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 = 7.5[/tex]

Answer :

We begin with the function

[tex]$$
f(x) = 4|x-5| + 3.
$$[/tex]

We are asked to solve for [tex]$x$[/tex] when [tex]$f(x)=15$[/tex]. That is, we need to solve

[tex]$$
4|x-5| + 3 = 15.
$$[/tex]

Step 1. Subtract [tex]$3$[/tex] from both sides:

[tex]$$
4|x-5| = 15 - 3 = 12.
$$[/tex]

Step 2. Divide both sides by [tex]$4$[/tex] to isolate the absolute value:

[tex]$$
|x-5| = \frac{12}{4} = 3.
$$[/tex]

Step 3. Use the definition of absolute value. The equation [tex]$|x-5| = 3$[/tex] implies two cases:

- Case 1: [tex]$x - 5 = 3$[/tex], which gives
[tex]$$
x = 5 + 3 = 8.
$$[/tex]
- Case 2: [tex]$x - 5 = -3$[/tex], which gives
[tex]$$
x = 5 - 3 = 2.
$$[/tex]

Thus, the solutions are [tex]$x = 2$[/tex] and [tex]$x = 8$[/tex].

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