1. Problem 10: Given $f(x)= \frac{x}{3x+2}$ and $g(x) = -\frac{4}{x}$, find $g \circ f$ and its domain. 2. To find $g(f(x))$, substitute $f(x)$ into $g$: $$g(f(x)) = g\left(\frac{x}{3x+2}\right) = -\frac{4}{\frac{x}{3x+2}} = -4 \cdot \frac{3x+2}{x} = -\frac{4(3x+2)}{x} = -\frac{12x+8}{x}$$ 3. Domain of $g \circ f$ is the set of all $x$ where $f(x)$ is in the domain of $g$ and $f(x)$ is defined. - $f(x)$ is defined when denominator $3x+2 \neq 0 \Rightarrow x \neq -\frac{2}{3}$. - $g(x)$ is defined when denominator $x \neq 0$, so $f(x) \neq 0$. 4. Find where $f(x) = 0$: $$\frac{x}{3x+2} = 0 \Rightarrow x=0$$ 5. So $x=0$ is excluded from the domain. 6. Therefore, domain of $g \circ f$ is all real $x$ except $x \neq -\frac{2}{3}$ and $x \neq 0$. --- 7. Problem 11: Given graph of $f(x)$, sketch the transformed graph $-f(x)-1$. 8. The transformation $-f(x)$ reflects the graph of $f(x)$ about the x-axis. 9. Then subtracting 1 shifts the graph down by 1 unit. 10. So to sketch $-f(x)-1$, reflect $f(x)$ over x-axis and move it down 1 unit. --- 11. Problem 12: Complete the statements: a) Domain of $\tan \theta$ is $\{\theta \mid \theta \neq \frac{\pi}{2} + k\pi, k \in \mathbb{Z}\}$, its Range is $(-\infty, \infty)$, and its Period is $\pi$. b) Domain of $\sin \theta$ is $(-\infty, \infty)$, its Range is $[-1,1]$, and its Period is $2\pi$. --- 12. Problem 13: To obtain the graph of $y = f(x + 19)$ from $y = f(x)$, shift the graph of $f(x)$ horizontally to the left by 19 units. --- Final answers: - $g \circ f(x) = -\frac{12x+8}{x}$ with domain $x \in \mathbb{R} \setminus \left\{0, -\frac{2}{3}\right\}$. - $-f(x)-1$ is the reflection of $f(x)$ about the x-axis shifted down by 1. - Domain and range of $\tan \theta$ and $\sin \theta$ as above. - $y = f(x+19)$ is $f(x)$ shifted left 19 units.