Function Notation: Evaluate and Interpret - College Algebra

Consider the function $f(x) = |x - 3| + 2$ . If $f(a) = f(b)$ where $a \neq b$ , and both $a$ and $b$ are in the domain $[-1, 7]$ , which of the following must be true?
Answer: $a + b = 6$ and one of $a, b$ is less than 3 while the other is greater than 3 The function $f(x) = |x - 3| + 2$ has its vertex at $x = 3$ , where it achieves its minimum value of $f(3) = 2$ . For $x < 3$ , we have $f(x) = -(x - 3) + 2 = -x + 5$ . For $x \geq 3$ , we have $f(x) = (x - 3) + 2 = x - 1$ . If $f(a) = f(b)$ with $a \neq b$ , then we need $|a - 3| + 2 = |b - 3| + 2$ , which simplifies to $|a - 3| = |b - 3|$ . This occurs when either $a - 3 = b - 3$ (giving $a = b$ , which contradicts $a \neq b$ ) or $a - 3 = -(b - 3)$ , which gives $a - 3 = -b + 3$ , so $a + b = 6$ . For this to happen with $a \neq b$ , one value must be on the left side of the vertex ( $< 3$ ) and the other on the right side ( $> 3$ ). For example, if $a = 1$ , then $f(1) = 2 + 2 = 4$ , and we need $f(b) = 4$ . From $b - 1 = 4$ , we get $b = 5$ . Indeed, $a + b = 1 + 5 = 6$ . Choice D is incorrect because while $|a - 3| = |b - 3|$ is necessary, it doesn't capture the full constraint that $a + b = 6$ .