Question 1 Suppose the functions \( f, g, h, r \) and \( \ell \) are defined as follows: \[ \begin{array}{l} f(x)=\frac{1}{3} \log _{3} \frac{1}{4}+\log _{3} x \\ g(x)=\sqrt{(x-3)^{2}} \\ h(x)=5 x-2 x^{2} \\ r(x)=2^{3 x+1}-2^{x-2} \\ \ell(x)=\frac{1}{\sqrt{x}} \end{array} \] 1.1 Write down \( D_{f} \), the domain of \( f \) and then solve the equation \( f(x)=-\log _{\frac{1}{3}} \sqrt[3]{x} \). 1.2 Write down \( D_{g} \) and then solve the equation \( g(x)=\frac{x}{2} \). 1.3 Write down \( D_{h} \) and then solve the inequality \( 2 \geq h(x) \). 1.4 Write down \( D_{r} \) and then solve the equation \( r(x)=0 \). 1.5 Write down \( D_{r \cdot \ell} \) without first calculating \( (r \cdot \ell)(x) \). 1.6 Write down \( D_{\frac{\ell}{g}} \) without first calculating \( \frac{\ell}{g}(x) \).

To tackle the questions step by step: 1.1 For the function \( f(x)=\frac{1}{3} \log _{3} \frac{1}{4}+\log _{3} x \), the domain \( D_f \) consists of all positive real numbers \( x \) since the logarithm function is defined only for positive values. Hence, \( D_f = (0, \infty) \). Now, to solve the equation \( f(x)=-\log _{\frac{1}{3}} \sqrt[3]{x} \), we can rewrite it as \( f(x) = \log_{3}(\frac{1}{\sqrt[3]{x}}) \). By equating and simplifying, we find the values of \( x \). 1.2 The function \( g(x)=\sqrt{(x-3)^{2}} \) is defined for all real numbers, so the domain \( D_g = (-\infty, \infty) \). To solve \( g(x)=\frac{x}{2} \), we notice that \( g(x) = |x - 3| \). Thus, we can set up cases based on the definition of the absolute value, leading to solutions for \( x \). 1.3 The function \( h(x)=5x-2x^{2} \), is a quadratic function, and its domain is also all real numbers \( D_h = (-\infty, \infty) \). To solve the inequality \( 2 \geq h(x) \), we set up the equation \( 2 = 5x - 2x^2 \) and find the critical points by solving for \( x \) and then determining the intervals where the inequality holds. 1.4 The function \( r(x)=2^{3x+1}-2^{x-2} \) is defined for all real numbers, so \( D_r = (-\infty, \infty) \). To solve \( r(x)=0 \), we set \( 2^{3x+1} = 2^{x-2} \). By equating the exponents of the base 2, we can solve for \( x \). 1.5 For \( D_{r \cdot \ell} \), we need the intersection of the domains of \( r(x) \) and \( \ell(x) \). Since \( \ell(x) = \frac{1}{\sqrt{x}} \) requires \( x > 0 \), the combined domain is \( D_{r \cdot \ell} = (0, \infty) \). 1.6 For \( D_{\frac{\ell}{g}} \), the function \( \frac{\ell}{g} = \frac{1/\sqrt{x}}{|x-3|} \) requires both \( \ell(x) \) and \( g(x) \) to be defined. Thus, \( x > 0 \) (from \( \ell(x) \)), and \( |x - 3| \neq 0 \) means \( x \neq 3 \). Therefore, \( D_{\frac{\ell}{g}} = (0, 3) \cup (3, \infty) \).