To determine the domain of [tex]\((b \circ a)(x)\)[/tex], we need to analyze the composite function [tex]\(b(a(x))\)[/tex].
1. First, consider the inner function [tex]\(a(x)\)[/tex]:
[tex]\[
a(x) = 3x + 1
\][/tex]
This is a linear function, and linear functions are defined for all real numbers. So, [tex]\(a(x)\)[/tex] is defined for all [tex]\(x \in (-\infty, \infty)\)[/tex].
2. Next, consider the outer function [tex]\(b(x)\)[/tex]:
[tex]\[
b(x) = \sqrt{x - 4}
\][/tex]
The function [tex]\(b(x)\)[/tex] involves a square root, which means the expression inside the square root, [tex]\(x - 4\)[/tex], must be non-negative for [tex]\(b(x)\)[/tex] to be defined. Hence, for [tex]\(b(x)\)[/tex] to be defined, we require:
[tex]\[
x - 4 \geq 0 \implies x \geq 4
\][/tex]
3. Determine the domain of [tex]\(b(a(x))\)[/tex]:
For the composite function [tex]\((b \circ a)(x)\)[/tex], or [tex]\(b(a(x))\)[/tex], to be defined, [tex]\(a(x)\)[/tex] must produce values that are within the domain of [tex]\(b(x)\)[/tex].
Substitute [tex]\(a(x)\)[/tex] into [tex]\(b(x)\)[/tex]:
[tex]\[
b(a(x)) = b(3x + 1) = \sqrt{(3x + 1) - 4} = \sqrt{3x - 3}
\][/tex]
For [tex]\(b(3x + 1)\)[/tex] to be defined:
[tex]\[
3x - 3 \geq 0 \implies 3x \geq 3 \implies x \geq 1
\][/tex]
4. Conclusion:
The domain of [tex]\((b \circ a)(x)\)[/tex] is the set of all [tex]\(x\)[/tex] values for which [tex]\(3x + 1 \geq 4\)[/tex], which simplifies to [tex]\(x \geq 1\)[/tex].
So, the correct domain of [tex]\((b \circ a)(x)\)[/tex] is:
[tex]\[
[1, \infty)
\][/tex]
The correct answer is:
[tex]\[
[1, \infty)
\][/tex]
