1. **State the problem:** We have three functions: - $f(x) = \sqrt{\frac{1}{4} - x^2}$ - $g(x) = \frac{5}{9}x + 3$ - $h(x) = \tan(x - 3)$ We need to find: (i) The domain and range of $f$ and $h$. 2. **Domain and range of $f(x)$:** - The function $f(x) = \sqrt{\frac{1}{4} - x^2}$ involves a square root, so the expression inside must be non-negative: $$\frac{1}{4} - x^2 \geq 0$$ - Rearranging: $$x^2 \leq \frac{1}{4}$$ - Taking square roots: $$-\frac{1}{2} \leq x \leq \frac{1}{2}$$ - So, the domain of $f$ is: $$\boxed{\left[-\frac{1}{2}, \frac{1}{2}\right]}$$ - To find the range, note that the square root outputs non-negative values. - The maximum value inside the root is when $x=0$: $$f(0) = \sqrt{\frac{1}{4} - 0} = \frac{1}{2}$$ - The minimum value is 0 when $x = \pm \frac{1}{2}$: $$f\left(\pm \frac{1}{2}\right) = 0$$ - Therefore, the range of $f$ is: $$\boxed{[0, \frac{1}{2}]}$$ 3. **Domain and range of $h(x)$:** - The function $h(x) = \tan(x - 3)$ is defined for all real $x$ except where the tangent function has vertical asymptotes. - Tangent is undefined where its argument equals $\frac{\pi}{2} + k\pi$, for any integer $k$. - So, the domain of $h$ is: $$\{x \in \mathbb{R} : x - 3 \neq \frac{\pi}{2} + k\pi, k \in \mathbb{Z}\}$$ - Or equivalently: $$x \neq 3 + \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$$ - The range of the tangent function is all real numbers: $$\boxed{(-\infty, \infty)}$$ **Final answers:** - Domain of $f$: $\left[-\frac{1}{2}, \frac{1}{2}\right]$ - Range of $f$: $[0, \frac{1}{2}]$ - Domain of $h$: $\{x \in \mathbb{R} : x \neq 3 + \frac{\pi}{2} + k\pi, k \in \mathbb{Z}\}$ - Range of $h$: $(-\infty, \infty)$