1. **Problem statement:** Find (i) the domain and range of functions $f(x) = \sqrt{\frac{1}{4} - x^2}$ and $h(x) = \tan(x - 3)$. 2. **Domain and range of $f$:** - The expression inside the square root must be non-negative: $$\frac{1}{4} - x^2 \geq 0$$ - Rearranged: $$x^2 \leq \frac{1}{4}$$ - So, $$-\frac{1}{2} \leq x \leq \frac{1}{2}$$ - Domain of $f$ is $$\left[-\frac{1}{2}, \frac{1}{2}\right]$$ - Range of $f$ is values of $f(x)$ for $x$ in domain: - Minimum inside root is 0, maximum is $\frac{1}{4}$, so $f(x)$ ranges from 0 to $\sqrt{\frac{1}{4}} = \frac{1}{2}$ - Range of $f$ is $$[0, \frac{1}{2}]$$ 3. **Domain and range of $h$:** - $h(x) = \tan(x - 3)$ is defined for all real $x$ except where $\cos(x - 3) = 0$ - $\cos(x - 3) = 0$ at $$x - 3 = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$$ - Domain of $h$ is $$\mathbb{R} \setminus \left\{3 + \frac{\pi}{2} + k\pi : k \in \mathbb{Z}\right\}$$ - Range of $h$ is all real numbers $$\mathbb{R}$$ because tangent takes all real values 4. **Composite functions:** - $(f \circ g)(x) = f(g(x)) = \sqrt{\frac{1}{4} - \left(\frac{5}{9}x + 3\right)^2}$ - $(f \circ h)(x) = f(h(x)) = \sqrt{\frac{1}{4} - \tan^2(x - 3)}$ - $(g \circ f)(x) = g(f(x)) = \frac{5}{9} \sqrt{\frac{1}{4} - x^2} + 3$ 5. **Inverse of $g$:** - Given $y = \frac{5}{9}x + 3$ - Solve for $x$: $$y - 3 = \frac{5}{9}x \implies x = \frac{9}{5}(y - 3)$$ - So, $$g^{-1}(x) = \frac{9}{5}(x - 3)$$ 6. **Limit verification using $\epsilon$-$\delta$ definition:** - Show that $$\lim_{x \to 5} (5x + 7) = 32$$ - For any $\epsilon > 0$, find $\delta > 0$ such that if $$|x - 5| < \delta$$ then $$|5x + 7 - 32| < \epsilon$$ - Simplify: $$|5x + 7 - 32| = |5x - 25| = 5|x - 5|$$ - To have $$5|x - 5| < \epsilon$$, choose $$\delta = \frac{\epsilon}{5}$$ - Then if $$|x - 5| < \delta$$, $$|5x + 7 - 32| < \epsilon$$ holds, verifying the limit. **Final answers:** - Domain of $f$: $$\left[-\frac{1}{2}, \frac{1}{2}\right]$$ - Range of $f$: $$[0, \frac{1}{2}]$$ - Domain of $h$: $$\mathbb{R} \setminus \left\{3 + \frac{\pi}{2} + k\pi : k \in \mathbb{Z}\right\}$$ - Range of $h$: $$\mathbb{R}$$ - $(f \circ g)(x) = \sqrt{\frac{1}{4} - \left(\frac{5}{9}x + 3\right)^2}$ - $(f \circ h)(x) = \sqrt{\frac{1}{4} - \tan^2(x - 3)}$ - $(g \circ f)(x) = \frac{5}{9} \sqrt{\frac{1}{4} - x^2} + 3$ - $g^{-1}(x) = \frac{9}{5}(x - 3)$ - Verified limit: $$\lim_{x \to 5} (5x + 7) = 32$$ using $\delta = \frac{\epsilon}{5}$.