Let's solve the equation \( f(x) = 39 \) given the function \( f(x) = 6|x-2| + 3 \).
To begin, we set the function equal to 39:
[tex]\[ 6|x-2| + 3 = 39 \][/tex]
First, isolate the absolute value term:
[tex]\[ 6|x-2| + 3 - 3 = 39 - 3 \][/tex]
[tex]\[ 6|x-2| = 36 \][/tex]
Next, divide both sides by 6:
[tex]\[ |x-2| = \frac{36}{6} \][/tex]
[tex]\[ |x-2| = 6 \][/tex]
The absolute value equation \( |x-2| = 6 \) means we have two cases to consider:
1. \( x - 2 = 6 \)
2. \( x - 2 = -6 \)
Solve each case separately:
Case 1: \( x - 2 = 6 \)
[tex]\[ x - 2 = 6 \][/tex]
[tex]\[ x = 6 + 2 \][/tex]
[tex]\[ x = 8 \][/tex]
Case 2: \( x - 2 = -6 \)
[tex]\[ x - 2 = -6 \][/tex]
[tex]\[ x = -6 + 2 \][/tex]
[tex]\[ x = -4 \][/tex]
So, the values of \( x \) that satisfy \( f(x) = 39 \) are:
[tex]\[ x = 8 \quad \text{and} \quad x = -4 \][/tex]
Therefore, the correct answer from the given options is:
[tex]\[ \boxed{x = 8, x = -4} \][/tex]
