1. Evaluate $\log_{\frac{1}{9}} \left( (\sqrt[3]{81})^2 \right)$. - First, simplify the inside: $\sqrt[3]{81} = 81^{\frac{1}{3}} = (3^4)^{\frac{1}{3}} = 3^{\frac{4}{3}}$. - Then, $(\sqrt[3]{81})^2 = \left(3^{\frac{4}{3}}\right)^2 = 3^{\frac{8}{3}}$. - The base of the log is $\frac{1}{9} = 9^{-1} = (3^2)^{-1} = 3^{-2}$. Use the change of base formula: $$\log_{3^{-2}} 3^{\frac{8}{3}} = \frac{\log_3 3^{\frac{8}{3}}}{\log_3 3^{-2}} = \frac{\frac{8}{3}}{-2} = -\frac{8}{6} = -\frac{4}{3}.$$ Answer: d. $-\frac{4}{3}$. 2. Find the x-intercepts of $y = g(x) - f(x)$ where $f(x) = x^2$ and $g(x) = x + 56$. - Set $g(x) - f(x) = 0$: $$x + 56 - x^2 = 0 \implies -x^2 + x + 56 = 0.$$ - Multiply both sides by $-1$: $$\cancel{-}x^2 + \cancel{x} + \cancel{56} = \cancel{0} \implies x^2 - x - 56 = 0.$$ - Factor: $$(x - 8)(x + 7) = 0$$ - Solutions: $$x = 8 \text{ or } x = -7.$$ Answer: c. $-7$ and $8$. 3. For linear $f(x) = ax + b$, $a \neq 0$, when does $f(f^{-1}(x)) = x$? - By definition, $f^{-1}$ is the inverse function of $f$. - Composing a function with its inverse always returns the input: $$f(f^{-1}(x)) = x$$ for all $x$ in the domain. Answer: a. always. 4. Find $g(f(-1))$ where $f(x) = \frac{1}{x}$ and $g(x) = \sqrt{x + 10}$. - Compute $f(-1) = \frac{1}{-1} = -1$. - Then $g(f(-1)) = g(-1) = \sqrt{-1 + 10} = \sqrt{9} = \pm 3$. Answer: c. $\pm 3$. 5. Find $f(g(4))$ where $f(x) = \log 4x$ and $g(x) = x^2 - 2$. - Compute $g(4) = 4^2 - 2 = 16 - 2 = 14$. - Then $f(g(4)) = \log 4 \times 14 = \log (56)$ but the options suggest $\log 4x$ means $\log(4x)$. - So $f(g(4)) = \log(4 \times 14) = \log 56$. - None of the options directly say $\log 56$, but using log properties: $$\log 56 = \log (4 \times 14) = \log 4 + \log 14.$$ Answer: d. $\log 4 + \log 14$. 6. Find the domain of $y = f(x) g(x)$ where $f(x) = \frac{1}{\log x}$ and $g(x) = \frac{1}{8 + x}$. - Domain restrictions: - $\log x$ is defined for $x > 0$. - $\log x \neq 0$ to avoid division by zero, so $x \neq 1$ because $\log 1 = 0$. - $8 + x \neq 0 \implies x \neq -8$. - Combine: $$\{x \in \mathbb{R} | x > 0, x \neq 1\}$$ Answer: c. $\{x \in \mathbb{R} | x > 0 \text{ and } x \neq 1\}$. Final answers: 15: d 16: c 17: a 18: c 19: d 20: c