College

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 = 75[/tex]

Answer :

- Set up the equation $4|x-5|+3 = 15$.
- Isolate the absolute value term: $|x-5| = 3$.
- Split the absolute value equation into two cases: $x-5 = 3$ and $x-5 = -3$.
- Solve for $x$ in each case, obtaining $x = 8$ and $x = 2$. 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 Term
Next, we subtract 3 from both sides of the equation: $$4|x-5| = 15 - 3$$ $$4|x-5| = 12$$

4. Simplifying the Equation
Now, we 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: $x-5 = 3$
Adding 5 to both sides, we get: $$x = 3 + 5$$ $$x = 8$$

Case 2: $x-5 = -3$
Adding 5 to both sides, we get: $$x = -3 + 5$$ $$x = 2$$

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 acceptable tolerances in manufacturing. For example, if a machine is designed to cut a metal rod to a length of 5 cm, but there is an acceptable tolerance of 0.3 cm, the actual length $x$ of the rod must satisfy the equation $|x - 5| \le 0.3$. Solving this inequality gives the range of acceptable lengths for the metal rod.