QUESTION 3 Suppose the functions \( f, g, h \) and \( l \) are defined as follows: \[ \begin{aligned} f(x)=4 x^{2}-5 x+1 \\ g(x)=2 \sqrt{2-\frac{x}{2}}-x\end{aligned} \] \[ \begin{aligned} h(x) & =-\frac{1}{2} x+3 \\ l(x) & =\log _{4}(x+3)-\log _{4}(x-2)\end{aligned} \] Write down \( D_{f} \) and solve the inequality \( f(x) \leq 0 \)

To address the problem, let's analyze each part step by step. ### 1. Determining the Domain \( D_f \) of \( f(x) \) The function \( f(x) = 4x^2 - 5x + 1 \) is a **quadratic polynomial**. Polynomial functions are defined for all real numbers because there are no restrictions like division by zero or taking the square root of a negative number. **Therefore, the domain \( D_f \) is:** \[ D_f = \mathbb{R} \] (This means \( f(x) \) is defined for all real numbers \( x \).) ### 2. Solving the Inequality \( f(x) \leq 0 \) We need to solve: \[ 4x^2 - 5x + 1 \leq 0 \] #### a. **Find the Roots of the Quadratic Equation:** First, solve the equation: \[ 4x^2 - 5x + 1 = 0 \] Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 4 \), \( b = -5 \), and \( c = 1 \). Calculate the discriminant: \[ \Delta = b^2 - 4ac = (-5)^2 - 4(4)(1) = 25 - 16 = 9 \] Now, find the roots: \[ x = \frac{5 \pm \sqrt{9}}{8} = \frac{5 \pm 3}{8} \] So, the two roots are: \[ x_1 = \frac{5 + 3}{8} = 1 \] \[ x_2 = \frac{5 - 3}{8} = \frac{1}{4} \] #### b. **Determine the Interval Where \( f(x) \leq 0 \):** The quadratic \( 4x^2 - 5x + 1 \) opens upwards (since the coefficient of \( x^2 \) is positive). Therefore, the expression \( 4x^2 - 5x + 1 \) is **less than or equal to zero** between its roots. **Thus, the solution to the inequality is:** \[ \frac{1}{4} \leq x \leq 1 \] ### **Final Answer:** - **Domain \( D_f \):** All real numbers \( \mathbb{R} \). - **Solution to \( f(x) \leq 0 \):** \( x \) is between \( \frac{1}{4} \) and \( 1 \), inclusive. Formally: \[ D_f = \mathbb{R} \] \[ \frac{1}{4} \leq x \leq 1 \]