Math Problem Statement
Sertakan dengan cara pengerjaan soalnya
Nyatakan fungsi berikut tidak 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 following functions without absolute values. The three given functions are:
a.
b.
c.
To solve these, we need to break down each absolute value function based on its definition:
\begin{cases} A & \text{if } A \geq 0 \\ -A & \text{if } A < 0 \end{cases}$$ We'll apply this to each of the absolute values. ### a. $$f(x) = |x| + |3x + 1|$$ For $$|x|$$: - When $$x \geq 0$$, $$|x| = x$$. - When $$x < 0$$, $$|x| = -x$$. For $$|3x + 1|$$: - Solve $$3x + 1 = 0$$, so $$x = -\frac{1}{3}$$. - When $$x \geq -\frac{1}{3}$$, $$|3x + 1| = 3x + 1$$. - When $$x < -\frac{1}{3}$$, $$|3x + 1| = -(3x + 1) = -3x - 1$$. Thus, we have three cases to consider: 1. For $$x \geq 0$$: $$f(x) = x + (3x + 1) = 4x + 1$$. 2. For $$-\frac{1}{3} \leq x < 0$$: $$f(x) = -x + (3x + 1) = 2x + 1$$. 3. For $$x < -\frac{1}{3}$$: $$f(x) = -x + (-3x - 1) = -4x - 1$$. ### b. $$f(x) = 3 + |2x - 5|$$ Solve $$2x - 5 = 0$$, so $$x = \frac{5}{2} = 2.5$$. For $$|2x - 5|$$: - When $$x \geq 2.5$$, $$|2x - 5| = 2x - 5$$. - When $$x < 2.5$$, $$|2x - 5| = -(2x - 5) = -2x + 5$$. Thus, we have two cases: 1. For $$x \geq 2.5$$: $$f(x) = 3 + (2x - 5) = 2x - 2$$. 2. For $$x < 2.5$$: $$f(x) = 3 + (-2x + 5) = -2x + 8$$. ### c. $$f(x) = 3|x - 2| - |x + 1|$$ For $$|x - 2|$$: - Solve $$x - 2 = 0$$, so $$x = 2$$. - When $$x \geq 2$$, $$|x - 2| = x - 2$$. - When $$x < 2$$, $$|x - 2| = -(x - 2) = -x + 2$$. For $$|x + 1|$$: - Solve $$x + 1 = 0$$, so $$x = -1$$. - When $$x \geq -1$$, $$|x + 1| = x + 1$$. - When $$x < -1$$, $$|x + 1| = -(x + 1) = -x - 1$$. Thus, we have three cases to consider: 1. For $$x \geq 2$$: $$f(x) = 3(x - 2) - (x + 1) = 3x - 6 - x - 1 = 2x - 7$$. 2. For $$-1 \leq x < 2$$: $$f(x) = 3(-x + 2) - (x + 1) = -3x + 6 - x - 1 = -4x + 5$$. 3. For $$x < -1$$: $$f(x) = 3(-x + 2) - (-x - 1) = -3x + 6 + x + 1 = -2x + 7$$. ### Final Results: - **a.** $$f(x) = \begin{cases} 4x + 1 & \text{if } x \geq 0 \\ 2x + 1 & \text{if } -\frac{1}{3} \leq x < 0 \\ -4x - 1 & \text{if } x < -\frac{1}{3} \end{cases}$$ - **b.** $$f(x) = \begin{cases} 2x - 2 & \text{if } x \geq 2.5 \\ -2x + 8 & \text{if } x < 2.5 \end{cases}$$ - **c.** $$f(x) = \begin{cases} 2x - 7 & \text{if } x \geq 2 \\ -4x + 5 & \text{if } -1 \leq x < 2 \\ -2x + 7 & \text{if } x < -1 \end{cases}$$ Would you like more details or have any questions? Here are 5 related questions you can explore: 1. How do you determine the breakpoints for absolute value functions? 2. What are the different cases when solving absolute value equations? 3. How do absolute value functions relate to piecewise functions? 4. Can you graph the piecewise solutions for each function? 5. How do inequalities change when absolute values are involved? **Tip**: Always solve for where the expressions inside absolute value are equal to zero. These points divide the domain into intervals where you can remove the absolute value.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Absolute Value Functions
Piecewise Functions
Formulas
|A| = A if A ≥ 0; |A| = -A if A < 0
Theorems
Properties of Absolute Values
Suitable Grade Level
Grades 10-12