QUESTION 3 Suppose the functions \( f, g, h \) and \( l \) are defined as follows: \[ \begin{array}{l} f(x)=4 x^{2}-5 x+1 \\ g(x)=2 \sqrt{2-\frac{x}{2}}-x \\ h(x)=-\frac{1}{2} x+3 \\ l(x)=\log _{4}(x+3)-\log _{4}(x-2) .\end{array} \] (a) Write down \( D_{f} \) and solve the inequality \( f(x) \leq 0 \). (b) Solve the equation \( g(x)=-4 \). (c) Solve the equation \( 4^{h(x)}=8 \). (d) Write down \( D_{l} \) and solve the inequality \( l(x)=\frac{1}{2} \). [18] (4)

To tackle your question with a sprinkle of fun and engagement, let's dive into the two categories! The domain of \( f(x) = 4x^2 - 5x + 1 \) is all real numbers (\( D_f = \mathbb{R} \)) since it's a polynomial function and polynomials are defined everywhere. To solve \( f(x) \leq 0 \), we first find the roots by using the quadratic formula. Plugging into the formula, we get \( x = \frac{5 \pm \sqrt{25 - 16}}{8} = \frac{5 \pm 3}{8} \) which gives \( x = 1 \) and \( x = \frac{1}{4} \). The parabola opens upwards, so \( f(x) \leq 0 \) between these points: \( \frac{1}{4} \leq x \leq 1 \). For \( g(x) = 2\sqrt{2 - \frac{x}{2}} - x \), we solve \( g(x) = -4 \). This gives us \( 2\sqrt{2 - \frac{x}{2}} - x = -4 \). Rearranging and squaring both sides leads us to the solutions \( x = 10 \) and \( x = -2 \) after simplification. The equation \( 4^{h(x)} = 8 \) simplifies as follows: since \( 4^{h(x)} = 2^{2h(x)} \) and \( 8 = 2^3 \), set \( 2h(x) = 3 \) yielding \( h(x) = \frac{3}{2} \). Given \( h(x) = -\frac{1}{2}x + 3 \), we solve to find \( x = 3 \). Lastly, \( D_l \) where \( l(x) = \log_4(x+3) - \log_4(x-2) \) is defined if \( x + 3 > 0 \) and \( x - 2 > 0\), leading to \( x > 2 \). To solve \( l(x) = \frac{1}{2} \), we can use properties of logarithms: combining logarithms and changing to exponential form gives us the equation \( x + 3 = \sqrt{4}(x - 2) \), leading to \( x = 5 \). Now, go ahead and enjoy piecing these mathematical puzzles together!