AP Precalculus Practice Test: Unit 2 Question #21 The Domain and Range of the Inverse of a Radical

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### AP Precalculus Practice Test: Unit 2, Question #21
**Topic:** The Domain and Range of the Inverse of a Radical Function

In this question, you are tasked with finding the **domain and range** of the **inverse** of a **radical function**. A radical function typically involves a square root or other roots, and its inverse will often have a restricted domain and range.

To solve this, we'll first examine the given radical function and then determine the domain and range of its inverse.

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### **Steps to Find the Domain and Range of the Inverse of a Radical Function:**

1. **Identify the given radical function:**
- Let the function be \( f(x) = \sqrt{x - h} + k \), where \( h \) and \( k \) are constants that affect the horizontal and vertical shifts of the graph.

2. **Find the inverse of the function:**
- To find the inverse, follow the steps:
1. Replace \( f(x) \) with \( y \):
\( y = \sqrt{x - h} + k \)
2. Switch \( x \) and \( y \):
\( x = \sqrt{y - h} + k \)
3. Solve for \( y \):
Subtract \( k \) from both sides:
\( x - k = \sqrt{y - h} \)
Square both sides:
\( (x - k)^2 = y - h \)
Solve for \( y \):
\( y = (x - k)^2 + h \)

3. **Determine the domain and range of the original function:**
- For the function \( f(x) = \sqrt{x - h} + k \), the domain is restricted by the square root, so:
\[
x - h \geq 0 \quad \text{or} \quad x \geq h
\]
Thus, the domain of the original function is \( [h, \infty) \).

- The range of the original function is all values greater than or equal to \( k \), so the range is \( [k, \infty) \).

4. **Find the domain and range of the inverse function:**
- The domain of the inverse function will be the range of the original function, and the range of the inverse function will be the domain of the original function.
- **Domain of the inverse:** The range of the original function \( f(x) = \sqrt{x - h} + k \) is \( [k, \infty) \), so the domain of the inverse is \( [k, \infty) \).
- **Range of the inverse:** The domain of the original function is \( [h, \infty) \), so the range of the inverse is \( [h, \infty) \).

---

### **Example Problem:**

Given the radical function \( f(x) = \sqrt{x - 3} + 2 \), find the domain and range of its inverse.

---

### **Solution Steps:**

1. **Start with the given radical function:**
\[
f(x) = \sqrt{x - 3} + 2
\]
- The domain of the original function is \( x \geq 3 \), so the domain is \( [3, \infty) \).
- The range of the original function is \( y \geq 2 \), so the range is \( [2, \infty) \).

2. **Find the inverse:**
- Replace \( f(x) \) with \( y \):
\[
y = \sqrt{x - 3} + 2
\]
- Switch \( x \) and \( y \):
\[
x = \sqrt{y - 3} + 2
\]
- Subtract 2 from both sides:
\[
x - 2 = \sqrt{y - 3}
\]
- Square both sides:
\[
(x - 2)^2 = y - 3
\]
- Solve for \( y \):
\[
y = (x - 2)^2 + 3
\]

3. **Determine the domain and range of the inverse:**
- The **domain** of the inverse function is the range of the original function: \( [2, \infty) \).
- The **range** of the inverse function is the domain of the original function: \( [3, \infty) \).

---

### **Final Answer:**

- The domain of the inverse is \( [2, \infty) \).
- The range of the inverse is \( [3, \infty) \).

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Nick Perich
Norristown Area High School
Norristown Area School District
Norristown, Pa

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