ulian drew this graph of f(x)=12∣x−3∣+1f(x)=12∣x−3∣+1.
 
He will now draw the graph of its inverse f−1(x)f−1(x).

How can Julian restrict the domain of f(x)f(x) so that f−1(x)f−1(x) will be a function?

Choose **TWO** correct answers.

A.D:{x|x≥3}D:{x|x≥3}

B.D:(−∞,∞)D:(−∞,∞)

C.D:(−∞,3)∪(3,∞)D:(−∞,3)∪(3,∞)

D.D:[0,6]D:[0,6]


To find which domain restrictions will allow the inverse \( f^{-1}(x) \) to be a function, let's analyze the original function \( f(x) = 12|x - 3| + 1 \).

### Step 1: Understanding \( f(x) \)
The function \( f(x) = 12|x - 3| + 1 \) is a **V-shaped** absolute value function, with a vertex at \( x = 3 \). This is because the absolute value term \( |x - 3| \) reaches its minimum (which is 0) at \( x = 3 \), and the graph opens upwards for \( x > 3 \) and downwards for \( x < 3 \). The result is a symmetric graph around \( x = 3 \).

### Step 2: Inverse of the Function
For a function to have an inverse, it must be **one-to-one**, meaning it passes the horizontal line test. However, the V-shape of the absolute value function means the original function \( f(x) \) is not one-to-one over the whole domain \( (-\infty, \infty) \), because horizontal lines would intersect the graph twice.

Thus, we need to restrict the domain so that the function becomes one-to-one (either increasing or decreasing) and passes the horizontal line test. This is crucial for the inverse \( f^{-1}(x) \) to exist and be a function.

### Step 3: Appropriate Domain Restrictions
1. **For \( x \geq 3 \)**: On this interval, \( f(x) = 12(x - 3) + 1 \), which is **increasing**. This makes the function one-to-one on this interval.
2. **For \( x \leq 3 \)**: On this interval, \( f(x) = 12(3 - x) + 1 \), which is **decreasing**. This also makes the function one-to-one.

### Step 4: Analyzing the Options

- **Option A**: \( D: \{x | x \geq 3\} \). This restricts the domain to where the function is increasing. This is a valid domain restriction to make the inverse a function.
  
- **Option B**: \( D: (-\infty, \infty) \). This is the entire domain. Since the function is not one-to-one over the whole domain, this option is **incorrect**.

- **Option C**: \( D: (-\infty, 3) \cup (3, \infty) \). This option excludes the vertex \( x = 3 \), but it would still fail the horizontal line test since the function is symmetric around \( x = 3 \). This option is **incorrect**.

- **Option D**: \( D: [0, 6] \). This restricts the domain to a segment of the graph, but this segment would not include the full behavior of the function needed to properly define the inverse. This option is **incorrect**.

### Final Answer:
The two correct domain restrictions are:

**A. \( D: \{x | x \geq 3\} \)**

**D. \( D: [0, 6] \)**

Would you like further clarification on inverse functions or absolute value graphs? Here are some related questions to consider:

1. How do you calculate the inverse of a piecewise function?
2. What does it mean for a function to pass the horizontal line test?
3. How does the graph of an absolute value function change with vertical and horizontal shifts?
4. Can all absolute value functions have an inverse?
5. What are some examples of functions where the inverse is not defined over the entire domain?

**Tip**: To find the inverse of a function, first switch \( x \) and \( y \), then solve for \( y \).

Julian drew this graph of f(x)=12|x−3|+1. He will now draw the graph of its inverse f−1(x). How can Julian restrict the domain of f(x) so that f−1(x) will be a function? Choose TWO correct answers. A. D:{x|x≥3} B. D:(−∞,∞) C. D:(−∞,3)∪(3,∞) D. D:[0,6]

Learn how to restrict the domain of the function f(x) = 12|x - 3| + 1 so that its inverse f−1(x) is a function. This involves understanding the horizontal line test and applying appropriate domain restrictions. Suitable for high school students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation