High School

Given the function [tex]f(x)=4|x-5|+3[/tex], for what values of [tex]x[/tex] is [tex]f(x)=15[/tex]?



A. [tex]x=2, x=8[/tex]



B. [tex]x=1.5, x=8[/tex]



C. [tex]x=2, x=7.5[/tex]



D. [tex]x=0.5, x=7.5[/tex]

Answer :

- Set up the equation $4|x-5|+3 = 15$.
- Isolate the absolute value term: $|x-5| = 3$.
- Split the equation into two cases: $x-5 = 3$ and $x-5 = -3$.
- Solve for $x$ in each case to find the solutions: $x = 2$ and $x = 8$. The final answer is $\boxed{x=2, x=8}$.

### Explanation
1. Understanding the Problem
We are given the function $f(x)=4|x-5|+3$, and we want to find the values of $x$ for which $f(x)=15$. This involves solving an absolute value equation.

2. Setting up the Equation
First, we set $f(x)$ equal to 15: $$4|x-5|+3 = 15$$

3. Isolating the Absolute Value
Next, we isolate the absolute value term. Subtract 3 from both sides of the equation:$$4|x-5| = 15 - 3$$$$4|x-5| = 12$$

4. Simplifying the Equation
Now, divide both sides by 4:$$|x-5| = \frac{12}{4}$$$$|x-5| = 3$$

5. Solving for x
To solve the absolute value equation $|x-5| = 3$, we consider two cases:

Case 1: $x-5 = 3$
Add 5 to both sides:$$x = 3 + 5$$$$x = 8$$

Case 2: $x-5 = -3$
Add 5 to both sides:$$x = -3 + 5$$$$x = 2$$

6. Final Answer
Therefore, the values of $x$ for which $f(x) = 15$ are $x = 2$ and $x = 8$.

### Examples
Absolute value equations are useful in many real-world scenarios, such as determining tolerances in manufacturing. For example, if a machine is supposed to cut a metal rod to a length of 5 cm, but it is acceptable for the length to be off by up to 0.3 cm, then the actual length $x$ must satisfy the equation $|x - 5| \le 0.3$. Solving this inequality tells us the acceptable range of lengths for the metal rod.

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