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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: $|x-5|=3$.
- Solve for two cases: $x-5=3$ and $x-5=-3$.
- The solutions are $x=2$ and $x=8$, so 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. Considering Two Cases
To solve the absolute value equation $|x-5|=3$, we consider two cases:

Case 1: $x-5=3$

Case 2: $x-5=-3$

6. Solving Case 1
For Case 1, we solve for $x$ by adding 5 to both sides:$$x-5=3$$$$x=3+5$$$$x=8$$

7. Solving Case 2
For Case 2, we solve for $x$ by adding 5 to both sides:$$x-5=-3$$$$x=-3+5$$$$x=2$$

8. 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 designed to produce parts that are 5 cm long, but a tolerance of 0.3 cm is allowed, the actual length $x$ of the part must satisfy the equation $|x-5| \le 0.3$. Solving this inequality helps determine the acceptable range of part lengths.

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