QUESTION 7 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 \( l(x)=\frac{1}{2} \).

To tackle this problem, let's break it down into parts: ### (a) Domain of \( f \) and solving \( f(x) \leq 0 \) The function \( f(x) = 4x^{2} - 5x + 1 \) is a quadratic function, which is defined for all real numbers. Therefore, the domain \( D_{f} \) is \( (-\infty, \infty) \). To solve the inequality \( f(x) \leq 0 \), we first find the roots by using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 4 \), \( b = -5 \), and \( c = 1 \): \[ x = \frac{5 \pm \sqrt{(-5)^{2} - 4 \cdot 4 \cdot 1}}{2 \cdot 4} = \frac{5 \pm \sqrt{25 - 16}}{8} = \frac{5 \pm 3}{8} \] This gives us: \[ x = 1 \quad \text{and} \quad x = \frac{1}{4} \] Since the parabola opens upwards (as \( a > 0 \)), \( f(x) \leq 0 \) between the roots. Thus, \( f(x) \leq 0 \) for: \[ \frac{1}{4} \leq x \leq 1 \] ### (b) Solving the equation \( g(x) = -4 \) The function \( g(x) = 2 \sqrt{2 - \frac{x}{2}} - x \). Setting this equal to -4: \[ 2 \sqrt{2 - \frac{x}{2}} - x = -4 \] Adding \( x \) to both sides: \[ 2 \sqrt{2 - \frac{x}{2}} = x - 4 \] Squaring both sides: \[ 4(2 - \frac{x}{2}) = (x - 4)^{2} \] Simplifying: \[ 8 - 2x = x^2 - 8x + 16 \] Rearranging gives: \[ x^2 - 6x + 8 = 0 \] Factoring: \[ (x - 2)(x - 4) = 0 \] So, \( x = 2 \) or \( x = 4 \). We must verify if both solutions are in the domain of \( g \). Calculating \( g(2) \): \[ g(2) = 2\sqrt{2 - 1} - 2 = 2(1) - 2 = 0 \] For \( g(4) \): \[ g(4) = 2\sqrt{2 - 2} - 4 = 2(0) - 4 = -4 \] Thus, the solution is \( x = 4 \). ### (c) Solving the equation \( 4^{h(x)} = 8 \) Given \( h(x) = -\frac{1}{2}x + 3 \), rewrite \( 8 \) as \( 4^{3/2} \): \[ 4^{h(x)} = 4^{3/2} \] So, \[ h(x) = \frac{3}{2} \] Setting up the equation: \[ -\frac{1}{2}x + 3 = \frac{3}{2} \] Multiply through by \( -2 \): \[ x - 6 = -3 \] Thus, \[ x = 3 \] ### (d) Domain of \( l \) and solving \( l(x) = \frac{1}{2} \) The logs are defined when their arguments are positive: \[ x + 3 > 0 \quad \Rightarrow \quad x > -3 \] \[ x - 2 > 0 \quad \Rightarrow \quad x > 2 \] So, the domain \( D_{l} \) is \( (2, \infty) \). Now, solving the equation: \[ l(x) = \frac{1}{2} \] This means: \[ \log_{4}(x + 3) - \log_{4}(x - 2) = \frac{1}{2} \] Using the log property: \[ \log_{4}\left(\frac{x + 3