4\) - \(f(3t) = 3(3t)^2 - 2 = 3(9t^2) - 2 = 27t^2 - 2\) 4. Problem 7(b): Find \(f(0), f(2), f(-2), f(3), f(\sqrt{2}), f(3t)\) for piecewise function \(f(x) = \begin{cases} \frac{1}{x}, & x > 3 \\ 2x, & x \leq 3 \end{cases}\). - \(f(0)\): Since \(0 \leq 3\), use \(2x\), so \(f(0) = 2(0) = 0\). - \(f(2)\): \(2 \leq 3\), so \(f(2) = 2(2) = 4\). - \(f(-2)\): \(-2 \leq 3\), so \(f(-2) = 2(-2) = -4\). - \(f(3)\): \(3 \leq 3\), so \(f(3) = 2(3) = 6\). - \(f(\sqrt{2})\): \(\sqrt{2} \approx 1.414 \leq 3\), so \(f(\sqrt{2}) = 2\sqrt{2}\). - \(f(3t)\): Value depends on \(3t\). - If \(3t > 3 \Rightarrow t > 1\), then \(f(3t) = \frac{1}{3t}\). - If \(3t \leq 3 \Rightarrow t \leq 1\), then \(f(3t) = 2(3t) = 6t\). 5. Problem 8(a): Find \(g(3), g(-1), g(\pi), g(-1.1), g(t^2 - 1)\) for \(g(x) = \frac{x + 1}{x - 1}\). - \(g(3) = \frac{3 + 1}{3 - 1} = \frac{4}{2} = 2\) - \(g(-1) = \frac{-1 + 1}{-1 - 1} = \frac{0}{-2} = 0\) - \(g(\pi) = \frac{\pi + 1}{\pi - 1}\) (exact form) - \(g(-1.1) = \frac{-1.1 + 1}{-1.1 - 1} = \frac{-0.1}{-2.1} \approx 0.0476\) - \(g(t^2 - 1) = \frac{(t^2 - 1) + 1}{(t^2 - 1) - 1} = \frac{t^2}{t^2 - 2}\) 6. Problem 8(b): Find \(g(3), g(-1), g(\pi), g(-1.1), g(t^2 - 1)\) for piecewise function \(g(x) = \begin{cases} \sqrt{x + 1}, & x \geq 1 \\ 3, & x < 1 \end{cases}\). - \(g(3)\): \(3 \geq 1\), so \(g(3) = \sqrt{3 + 1} = \sqrt{4} = 2\) - \(g(-1)\): \(-1 < 1\), so \(g(-1) = 3\) - \(g(\pi)\): \(\pi \approx 3.1415 \geq 1\), so \(g(\pi) = \sqrt{\pi + 1}\) - \(g(-1.1)\): \(-1.1 < 1\), so \(g(-1.1) = 3\) - \(g(t^2 - 1)\): depends on \(t^2 - 1\) - If \(t^2 - 1 \geq 1 \Rightarrow t^2 \geq 2\), then \(g(t^2 - 1) = \sqrt{t^2 - 1 + 1} = \sqrt{t^2} = |t|\) - If \(t^2 - 1 < 1 \Rightarrow t^2 < 2\), then \(g(t^2 - 1) = 3\) Final answers are summarized above with detailed steps and domain considerations.