https://file.aiquickdraw.com/mathbot/file/1727336246428yj89ryll.jpg

The problem asks to express the given functions in their absolute value form. Let's analyze each function:

### a. \( f(x) = |x| + |3x + 1| \)

This function involves two absolute values. To handle this, we analyze the critical points where each absolute value changes its expression. 

1. **Find the critical points** by setting the arguments inside the absolute values equal to zero:
   - \( x = 0 \) (for \( |x| \))
   - \( 3x + 1 = 0 \Rightarrow x = -\frac{1}{3} \) (for \( |3x + 1| \))

2. **Determine intervals** based on these points:
   - \( x < -\frac{1}{3} \)
   - \( -\frac{1}{3} \leq x < 0 \)
   - \( x \geq 0 \)

3. **Express \( f(x) \) in each interval:**
   - For \( x < -\frac{1}{3} \): \( f(x) = -x - (3x + 1) = -4x - 1 \)
   - For \( -\frac{1}{3} \leq x < 0 \): \( f(x) = -x + (3x + 1) = 2x + 1 \)
   - For \( x \geq 0 \): \( f(x) = x + (3x + 1) = 4x + 1 \)

Thus, the function \( f(x) = |x| + |3x + 1| \) can be expressed as:
\[
f(x) = 
\begin{cases} 
-4x - 1 & \text{for } x < -\frac{1}{3} \\
2x + 1 & \text{for } -\frac{1}{3} \leq x < 0 \\
4x + 1 & \text{for } x \geq 0 
\end{cases}
\]

### b. \( f(x) = 3 + |2x - 5| \)

1. **Find the critical point**:
   - \( 2x - 5 = 0 \Rightarrow x = \frac{5}{2} \)

2. **Determine intervals**:
   - \( x < \frac{5}{2} \)
   - \( x \geq \frac{5}{2} \)

3. **Express \( f(x) \) in each interval:**
   - For \( x < \frac{5}{2} \): \( f(x) = 3 - (2x - 5) = 8 - 2x \)
   - For \( x \geq \frac{5}{2} \): \( f(x) = 3 + (2x - 5) = 2x - 2 \)

Thus, the function \( f(x) = 3 + |2x - 5| \) can be expressed as:
\[
f(x) = 
\begin{cases} 
8 - 2x & \text{for } x < \frac{5}{2} \\
2x - 2 & \text{for } x \geq \frac{5}{2} 
\end{cases}
\]

### c. \( f(x) = 3|x - 2| - |x + 1| \)

1. **Find the critical points**:
   - \( x - 2 = 0 \Rightarrow x = 2 \)
   - \( x + 1 = 0 \Rightarrow x = -1 \)

2. **Determine intervals**:
   - \( x < -1 \)
   - \( -1 \leq x < 2 \)
   - \( x \geq 2 \)

3. **Express \( f(x) \) in each interval:**
   - For \( x < -1 \): 
     \[
     f(x) = 3(-(x - 2)) - (-(x + 1)) = 3(2 - x) + (x + 1) = 7 - 2x
     \]
   - For \( -1 \leq x < 2 \): 
     \[
     f(x) = 3(x - 2) - (-(x + 1)) = 3(x - 2) + (x + 1) = 4x - 5
     \]
   - For \( x \geq 2 \): 
     \[
     f(x) = 3(x - 2) - (x + 1) = 3x - 6 - (x + 1) = 2x - 7
     \]

Thus, the function \( f(x) = 3|x - 2| - |x + 1| \) can be expressed as:
\[
f(x) = 
\begin{cases} 
7 - 2x & \text{for } x < -1 \\
4x - 5 & \text{for } -1 \leq x < 2 \\
2x - 7 & \text{for } x \geq 2 
\end{cases}
\]

Let me know if you want any additional details or have questions!

### Related Questions:
1. How do we graph these piecewise absolute value functions?
2. What are the slopes of each linear piece within the given intervals?
3. How can we determine the minimum or maximum values of these functions?
4. What changes if a constant is added or subtracted outside the absolute value?
5. How does shifting the arguments inside the absolute values affect the critical points?

### Tip:
When analyzing absolute value expressions, always identify the points where each absolute value term changes sign to define the intervals clearly.

Nyatakan fungsi berikut dalam nilai mutlak: a. f(x) = |x| + |3x + 1| b. f(x) = 3 + |2x - 5| c. f(x) = 3|x - 2| - |x + 1|

This exercise covers how to express functions involving absolute values in a piecewise form. The problems include functions like f(x) = |x| + |3x + 1|, f(x) = 3 + |2x - 5|, and f(x) = 3|x - 2| - |x + 1|. Ideal for students in Grades 10-12, the solutions show how to break down functions based on their critical points.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation