1. **Problem statement:** (a)(i) Find the domain and range of functions $f(x) = \sqrt{\frac{1}{4} - x^2}$ and $h(x) = \tan(x - 3)$. (a)(ii) Find the composite functions $(f \circ g)(x)$, $(f \circ h)(x)$, and $(g \circ f)(x)$ where $g(x) = \frac{5}{9}x + 3$. (a)(iii) Find the inverse function $g^{-1}(x)$. (b) Use the formal $(\epsilon - \delta)$ definition of limit to verify $\lim_{x \to 5} (5x + 7) = 32$. 2. **Domain and range of $f$:** - The function $f(x) = \sqrt{\frac{1}{4} - x^2}$ requires the radicand to be non-negative: $$\frac{1}{4} - x^2 \geq 0 \implies x^2 \leq \frac{1}{4} \implies -\frac{1}{2} \leq x \leq \frac{1}{2}$$ - So, domain of $f$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$. - Range of $f$ is values of $f(x)$ for $x$ in domain. Since $f(x)$ is a square root, minimum is 0 (at $x=\pm \frac{1}{2}$), maximum is $\sqrt{\frac{1}{4} - 0} = \frac{1}{2}$ at $x=0$. - Therefore, 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 $x - 3 = \frac{\pi}{2} + k\pi$, $k \in \mathbb{Z}$. - So 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. **Formal $(\epsilon - \delta)$ proof for limit:** - Claim: $\lim_{x \to 5} (5x + 7) = 32$. - For any $\epsilon > 0$, we want $|5x + 7 - 32| < \epsilon$ whenever $|x - 5| < \delta$. - Simplify inside absolute value: $$|5x + 7 - 32| = |5x - 25| = 5|x - 5|$$ - To ensure $5|x - 5| < \epsilon$, choose $\delta = \frac{\epsilon}{5}$. - Then if $|x - 5| < \delta$, we have $|5x + 7 - 32| < \epsilon$. - This completes the proof. **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 $\lim_{x \to 5} (5x + 7) = 32$ using $(\epsilon - \delta)$ definition.