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 rules:** - Domain: set of all $x$ values for which the function is defined. - Range: set of all possible output values. - For $f(x) = \sqrt{\frac{1}{4} - x^2}$, the expression inside the square root must be $\geq 0$. - For $h(x) = \tan(x - 3)$, domain excludes points where $\cos(x-3) = 0$. 3. **(a)(i) Domain and range:** - Domain of $f$: solve $\frac{1}{4} - x^2 \geq 0$. $$x^2 \leq \frac{1}{4}$$ $$-\frac{1}{2} \leq x \leq \frac{1}{2}$$ - Range of $f$: since square root outputs non-negative values and max inside root is $\frac{1}{4}$, $$\text{range} = [0, \sqrt{\frac{1}{4}}] = [0, \frac{1}{2}]$$ - Domain of $h$: all real numbers except where $\cos(x-3) = 0$. $$x - 3 = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$$ $$x = 3 + \frac{\pi}{2} + k\pi$$ - Range of $h$: $(-\infty, \infty)$ because tangent takes all real values. 4. **(a)(ii) 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. **(a)(iii) Inverse of $g(x)$:** - Start with $y = \frac{5}{9}x + 3$ - Solve for $x$: $$y - 3 = \frac{5}{9}x$$ $$x = \frac{9}{5}(y - 3)$$ - So, $$g^{-1}(x) = \frac{9}{5}(x - 3)$$ 6. **(b) Formal $(\epsilon - \delta)$ limit verification:** - Given $\lim_{x \to 5} (5x + 7) = 32$. - For any $\epsilon > 0$, find $\delta > 0$ such that if $|x - 5| < \delta$, then $$|5x + 7 - 32| < \epsilon$$ - Simplify inside absolute value: $$|5x + 7 - 32| = |5x - 25| = 5|x - 5|$$ - To ensure $5|x - 5| < \epsilon$, choose $$\delta = \frac{\epsilon}{5}$$ - Therefore, if $|x - 5| < \delta$, then $$|5x + 7 - 32| < \epsilon$$ - This completes the proof. **Final answers:** - Domain of $f$: $[-\frac{1}{2}, \frac{1}{2}]$ - Range of $f$: $[0, \frac{1}{2}]$ - Domain of $h$: $\mathbb{R} \setminus \{3 + \frac{\pi}{2} + k\pi : k \in \mathbb{Z}\}$ - Range of $h$: $(-\infty, \infty)$ - $(f \circ g)(x) = \sqrt{\frac{1}{4} - \left(\frac{5}{9}x + 3\right)^2}$ - $(f \circ h)(x) = \sqrt{\frac{1}{4} - \tan^2(x - 3)}$ - $(g \circ f)(x) = \frac{5}{9} \sqrt{\frac{1}{4} - x^2} + 3$ - $g^{-1}(x) = \frac{9}{5}(x - 3)$ - Limit verified with $\delta = \frac{\epsilon}{5}$ for $\lim_{x \to 5} (5x + 7) = 32$.