For questions 1 through 4, The functions \( p \) and \( r \) are shown. \[ p(x)=x^{2}-3 x+5 \quad r(x)=3 x-1 \] 1. Find a new function, \( c(x) \), that is defined by \( p(r(x)) \). 2. Determine the domain of \( c(x) \). Determine the range of \( c(x) \). Find \( p\left(r\left(\frac{1}{2}\right)\right) \).

Let's address each question step by step. --- ### **1. Find a new function, \( c(x) \), that is defined by \( p(r(x)) \).** Given: \[ p(x) = x^{2} - 3x + 5 \quad \text{and} \quad r(x) = 3x - 1 \] To find \( c(x) = p(r(x)) \), substitute \( r(x) \) into \( p(x) \): \[ c(x) = p(r(x)) = p(3x - 1) = (3x - 1)^{2} - 3(3x - 1) + 5 \] Expand and simplify: \[ \begin{align*} (3x - 1)^{2} &= 9x^{2} - 6x + 1 \\ -3(3x - 1) &= -9x + 3 \\ \end{align*} \] Combine all terms: \[ c(x) = 9x^{2} - 6x + 1 - 9x + 3 + 5 = 9x^{2} - 15x + 9 \] **Final Function:** \[ c(x) = 9x^{2} - 15x + 9 \] --- ### **2. Determine the domain of \( c(x) \).** Both \( p(x) \) and \( r(x) \) are polynomials: - \( r(x) = 3x - 1 \) is defined for all real numbers. - \( p(x) = x^{2} - 3x + 5 \) is also defined for all real numbers. Since the composition of two functions that are both defined for all real numbers is also defined for all real numbers: **Domain of \( c(x) \):** \[ \text{All real numbers, } \mathbb{R} \] --- ### **3. Determine the range of \( c(x) \).** The function \( c(x) = 9x^{2} - 15x + 9 \) is a quadratic function. Since the coefficient of \( x^{2} \) is positive (9), the parabola opens upwards, and thus the function has a minimum value. **Finding the vertex (minimum point):** The vertex of a quadratic function \( ax^{2} + bx + c \) occurs at: \[ x = -\frac{b}{2a} \] For \( c(x) \): \[ a = 9, \quad b = -15 \\ x = -\frac{-15}{2 \times 9} = \frac{15}{18} = \frac{5}{6} \] **Calculating \( c\left(\frac{5}{6}\right) \):** \[ c\left(\frac{5}{6}\right) = 9\left(\frac{5}{6}\right)^{2} - 15\left(\frac{5}{6}\right) + 9 \\ = 9\left(\frac{25}{36}\right) - \frac{75}{6} + 9 \\ = \frac{225}{36} - \frac{75}{6} + 9 \\ = \frac{25}{4} - \frac{25}{2} + 9 \\ = \frac{25 - 50 + 36}{4} \\ = \frac{11}{4} \] **Range of \( c(x) \):** \[ c(x) \geq \frac{11}{4} \] \[ \text{In interval notation: } \left[ \frac{11}{4}, \infty \right) \] --- ### **4. Find \( p\left(r\left(\frac{1}{2}\right)\right) \).** First, compute \( r\left(\frac{1}{2}\right) \): \[ r\left(\frac{1}{2}\right) = 3 \times \frac{1}{2} - 1 = \frac{3}{2} - 1 = \frac{1}{2} \] Next, compute \( p\left(\frac{1}{2}\right) \): \[ p\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{2} - 3 \times \frac{1}{2} + 5 = \frac{1}{4} - \frac{3}{2} + 5 \] \[ = \frac{1}{4} - \frac{6}{4} + \frac{20}{4} = \frac{1 - 6 + 20}{4} = \frac{15}{4} \] **Final Result:** \[ p\left(r\left(\frac{1}{2}\right)\right) = \frac{15}{4} \] --- ### **Summary of Answers:** 1. **Function \( c(x) \):** \[ c(x) = 9x^{2} - 15x + 9 \] 2. **Domain of \( c(x) \):** \[ \mathbb{R} \] 3. **Range of \( c(x) \):** \[ \left[ \frac{11}{4}, \infty \right) \] 4. **Value of \( p\left(r\left(\frac{1}{2}\right)\right) \):** \[ \frac{15}{4} \]