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 $$ p\left(r\left(\frac{1}{2}\right)\right) $$.

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Let's address each question step by step.

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### **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:**
$$
c(x) = 9x^{2} - 15x + 9
$$

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### **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 $$ c(x) $$:**
$$
\text{All real numbers, } \mathbb{R}
$$

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### **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 $$ c(x) $$:**
$$
c(x) \geq \frac{11}{4}
$$
$$
\text{In interval notation: } \left[ \frac{11}{4}, \infty \right)
$$

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### **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:**
$$
p\left(r\left(\frac{1}{2}\right)\right) = \frac{15}{4}
$$

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### **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 $$ p\left(r\left(\frac{1}{2}\right)\right) $$:**
   $$
   \frac{15}{4}
   $$
1. $$ c(x) = 9x^{2} - 15x + 9 $$ 2. Domain: All real numbers 3. Range: $$ \left[ \frac{11}{4}, \infty \right) $$ 4. $$ p\left(r\left(\frac{1}{2}\right)\right) = \frac{15}{4} $$

For questions 1 through 4, The functions and are shown.

Find a new function, , that is defined by .

Determine the domain of .

Determine the range of .

Find .

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For questions 1 through 4, The functions and are shown. Find a new function, , that is defined by . Determine the domain of . Determine the range of . Find .