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Given the function [tex]$f(x) = 6|x - 2| + 3[tex]$[/tex], for what values of [tex]$[/tex]x[tex]$[/tex] is [tex]$[/tex]f(x) = 39$[/tex]?

A. [tex]$x = -8, x = 4$[/tex]
B. [tex]$x = 8, x = -4$[/tex]
C. [tex]$x = 9, x = -4$[/tex]
D. [tex]$x = 8, x = -10$[/tex]

Sagot :

GinnyAnswer
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]
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