Math Problem Statement

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|

Solution

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

a. f(x)=∣x∣+∣3x+1∣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=0x = 0 (for ∣x∣|x|)
    • 3x+1=0⇒x=−133x + 1 = 0 \Rightarrow x = -\frac{1}{3} (for ∣3x+1∣|3x + 1|)

  2. Determine intervals based on these points:

    • x<−13x < -\frac{1}{3}
    • −13≤x<0-\frac{1}{3} \leq x < 0
    • x≥0x \geq 0

  3. Express f(x)f(x) in each interval:

    • For x<−13x < -\frac{1}{3}: f(x)=−x−(3x+1)=−4x−1f(x) = -x - (3x + 1) = -4x - 1
    • For −13≤x<0-\frac{1}{3} \leq x < 0: f(x)=−x+(3x+1)=2x+1f(x) = -x + (3x + 1) = 2x + 1
    • For x≥0x \geq 0: f(x)=x+(3x+1)=4x+1f(x) = x + (3x + 1) = 4x + 1

Thus, the function f(x)=∣x∣+∣3x+1∣f(x) = |x| + |3x + 1| can be expressed as:

\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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Absolute Value
Piecewise Functions

Formulas

|x| = x if x >= 0, -x if x < 0

Theorems

Piecewise Definition of Absolute Value Functions

Suitable Grade Level

Grades 10-12

Related Recommendation

Breaking Down Absolute Value Functions | Algebra Practice Problem

Piecewise-defined function for f(x) = |3x + 4|

How to Write f(x) = |x + 1| as a Piecewise Function

Piecewise-Defined Function for Absolute Value: f(x) = |4x + 1|

Piecewise-Defined Absolute Value Function: f(x) = |5x + 4|